Alternating Current

 Marked Questions can be used as Revision Questions.

 fpfUgr iz'u nksgjkus ;ksX; iz'u gSA

PART - I : SUBJECTIVE QUESTIONS

Hkkx - I : fo"k;kRed iz'u ¼SUBJECTIVE QUESTIONS½
Section (A) : Average, peak and RMS value
[k.M (A) : vkSlr]'kh"kZ rFkk oxZek/; ewy eku
A-1. The household supply of electricity is at 220 V rms value and 50 Hz .Calculate the peak voltage and the
minimum possible time in which the voltage can change from the rms value to zero.

Ans. 220 2 V, 2.5 ms

Sol. VO = 2 Vrms = 220 2
V VO
V = O  = VO sin t 
2 2

  2f × t = t = 2.5 ms
4
Hindi ?kjksa esa 220 V oxZek/; ewy eku rFkk 50 Hz dh fo|qr vkiwrhZ dh tkrh gS rks 'kh"kZ foHko Kkr djks rFkk foHko dks
oxZek/; ewy eku ls 'kwU; rd ifjofrZr gksus esa U;wure fdruk le; yxsxkA
Ans: 220 2 V, 2.5 ms
Sol. VO = 2 Vrms = 220 2
V VO
V = O  = VO sin t 
2 2

  2f × t = t = 2.5 ms
4

A-2. In a LR circuit discharging current is given by  = 0 e–t/ where  is the time constant of the circuit find
the rms current for the period t = 0 to t = 
o
Ans. (e2  1) / 2
e
2
 – t
 
o  O   .dt
  e

Sol. 2
rms =

 Irms= O
e
e 2

–1 /2

Hindi
A-2. LR ifjiFk esa çokfgr fujkos'ku /kkjk  = o e–t/ ls O;Dr dh tkrh gS] tgka  ifjiFk dk le; fu;rkad gSA rks
t = 0 ls t = ds fy, oxZek/; ewy /kkjk Kkr djksA
o
Ans. (e2  1)/ 2
e
2
 – t
 
o  O e   .dt 
Sol. 2
rms =

= O
e
e 2
–1 /2

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Alternating Current
A-3. If a direct current of value ‘a’ ampere is superimposed on an alternating current  = b sin t flowing
through a wire, what is the effective(rms) value of the resulting current in the circuit?

dc

i
a i ac
b
(0, 0)
(0, 0) t t
+
1/ 2
 1 
Ans.  eff  a2  b2 
 2 
1
T 2
2
  (a  b sin t) dt  1/ 2

Sol. rms  o  =   a2  1 b2 

 T  eff  2 

 
 
Hindi
A-3. ,d rkj esa çokfgr çR;korhZ /kkjk  = b sin t ij 'a' ,sfEi;j okyh fn"V /kkjk v/;kjksfir dh tkrh gS rks ifjiFk esa
ifj.kkeh /kkjk dk çHkkoh (oxZek/; ewy) eku D;k gS ?
dc

i
a i ac
b
(0, 0)
(0, 0) t t
+
1/ 2
 1 
Ans.  eff  a2  b2 
 2 
1
T 2
2
  (a  b sin t) dt  1/ 2

Sol. rms  o  =   a2  1 b2 

 T  eff  2 

 
 

A-4. Find the average for the saw-tooth voltage of peak value V0 from t=0 to t=2T as shown in figure.

Ans. 0
Sol. For t = 0 to t = 2T here area under the curve is zero so < V > = 0

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Alternating Current
A-4. fp=k esa n'kkZ;s] 'kh"kZ eku V0 okys f=kHkqtkdkj foHko ¼vkjknkr½ dk t = 0 ls t = 2T ds fy, vkSlr eku Kkr djks &

Ans. 0
Sol. t = 0 ls t = 2 T rd ;gkW oØ ls f?kjks {ks=kQy 'kwU; gS vr% < V > = 0

Section (B) : Power consumed in an ac circuit

[k.M (B) : çR;korhZ /kkjk ifjiFk esa 'kfDr O;;
B-1. A bulb is designed to operate at 12 volts constant direct current. If this bulb is connected to an
alternating current source and gives same brightness. What would be the peak voltage of the source ?
Ans: 12 2 volts
Sol. VO = Vrms × 2 = 12 2
Hindi

B-1. ,d fo|qr cYc 12 oksYV vpj fn"V /kkjk L=kksr ij dk;Z djrk gSA bl cYc dks çR;korhZ lzksr ls tksM+us ij ;g mlh
ds leku ped nsrk gS rks lzksr dk 'kh"kZ foHko D;k gksxk\
Ans: 12 2 volts
Sol. VO = Vrms × 2 = 12 2

B-2. A resistor of resistance 100  is connected to an AC source  = (12V) sin (250 s – 1)t. Find the power
consumed by the bulb.
Ans: 0.72 W
2 2 144
Sol. P = Vrms Irms cos  = rms × R = rms = = 0.72 W
R 2  100
Hindi

B-2. 100  dk çfrjks/k  = (12V) sin (250 s – 1)t ds çR;korhZ /kkjk lzksr ls tqM+k gSA rks cYc }kjk O;f;r 'kfDr Kkr
djks\
Ans: 0.72 W
2
2 rms 144
Sol. P = Vrms Irms cos  = rms ×R= = = 0.72 W
R 2  100

B 3. In an ac circuit the instantaneous values of current and applied voltage are respectively i = 2(Amp) sin

(250s–1)t and  = (10V) sin [(250 s–1)t + ]. Find the instantaneous power drawn from the source at
3
2
t = ms and its average value.
3
Ans. 10 W, 5 W

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Alternating Current
2
Sol. P = i  = [2 sin 250  t] [10 sin (250t + /3)] at t = × 10–3  P = 10 watt
3
2 10
< P > = irms rms cos  = × × cos/3
2 2
Hindi
– 1
B 3. çR;korhZ /kkjk ifjiFk esa /kkjk rFkk vkjksfir foHko ds rkR{kf.kd eku Øe'k% i = 2(Amp) sin (250 s )t rFkk  =
 2
(10V) sin [(250s–1)t + ] gSA rks t = ms le; ij lzksr }kjk nh xbZ rkR{kf.kd 'kfDr rFkk bldk vkSlr eku
3 3
Kkr djksA
Ans: 10 W, 5 W
2
Sol. P = i  = [2 sin 250  t] [10 sin (250t + /3)] t= × 10–3 ij  P = 10 watt
3
2 10
< P > = irms rms cos  = × × cos/3
2 2

Section (C) : AC source with R, L, C connected in series

[k.M (C) : R, L, C Js.khØe ds lkFk çR;korhZ /kkjk lzksr

C-1. The dielectric strength of air is 3.0 × 106 V/m. A parallel plate air capacitor has area 20 cm2 and plate
separation 2 mm. Find the maximum rms voltage of an AC source which can be safely connected to
this capacitor .
Ans. 3.0 kV
VO
Sol. VO = 3 × 106 × 2 × 10–3 Vrms = = 3 kV
2
C-1. ,d lekUrj ok;q iV~V la/kkfj=k dk {ks=kQy 20 cm2 rFkk IysVksa ds e/; dh nwjh 2 mm gSA ok;q dh ijkoS|qr lkeF;Z
3.0 × 106 V/m gSA la/kkfj=k dks çR;korhZ /kkjk lzksr ls tksM+us ij la/kkfj=k lqjf{kr jgs blds fy, AC lzksr dk
vf/kdre oxZek/; ewy foHko Kkr djksA
Ans.: 3.0 kV
2 × 10–3 VO
Sol. VO = 3 × 106 × Vrms = = 3 kV
2
C-2. An electric bulb is designed to consume 55 W when operated at 110 volts. It is connected to a
220 V, 50 Hz line through a choke coil in series. What should be the inductance of the coil for which the
bulb gets correct voltage ?
2.2 3 7 3
Ans. =1.2 H = H
 10
Sol. [(L) irms]2 + (110)2 = (220)2
55
here  = 2 × 50 and irms =
110
L = 1.2 H
Hindi
C-2. ,d fo|qr cYc ij 110 oksYV vkjksfir djus ij ;g 55 W 'kfDr O;; djrk gSA bls 220 V, 50 Hz L=kksr ls pksd
dq.Myh }kjk Js.kh esa tksM+k tkrk gSA cYc ij lgh foHko çkIr djus ds fy, dq.Myh dk çsjdRo fdruk gksuk pkfg,\
2.2 3 7 3
Ans. =1.2 H = H
 10

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Alternating Current
Sol. [(L) irms]2 + (110)2 = (220)2
55
tgk¡  = 2 × 50 rFkk irms =
110
L = 1.2 H

C 3. A resistor, a capacitor and an inductor (R = 300  C = 20 µF, L = 1.0 henry) are connected in series
50
with an AC source of, Erms = 50 V and  = Hz. Find (a) the rms current in the circuit and (b) the rms

potential differences across the capacitor, the resistor and the inductor.

Ans: (a) 0.1 A (b) 50 V, 30 V, 10 V (Note that the sum of the rms potential differences across the three
elements is greater than the rms voltage of the source.)
2
Erms  1 
Sol. (a) Irms = where Z = R 2   L – 
z   c
(b) VR rms = Irms R = 30, VL rms = Irms (L) = 10
 1 
VC rms = Irms   = 50
 C 
Hindi
C 3. ,d ifjiFk esa izfrjks/k] la/kkfj=k ,oa izsj.k dq.Myh (R = 300  C = 20 µF, L = 1.0 gsujh) Js.khØe esa ,d izR;korhZ
50
L=kksr, Erms = 50 V rFkk  = Hz ls tqM+s gks rks (a) ifjiFk esa oxZek/; ewy /kkjk rFkk (b) la/kkfj=k, çfrjks/k rFkk

çsjdRo ds fljksa ij oxZek/; ewy foHkokUrj Kkr djksA
Ans: (a) 0.1 A (b) 50 V, 30 V, 10 V (uksV% rhuksa vo;oksa ij oxZek/; ewy foHkokUrj dk ;ksx lzksr ds oxZek/; ewy
oksYVst ls vf/kd gksrk gS)
2
Erms  1 
Sol. (a) Irms = where Z = R 2   L – 
z  c 
(b) VR rms = Irms R = 30, VL rms = Irms (L) = 10
 1 
VC rms = Irms   = 50
 C 

C-4. Consider the situation of the previous problem calculate the average electric field energy stored in the
capacitor and the average magnetic field energy stored in the inductor coil.
Ans. 25 mJ, 5mJ
T
C V 2 dt
1 O 1 2
Sol. < UC > = = C Vrms = 25 mJ
2 T 2
1 2 1
< UL > = L rms = × (1) × (0.1)2 = 5 mJ
2 2
Hindi
C-4. mijksDr ç'u esa la/kkfj=k esa laxzfgr vkSlr fo|qr {ks=k ÅtkZ rFkk izsj.k dq.Myh esa laxzfgr vkSlr pqEcdh; {ks=k ÅtkZ
Kkr djksA
Ans. 25 mJ, 5mJ

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Alternating Current
T
C V 2 dt
1 O 1 2
Sol. < UC > = = C Vrms = 25 mJ
2 T 2
1 2 1
< UL > = L rms = × (1) × (0.1)2 = 5 mJ
2 2

50
C-5. A 20 volts 5 watt lamp (lamp to be treated as a resistor) is used on AC mains of 200 volts and 11

c.p.s. Calculate the (i) capacitance of the capacitor, or inductance of the inductor, to be put in series to
run the lamp. (ii) How much pure resistance should be included in place of the above device so that the
lamp can run on its rated voltage. (iii) which is more economical (the capacitor, the inductor or the
resistor).
125
Ans. (i) F or 2.4 H (ii) 720 (iii) It will be more economical to use inductance or capacitance in series
33
with the lamp to run it as it.It consumes no power while there would be dissipation of power when
resistance is inserted in series with the lamp.
2
 1  5 50
Sol. (i)   rms  + [20]2 = [200]2 where Irms = &  = 2 × 11
 C  20 
or [(L) rms]2 + (20)2 = 2002
(ii) Irms × R + 20 = 200
(iii) does not loss in C and L.

Hindi
50
C-5. 200 oksYV rFkk 11 pDdj çfr lSd.M vko`fr dh eq[; çR;korhZ /kkjk ls 20 oksYV rFkk 5 okWV dk ysEi tqM+k gS

¼ysEi 'kq) izfrjks/kh ifjiFk j[krk gS½ A rks Kkr djks \ (i) ysEi dks dk;Z djus ds fy, Js.khØe eas tqM+s la/kkfj=k dh
/kkfjrk, ;k dq.Myh dk çsjdRo, (ii) mijksDr midj.k ds LFkku ij fdruk çfrjks/k tksM+k tk;s ftlls ysEi Loa; ds
vafdr foHko ij dk;Z dj ldsA (iii) mijksDr esa ls dkSulh O;oLFkk ¼la/kkfj=k] çsjdRo ;k çfrjks/k½ ferO;;h ¼de
ykxr½ gksxhA
125
Ans. (i) F ;k 2.4 H (ii) 720 (iii) ySEi ds lkFk Js.kh Øe esa çsjdRo vFkok la/kkfj=k dks tksM+dj mi;ksx esa ysuk
33
T;knk ferO;;h gksxk D;ksafd blesa dksbZ 'kfDr {k; ugha gksxk ysfdu tc ySEi ds lkFk çfrjks/k tksM+k tkrk gS rks mlesa
'kfDr {k; gksrk gSA
2
 1  5 50
Sol. (i)   rms  + [20]2 = [200]2 where Irms = &  = 2 × 11
 C  20 
;k [(L) rms]2 + (20)2 = 2002
(ii) Irms × R + 20 = 200
(iii) does not loss in C and L.
C rFkk L esa gkuh ugh gksrh gSA

3
C-6. A circuit has a resistance of 50 ohms and an inductance of henry. It is connected in series with a

40
condenser of F and AC supply voltage of 200 V and 50 cycles/sec. Calculate [16JP120249]

(i) the impedance of the circuit,
(ii) the p.d. across inductor coil and condenser.
(iii) Power factor
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Alternating Current
1
Ans. Z = 50 2 ohm, VC = 500 2 volt and VL= 600 2 volt,
2
2
2  1 
Sol. (i) z = R   L – 
 C 
200 1 200
(ii) VL = (L) × = 600 2 and VC =  = 500 2
Z C Z
R 50 1
(iii) cos  = = =
Z 50 2 2
Hindi
3 40
C-6. ,d ifjiFk esa 50 vkse dk çfrjks/k rFkk gsujh dk çsjdRo gSA ;g F la/kkfj=k rFkk 200 V, 50 pDdj
 
/lSd.M ds çR;korhZ L=kksr ds lkFk Js.khØe esa tqM+s gS rks x.kuk djks &
(i) ifjiFk dh çfrck/kk
(ii) çsj.k dq.Myh rFkk la/kkfj=k ds fljksa ij foHkokUrjA
(iii) 'kfDr xq.kkad
1
Ans. Z = 50 2 ohm, VC = 500 2 volt rFkk VL= 600 2 volt,
2
2
 1 
Sol. (i) z = R 2   L – 
 C 
200 1 200
(ii) VL = (L) × = 600 2 and VC =  = 500 2
Z C Z
R 50 1
(iii) cos  = = =
Z 50 2 2

C-7. A coil draws a current of 1.0 ampere and a power of 100 watt from an A.C. source of 110 volt and
5 22
hertz. Find the inductance and resistance of the coil.

21
Ans. H, 100 
22
2
Sol. rms × R = 100 ........(1)
2
(1) × R = 100    R = 100 
21
Vrms = Irms Z 110 = Irms Z 110 = 1 × Z = R 2  (L)2 L= H
22
Hindi
5 22
C-7. ,d dq.Myh 110 volt rFkk gVZt okys çR;korhZ /kkjk lzksr ls 1.0 ,sfEi;j dh /kkjk rFkk 100 okWV 'kfDr ysrh

gSA rks dq.Myh dk çsjdRo rFkk çfrjks/k Kkr djks \
21
Ans. H, 100 
22
2
Sol. rms × R = 100 ........(1)
2
(1) × R = 100    R = 100 
21
Vrms = Irms Z 110 = Irms Z 110 = 1 × Z = R 2  (L)2 L= H
22
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Alternating Current
C-8. An inductor 2/ Henry, a capacitor 100/ µF and a resistor 75  are connected in series across a
source of emf V = 10 sin 100  t. Here t is in second. (a) find the impedance of the circuit.(b) find the
energy dissipated in the circuit in 20 minutes.
Ans. 125 , 288 J
2
 1 
Sol. (a) Z = R 2   L –  = 125
 C
2
2
 10 / 2 
(b) H = rms Rt =   × R × 20 × 60 = 288
 Z 
HINDI
C-8. V = 10 sin 100 t fo|qr okgd cy okys lzksr ls 2/ gsujh çsjdRo, 100/ µF la/kkfj=k rFkk 75  çfrjks/k Js.kh Øe
esa tqM+s gS ;gk¡ t lSd.M esa gS rks (a) ifjiFk dh çfrck/kk Kkr djks (b) 20 feuV esa ifjiFk esa {kf;r ÅtkZ Kkr djks \
Ans. 125 , 288 J
2
 1 
Sol. (a) Z = R 2   L –  = 125
 C 
2
 10 / 2 
(b) H = R t =   × R × 20 × 60 = 288
 Z 

Section (D) : resonance

[k.M (D) : vuqukn
D-1. A series circuit consists of a resistance, inductance and capacitance. The applied voltage and the
current at any instant are given by
E = 141.4 cos (5000 t – 10º)
and  = 5 cos (5000 t – 370º)
The inductance is 0.01 henry. Calculate the value of capacitance and resistance.
141.4
Ans. 4 F, R = 
5
1 1 1
Sol. Here phase difference  = 360º &  = 5000 at Resonance C = 2 = 
 L (5000)2 0.01
V0 141.4
R= 
0 5
Hindi
D-1. ,d Js.kh ifjiFk esa çfrjks/k] çsjdRo rFkk la/kkfj=k tqM+s gSA fdlh {k.k vkjksfir foHko rFkk /kkjk dks fuEu }kjk O;Dr
djrs gS
E = 141.4 cos (5000 t – 10º)
rFkk  = 5 cos (5000 t – 370º)
çsjdRo 0.01 gsujh gks rks /kkfjrk dk eku rFkk izfrjks/k dk eku Kkr djksA
141.4
Ans. 4 F, R = 
5
1 1 1
Sol. ;gk¡ dykUrj  = 360º vkSj vuqukn C = =  ij  = 5000 gSA
2L (5000)2 0.01
V0 141.4
R= 
0 5

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Alternating Current
D-2. An inductance of 2.0 H, a capacitance of 18 µF and a resistance of 10 k are connected to an AC
source of 20 V with adjustable frequency (a) What frequency should be chosen to maximise the
current(RMS) in the circuit? (b) What is the value of this maximum current (RMS) ?
250
Ans: (a) Hz (b) 2 mA
3
1 20 20
Sol. (a) f = (b) Irms (max) = = = 2 mA.
2 LC R 10  103
Hindi
D-2. ,d 2.0 H dk çsjdRo,18 µF dk la/kkfj=k rFkk 10 k dk çfrjks/k lek;ksftr vko`fr (adjustable frequency) okys
20 V ds çR;korhZ /kkjk lzksr ls tqM+s gS rks (a) ifjiFk esa vf/kdre /kkjk (RMS) çokfgr djus ds fy, p;fur vko`fr
dk eku D;k gksuk pkfg,? (b) vf/kdre /kkjk (RMS) dk eku D;k gS ?
250
Ans: (a) Hz (b) 2 mA
3
1 20 20
Sol. (a) f = (b) Irms (max) = = = 2 mA.
2 LC R 10  103

D-3. An inductor-coil, a capacitor are connected in series with an AC source of rms voltage 24 V. When the
frequency of the source is varied a maximum rms current of 6.0 A is observed. If this inductor coil is
connected to a DC source of 12 V and having internal resistance 4.0  what will be the current in
steady state?
Ans : 1.5 A
24 12
Sol. R= = 4   I= = 1.5 Amp.
6 44
Hindi
D-3. ,d çsj.k dq.Myh rFkk ,d la/kkfj=k çR;korhZ /kkjk lzksr ls Js.kh Øe esa tqM+s gSA çR;korhZ /kkjk lzksr dk oxZek/; ewy
foHko 24 V gSA tc lzksr dh vko`fr ifjofrZr dh tkrh gS rks vf/kdre oxZek/; ewy /kkjk 6.0 A çsf{kr dh tkrh gSA
;fn izsj.k dq.Myh dks 12 V ds fn"V/kkjk L=kksr rFkk 4.0 vkUrfjd çfrjks/k okyh cSVjh ls tksM+ fn;k tk;s rks LFkkbZ
voLFkk /kkjk D;k gksxh\
Ans : 1.5 A
24 12
Sol. R= = 4   I= = 1.5 Amp.
6 44

D-4. An electro magnetic wave of wavelength 300 metre can be transmitted by a transmission centre. A
condenser of capacity 2.5 F is available. Calculate the inductance of the required coil for a resonant
circuit.Use 2=10.
Ans. 1×10–8 henry
1 3  108 2   3  108
Sol. L = 2 here  = 2 f = 2 =
C  300
HINDI
D-4. ,d lapkj dsUnz ls 300 ehVj rjaxnS/;Z okyh fo|qr pqEcdh; rjax lapfjr dh tkrh gSA 2.5 F /kkfjrk okyk la/kkfj=k
miyC/k gSA vuqukn ifjiFk ds fy, vko';d dq.Myh dk çsjdRo Kkr djksA 2=10 dk mi;ksx djksA
Ans. 1×10–8 gsujh
1 3  108 2  3  108
Sol. L= here ;gk¡  = 2 f = 2 =
2 C  300

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Alternating Current
Section (E) : Transformer
[k.M (E) : VªkalQkeZj
E-1 A transformer has 50 turns in the primary and 100 turns in the secondary. If the primary is connected to
a 220 V DC supply, what will be the voltage across the secondary ?
Ans: zero
Sol. Transfoormer does not work on D.C.
Hindi
E-1 ,d VªkalQkeZj dh çkFkfed dq.Myh esa 50 ?ksjs rFkk f}rh;d dq.Myh esa 100 ?ksjs gSA ;fn çkFkfed dq.Myh 220 V dh
fn"V/kkjk lzksr ls tqM+h gks rks f}rh; dq.Myh ds ifjr% foHko D;k gksxk \
Ans: 'kwU;
Sol. VªkalQkeZj D.C. ij dk;Z ugh djrk gS

N2
E-2. In a transformer ratio of secondary turns (N2) and primary turns (N1) i.e.  4 . If the voltage applied
N1
in primary is 200 V, 50 Hz, find (a) voltage induced in secondary (b) If current in primary is 1A, find the
current in secondary if the transformer is (i) ideal and (ii) 80% efficient and there is no flux leakage.

Ans. (a) 800 V (b) (i) 0.25 A (ii) 0.2 A.

V N2
Sol. (a) 2  =4  V2 = 200 × 4 = 800
V1 N1
2 N1 1 1
(b) (i)  =  I2 = = 0.25 A
1 N2 4 4
(ii) I1 V1 × 0.80 = V2 I2  I2 = 0.8 × 0.25 = 0.2 A
Hindi
N2
E-2. ,d VªkalQkeZj esa f}rh;d dq.Myh esa ?ksjksa (N2) rFkk çkFkfed dq.Myh esa ?ksjkas (N1) dk vuqikr vFkkZr~ 4 gSA
N1
çkFkfed dq.Myh esa vkjksfir foHko 200 V, 50 Hz gS rks (a) f}rh;d dq.Myh esa çsfjr foHko Kkr djks ? (b) ;fn
çkFkfed esa /kkjk 1A, gks rks f}rh;d dq.Myh esa /kkjk Kkr djks \ ;fn VªkalQkeZj (i) vkn'kZ gks rFkk (ii) n{krk 80%
gks] dksbZ ¶yDl {k; ugha gSA
Ans. (a) 800 V (b) (i) 0.25 A (ii) 0.2 A.
V N2
Sol. (a) 2  =4  V2 = 200 × 4 = 800
V1 N1
2 N1 1 1
(b) (i)  =  I2 = = 0.25 A
1 N2 4 4
(ii) I1 V1 × 0.80 = V2 I2  I2 = 0.8 × 0.25 = 0.2 A

PART - II : ONLY ONE OPTION CORRECT TYPE

Hkkx - II : dsoy ,d lgh fodYi çdkj (ONLY ONE OPTION CORRECT TYPE)
Section (A) : Average, peak and RMS values and RMS values
[k.M (A) % vkSlr]'kh"kZ rFkk oxZek/; ewy eku
A-1. r.m.s. value of current i = 3 + 4 sin ( t + /3) is:
5 7
(A) 5 A (B*) 17 A (C) A (D) A
2 2

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Alternating Current
1
T 2 2
  i dt  T 2
1/ 2

Sol. Irms =  O  =  [3  4 sin(t   / 3)] dt   17 .

 T   
0 T 
 
 
A-1. /kkjk i = 3 + 4 sin ( t + /3) dk oxZ ek/;ewy eku gksxk&
17 A 5 7
(A) 5 A (B*) (C) A (D) A
2 2
1
T 2 2
  i dt  1/ 2
 O   T [3  4 sin(t   / 3)]2 
Sol. Irms = =  dt   17 .
 T  T
  0 
 

A-2. The peak value of an alternating e.m.f given by E = E0 cos t, is 10 volt and frequency is 50 Hz. At time
t = (1/600) sec, the instantaneous value of e.m.f is :
(A) 10 volt (B*) 5 3 volt (C) 5 volt (D) 1 volt
 1 
Sol. E = 10 cos  2  50   =5 3
 600 
Hindi çR;korhZ fo-ok-cy E = E0 cos t dk 'kh"kZ eku 10 volt rFkk vko`fr 50 Hz gSA le; t = (1/600) sec ij fo-ok-cy
dk rkR{kf.kd eku gS :
(A) 10 volt (B*) 5 3 volt (C) 5 volt (D) 1 volt
 1 
Sol. E = 10 cos  2  50  =5 3
 600 

A-3. The voltage of an AC source varies with time according to the equation, V = 100 sin 100  t cos 100 t.
Where t is in second and V is in volt. Then :
(A) the peak voltage of the source is 100 volt
(B) the peak voltage of the source is (100/ 2 ) volt
(C*) the peak voltage of the source is 50 volt
(D) the frequency of the source is 50 Hz
Sol. V = 100 sin100 t cos 100  t
V = 50 sin 200  t
here VO = 50 &  = 200  f = 100 Hz
Hindi
çR;korhZ /kkjk lzksr dk foHko ] le; ds lkFk lehdj.k V = 100 sin 100  t cos 100 t }kjk ifjofrZr gksrk gS ]
tgk¡ t lSd.M rFkk V oksYV esa gS rks :
(A) lzksr dk 'kh"kZ foHko 100 oksYV gSA

(B) lzksr dk 'kh"kZ foHko (100/ 2 ) oksYV gSA

(C*) lzksr dk 'kh"kZ foHko 50 oksYV gSA
(D) lzksr dh vko`fr 50 Hz gSA
Sol. V = 100 sin100 t cos 100  t
V = 50 sin 200  t
tgk¡ VO = 50 &  = 200  f = 100 Hz

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Alternating Current
A-4. An alternating voltage is given by : e = e1 sint + e2 cost. Then the root mean square value of
voltage is given by :
e1 e2 e12  e22
(A) e12  e22 (B) e1 e2 (C) (D*)
2 2
T
(e1 sin t  e2 cos t)2 dt
Sol. V2rms = 
0 T
e12  e22
=
2
2
where  = .
T
Hindi
çR;korhZ /kkjk foHko : e = e1 sint + e2 cost gS rks foHko dk oxZek/; ewy eku gS :
e1 e2 e12  e22
(A) e12  e22 (B) e1 e2 (C) (D*)
2 2
T
(e1 sin t  e2 cos t)2 dt
Sol. V2rms = 
0 T
e12  e22
=
2
2
tgk¡ = .
T

2t
A-5. An AC voltage is given by : E = E0 sin
T
Then the mean value of voltage calculated over time interval of T/2 seconds :
(A) is always zero (B) is never zero (C) is (2E0/) always (D*) may be zero
Sol.

If net area of E – t curve is zero for given interval then average value will be zero.
Hindi
,d çR;korhZ foHko fuEu }kjk fn;k tkrk gS &
2t
E = E0 sin
T
rks T/2 lSd.M le;kUrjky ds fy, x.kuk djus ls foHko dk vkSlr eku :
(A) ges'kk 'kwU; gksxkA (B) dHkh 'kwU; ugha gksxkA (C) ges'kk (2E0/) gksxkA (D*) 'kwU; gks ldrk gSA
Sol.

;fn E – t oØ dk {ks0Q0 'kqU; gks ¼fn;s x;s vUrjky esa½ rks E dk vkSlr eku 'kqU; gksxkA

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Alternating Current
 
A-6. An AC voltage of V = 220 2 sin  100t   is applied across a DC voltmeter, its reading will be:
 2

(A) 220 2 V (B) 2 V (C) 220 V (D*) zero

Sol. D.C. Voltmeter measures Average value only
 
Hindi V = 220 2 sin  100t   çR;korhZ /kkjk foHko dks fn"V /kkjk oksYVehVj ij yxk;k tkrk gSA oksYVehVj dk
 2
ikB;kad gksxk %
(A) 220 2V (B) 2V (C) 220 V (D*) 'kwU;
Sol. D.C. oksYV ehVj dsoy vkSlr foHkokUrj ekirk gSA

Section (B) : Power consumed in an AC circuit

[k.M (B) % çR;korhZ /kkjk ifjiFk esa 'kfDr O;;

B-1. The average power delivered to a series AC circuit is given by (symbols have their usual meaning) :
(A) Erms rms (B*) Erms rms cos  (C) Erms rms sin  (D) zero
Hindi
Js.kh çR;korhZ /kkjk ifjiFk dks nh xbZ vkSlr 'kfDr gksrh gS ¼tgk¡ ladsrksa dk vFkZ lkekU; gS½ %
(A) Erms rms (B*) Erms rms cos  (C) Erms rms sin  (D) 'kwU;

B-2. Energy dissipates in LCR circuit in :

(A) L only (B) C only (C*) R only (D) all of these
Hindi
LCR ifjiFk esa ÅtkZ gkfu gksrh gS :
(A) dsoy L esa (B) dsoy C esa (C*) dsoy R esa (D) mijksDr lHkh ij

B-3. The potential difference V across and the current  flowing through an instrument in an AC circuit are
given by :
V = 5 cos t volt
 = 2 sin t Amp.
The power dissipated in the instrument is :
(A*) zero (B) 5 watt (C) 10 watt (D) 2.5 watt
Sol. Pav = vrms Irms cos  
Here  = 90º so Pav = 0
Hindi
çR;korhZ /kkjk ifjiFk esa fdlh ;a=k ds fljksa ij foHkokUrj V rFkk izokfgr /kkjk fuEu }kjk fn;s tkrs gSA
V = 5 cos t volt
 = 2 sin t Amp.
;a=k esa O;; 'kfDr gksxh &
(A*) 'kwU; (B) 5 okWV (C) 10 okWV (D) 2.5 okWV
Sol. Pav = vrms Irms cos 
;gk¡  = 90º vr% Pav = 0

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Alternating Current
B-4. A direct current of 2 A and an alternating current having a maximum value of 2 A flow through two
identical resistances. The ratio of heat produced in the two resistances in the same time interval will be:
(A) 1 : 1 (B) 1 : 2 (C*) 2 : 1 (D) 4 : 1
2
HD.C. R
Sol. = 2 =2
HA.C. rmsR
Hindi nks leku çfrjks/kksa ls 2 A dh fn"V /kkjk rFkk 2 A vf/kdre eku okyh çR;korhZ /kkjk çokfgr dh tkrh gS rks
leku le; varjky esa] nksuksa çfrjks/kksa esa mRiUu Å"ekvksa dk vuqikr gksxk &
(A) 1 : 1 (B) 1 : 2 (C*) 2 : 1 (D) 4 : 1
HD.C.  2R
Sol. = 2 =2
HA.C. rmsR

B-5. A sinusoidal AC current flows through a resistor of resistance R. If the peak current is p, then average
power dissipated is :
,d çfrjks/k R ls T;koØh; çR;korhZ /kkjk çokfgr gks jgh gSA ;fn 'kh"kZ /kkjk p gks rks vkSlr 'kfDr O;; gS &
1 2 4 2 1 2
(A) Ip2R cos (B*) Ip R (C) Ip R (D) Ip R
2  2
2
    2R
Sol. <P> = rms R =  P  R  P
 2 2

B-6. What is the rms value of an alternating current which when passed through a resistor produces heat,
which is thrice that produced by a D.C. current of 2 ampere in the same resistor in the same time
interval?
(A) 6 ampere (B) 2 ampere (C*) 2 3 ampere (D) 0.65 ampere
,d çfrjks/k ls çR;korhZ /kkjk çokfgr dh tkrh gS rks blls mRiUu Å"ek] mlh çfrjks/k ls 2 ,sfEi;j dh fn"V /kkjk
leku le; rd çokfgr djus ij mRiUu Å"ek dh rhu xquh gS rks çR;korhZ /kkjk dk oxZek/; ewy eku D;k gS?
(A) 6 ,fEi;j (B) 2 ,fEi;j (C*) 2 3 ,fEi;j (D) 0.65 ,fEi;j
Sol. P = 2rms R = [(2)2 R ] × 3  rms = 2 3 A

B-7. A resistor and a capacitor are connected to an AC supply of 200 volt, 50 Hz in series. The current in the
circuit is 2 ampere. If the power consumed in the circuit is 100 watt, then the resistance in the circuit is:
,d çfrjks/k rFkk la/kkfj=k 200 volt, 50 Hz vko`fr ds çR;korhZ /kkjk L=kksr ds lkFk Js.khØe esa tqM+s gSA ifjiFk esa
2 ,sfEi;j dh /kkjk çokfgr gks jgh gSA ;fn ifjiFk esa O;; 'kfDr 100 watt, gks rks ifjiFk esa çfrjks/k gksxkA &
(A) 100  (B*) 25  (C) 125  75  (D) 400 
2
Sol.  R = 100
100 100
R= 2 = = 25.
 (2)2
B-8. The impedance of a series circuit consists of 3 ohm resistance and 4 ohm reactance. The power factor
of the circuit is :
,d ifjiFk dh çfrck/kk esa 3 vkse çfrjks/k rFkk 4 vkse çfr?kkr gS rks ifjiFk dk 'kfDr xq.kkad gS :
(A) 0.4 (B*) 0.6 (C) 0.8 (D) 1.0
x 4
Sol. tan  = =
R 3
3
cos  = = 0.6
5

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Alternating Current
B-9. A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage
V = 10 sin (100 t). The peak current in the circuit will be :
(A) 2 amp (B) 1 amp (C) 10 amp (D*) 20 amp
V 10
Sol. IO = 0 =
L 100  5  103
Hindi
5.0 mH çsjdRo rFkk ux.; çfrjks/k okyh dq.Myh V = 10 sin (100 t) okys çR;korhZ foHko ls tqM+h gS rks ifjiFk esa
'kh"kZ /kkjk gksxh &
(A) 2 amp (B) 1 amp (C) 10 amp (D*) 20 amp
V0 10
Sol. IO = =
L 100  5  103

B-10. An electric bulb and a capacitor are connected in series with an AC source. On increasing the
frequency of the source, the brightness of the bulb :
(A*) increase (B) decreases
(C) remains unchanged (D) sometimes increases and sometimes decreases
,d çR;korhZ /kkjk lzksr ls fo|qr cYc rFkk la/kkfj=k] Js.khØe esa tqM+sa gSA lzksr dh vko`fr c<+kus ij cYc dh ped :
(A*) c<+sxh (B) ?kVsxh
(C) vifjofrZr jgsxh (D) dqN le; rd c<+sxh rFkk dqN le; rd ?kVsxh
1
Sol. XC = will decrease if we increase frequency then z will decrease so current will increase & intensity
c
will increase.
1
lzksr dh vko`fÙk c<+kus esa XC = ?kVrk gSA vr% z ?kVsxkA vr% ped c<+sxhA
c

B-11. By what percentage the impedance in an AC series circuit should be increased so that the power factor
changes from (1/2) to (1/4) (when R is constant) ?
'kfDr xq.kkad dks (1/2) ls (1/4) rd ifjofrZr djus ds fy, çR;korhZ /kkjk ifjiFk dh çfrck/kk esa fdrus çfr'kr ls
o`f) djuh gksxh (R vpj j[kk x;k gS)?
(A) 200% (B*) 100% (C) 50% (D) 400%
R
Sol. cos =
z
z' z
% change = × 100 = 100% .
z

B-12. If the frequency of the source e.m.f. in an AC circuit is n, the power varies with a frequency :
,d çR;korhZ /kkjk ifjiFk esa lzksr fo-ok-cy dh vko`fr n gS rks 'kfDr fdl vko`fr ls ifjofrZr gksxh &
(A) n (B*) 2 n (C) n/2 (D) zero 'kwU;
Sol.

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Alternating Current
Section (C) : AC source with R, L, C connected in series
R, L, C Js.khØe esa çR;korhZ /kkjk lzksr ls tqM+s gS
C-1. A 0.21-H inductor and a 88- resistor are connected in series to a 220-V, 50-Hz AC source. The
current in the circuit and the phase angle between the current and the source voltage are respectively.
(Use = 22/7)
0.21-H dk çsjdRo rFkk 88- dk çfrjks/k 220-V rFkk 50-Hz okys çR;korhZ /kkjk lzksr ls tqM+k gS rks ifjiFk esa /kkjk
rFkk lzksr foHko ,oa /kkjk ds e/; dykUrj Øe'k% gksaxs ¼= 22/7 dk mi;ksx djsa ½
(A*) 2 A, tan–1 3/4 (B) 14.4 A, tan–1 7/8 (C) 14.4 A, tan–1 8/7 (D) 3.28 A, tan–1 2/11
V Vrms
Sol. rms = rms = = 2A
Z R  (L)2
2

L 66 3
tan = = = .
R 88 4

C-2. A 100 volt AC source of angular frequency 500 rad/s is connected to a LCR circuit with L = 0.8 H,
C = 5 F and R = 10 , all connected in series. The potential difference across the resistance is
LCR ifjiFk ls 100 oksYV rFkk 500 jSfM;u@lSd.M vko`fr dk çR;korhZ /kkjk lzksr tqM+k gS] tgka L = 0.8 H, C = 5
F rFkk R = 10  lHkh Js.khØe esa tqM+s gS rks çfrjks/k ds fljksa ij foHkokUrj gS &
100
(A) volt (B*) 100 volt (C) 50 volt (D) 50 3
2
V 100
Sol. rms = rms =
Z  1 
2

R 2   L 
 C 
P.d. across resistance çfrjks/k dk foHkokUrj = R rms = 100 volt.

C-3. A pure resistive circuit element X when connected to an AC supply of peak voltage 200 V gives a peak
current of 5 A which is in phase with the voltage. A second circuit element Y, when connected to the
same AC supply also gives the same value of peak current but the current lags behind by 90°. If the
series combination of X and Y is connected to the same supply, what will be the rms value of current ?
,d 'kq) çfrjks/kh ifjiFk vo;o X ds lkFk 200 V 'kh"kZ foHko okyk çR;korhZ /kkjk lzksr tqM+k gSA blesa 5 A 'kh"kZ
/kkjk çokfgr gks jgh gS] tks foHko ls leku dyk esa gSA nwljs ifjiFk vo;o Y dks mlh çR;korhZ /kkjk lzksr ls tksM+us
ij mruh gh 'kh"kZ /kkjk çokfgr gksrh gS] ijUrq ;g /kkjk 90° i'pxkeh gSA ;fn X rFkk Y ds Js.kh la;kstu dks mlh
L=kksr ls tksM+ fn;k tk;s rks /kkjk dk oxZek/; ewy eku D;k gksxk \
10 5 5
(A) amp (B) amp (C*) amp (D) 5 amp
2 2 2
V 200
Sol. R= 0 = = 40  (For circuit x) (ifjiFk x ds fy,)
0 5
V0
XL = = 40  (For circuit y) (ifjiFk y ds fy,)
0
If x & y are in series ;fn x rFkk y Js.kh esa gS rks
200 5 0 5
= = Amp.  rms = = amp.
40  2 2 2 2

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Alternating Current
C-4. In an AC circuit, a resistance of R ohm is connected in series with an inductance L. If phase angle
between voltage and current be 45°, the value of inductive reactance will be.
(A) R/4 (B) R/2
(C*) R (D) cannot be found with the given data
çR;korhZ /kkjk ifjiFk esa çfrjks/k R vkse ] çsjdRo L ds lkFk Js.kh esa tqM+k gSA ;fn foHko rFkk /kkjk ds e/; dyk dks.k
45° gks rks çsjdRoh; çfr?kkr gksxk &
(A) R/4 (B) R/2
(C*) R (D) fn;s x;s vkadM+ks ls Kkr ugha fd;k tk ldrkA
L
Sol. tan = tan45º =
R
XL = L = R.

C-5. In an AC circuit the potential differences across an inductor and resistor joined in series are respectively
16 V and 20 V. The total potential difference across the circuit is
,d çR;korhZ /kkjk ifjiFk esa Js.khØe esa tqM+s çsjdRo rFkk çfrjks/k ds fljksa ij foHkokUrj Øe'k% 16 V rFkk 20 V gS rks
ifjiFk ds fljksa dk dqy foHkokUrj gS
(A) 20 V (B*) 25.6 V (C) 31.9 V (D) 53.5 V
2 2 2 2
Sol. Vnet = V V
R L = (20)  (16) = 25.6.

C-6. An AC voltage source V = 200 2 sin 100 t is connected across a circuit containing an AC ammeter (it
reads rms value) and capacitor of capacity 1 F. The reading of ammeter is :
,d V = 200 2 sin 100 t foHko ds çR;korhZ /kkjk lzksr ls çR;korhZ /kkjk vehVj ¼tks dsoy rms eku i<+rk gS½ rFkk
1 F dk la/kkfj=k tqM+k gS rks vehVj dk ikB;kad gS :
(A) 10 mA (B*) 20 mA (C) 40 mA (D) 80 mA
200 2
Sol. = = 200 × C = 20 mA.
 XC   2

C-7. When 100 V DC is applied across a solenoid, a steady current of 1 A flows in it. When 100 V AC is
applied across the same solenoid, the current drops to 0.5 A. If the frequency of the AC source is
150/ 3  Hz, the impedance and inductance of the solenoid are :
(A) 200  and 1/3 H (B) 100  and 1/16 H
(C*) 200  and 1.0 H (D) 1100  and 3/117 H
,d ifjukfydk ij 100 V dk fn"V foHko vkjksfir gS] blesa 1 A dh fLFkj /kkjk çokfgr gks jgh gSA ;fn blh
ifjukfydk ij 100 V dk çR;korhZ /kkjk L=kksr vkjksfir dj fn;k tk;s rks blesa /kkjk 0.5 A rd de gks tkrh gSA
;fn çR;korhZ /kkjk lzksr dh vko`fr 150/ 3  Hz gS rks ifjukfydk dh çfrck/kk rFkk çsjdRo gS &
(A) 200  rFkk 1/3 H (B) 100  rFkk 1/16 H
(C*) 200  rFkk 1.0 H (D) 1100  rFkk 3/117 H
100 Z 2  R2
Sol. R= = 100  x=
1
100 x
Z= = 200  L= = 1H.
0.5 

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Alternating Current
C-8. If in a series LCR AC circuit, the rms voltage across L, C and R are V1, V2 and V3 respectively, then the
voltage of the source is always :
(A) equal to V1 + V2 + V3 (B) equal to V1 – V2 + V3
(C) more than V1 + V2 + V3 (D*) none of these is true
LCR Js.kh çR;korhZ /kkjk ifjiFk esa L, C rFkk R ds fljksa ij foHko Øe'k% V1, V2 rFkk V3 gS rks lzksr ds fljksa ij foHko
ges'kk &
(A) V1 + V2 + V3 ds leku gSA (B) V1 – V2 + V3 ds leku gSA
(C) V1 + V2 + V3 ls vf/kdA (D*) mijksDr esa ls dksbZ Hkh lR; ugha gSA
Sol. Voltage of source is always less than (V1 + V2 + V3), Vnet = V12  V22  V32
lzksr dh oksYVrk lnSo (V1 + V2 + V3) ls de gksrh gSA

Section (D) : resonance vuqukn

D-1. The value of power factor cos in series LCR circuit at resonance is :
LCR Js.kh ifjiFk esa vuqukn ij 'kfDr xq.kkad cos dk eku gS &
(A) zero 'kqU; (B*) 1 (C) 1/2 (D) 1/2 ohm
Sol. At resonance vuqukn dh fLFkfr ij xL = xC
So, blfy, z = R,  cos = 1

D-2. A series LCR circuit containing a resistance of 120 ohm has angular resonance frequency
4 × 103 rad s–1. At resonance, the voltage across resistance and inductance are 60V and
40 V respectively. The values of L and C are respectively :
,d Js.kh LCR ifjiFk esa 120 vkse dk çfrjks/k tqM+k gS rFkk bldh dks.kh; vuquknh vko`fr 4 × 103 jsfM;u@lS0 gSA
vuqukn ij çfrjks/k rFkk çsjdRo ds fljksa ij foHko Øe'k% 60V rFkk 40 V gS rks L rFkk C ds eku Øe'k% gS :
(A*) 20 mH, 25/8 F (B) 2mH, 1/35 F (C) 20 mH, 1/40 F (D) 2mH, 25/8 nF
60 1
Sol. rms = = Amp.
120 2
VL = rms × (L)
1
40 = × (40 × 103) × L
2
L = 20 mH
 1 
At resonance vuqukn ij VC = Irms   = VL
 c 
1 1 1 25
C= × 3
× C= F.
2 4  10 40 8

D-3. In an LCR circuit, the capacitance is made one-fourth, when in resonance. Then what should be the
change in inductance, so that the circuit remains in resonance ?
LCR ifjiFk esa vuqukn ij /kkfjrk ,d pkSFkkbZ dj nh tkrh gS rks ifjiFk esa vuqukn cuk;s j[kus ds fy, çsjdRo esa
fdruk ifjorZu gksuk pkfg, ?
(A*) 4 times (B) 1/4 times (C) 8 times (D) 2 times
(A*) 4 xquk (B) 1/4 xquk (C) 8 xquk (D) 2 xquk
1
Sol. At resonance vuqukn ij L =
C
1
L .
C

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Alternating Current
D-4. A resistor R, an inductor L and a capacitor C are connected in series to an oscillator of frequency . If
the resonant frequency is r, then the current lags behind voltage, when :
 vko`fr okys nksfy=k ls çfrjks/k R, çsjdRo L rFkk la/kkfj=k C Js.khØe esa tqM+s gSA ;fn vuquknh vko`fr r gks ] rks
/kkjk foHko ls i'pxkeh gksxh tc :
(A)  = 0 (B)  < r (C)  = r (D*)  > r
Sol. Current lags behind voltage. vr% /kkjk foHko ls ihNs gksxhA
If ;fn XL > XC
1 1
  2L   
2C 2 LC

1
But as r =
2 LC
therefore,  > r

D-5. A resistor R, an inductor L, a capacitor C and voltmeters V1, V2 and V3 are connected to an oscillator in
the circuit as shown in the adjoining diagram. When the frequency of the oscillator is increased, upto
resonance frequency, the voltmeter reading (at resonance frequency) is zero in the case of :
fp=kkuqlkj çfrjks/k R, çsjdRo L, la/kkfj=k C rFkk oksYVehVj V1, V2 rFkk V3 nksfy=k ds lkFk fp=kkuqlkj ifjiFk esa tqM+s
gSA tc nksfy=k dh vko`fr vuquknh vko`fr rd c<+k;h tkrh gS rks fdl oksYVehVj dk ikB;kad vuqukn vko`fr ij
'kwU; gksxk &

(A) voltmeter V1 (B*) voltmeter V2

(C) voltmeter V3 (D) all the three voltmeters
(A) oksYVehVj V1 (B*) oksYVehVj V2
(C) oksYVehVj V3 (D) rhuksa oksYVehVj esa
Sol. At resonance voltages across C and L are in opposite phase so net voltage will be zero.
vuqukn dh fLFkfr esa C rFkk L ij foHko foijhr dyk eas gksrs gS
So, vr% V2 = 0.

D-6. In the series LCR circuit as shown in figure, the voltmeter and ammeter readings are :
fp=kkuqlkj LCR Js.kh ifjiFk esa, oksYVehVj rFkk vehVj ds ikB;kad gS :

(A*) V = 100 volt,  = 2 amp (B) V = 100 volt,  = 5 amp

(C) V = 1000 volt,  = 2 amp (D) V = 300 volt,  = 1 amp

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Alternating Current
Sol. At resonance vuqukn ij (VC = VL)
V = rms × R
Vrms
= ×R (here z = R) (;gka z = R)
Z
100
V = Vrms = 100 volt & rms = = 2 Amp.
50

Section (E) : Transformer VªkalQkeZj

E-1. A power transformer (step up) with an 1 : 8 turn ratio has 60 Hz, 120 V across the primary; the load in
the secondary is 104 . The current in the secondary is
,d 'kfDr VªkalQkeZj (mPpk;h) esa ?ksjksa dk vuqikr 1 : 8 gSA blds çkFkfed fljksa ij 60 Hz, 120 V rFkk f}rh;d
fljksa ij yksM çfrjks/k 104  tksM+rs gS rks f}rh;d dq.Myh esa /kkjk gS &
(A) 96 A (B) 0.96 A (C) 9.6 A (D*) 96 mA
V2 N 8
Sol. = 2 =
V1 N1 1
V2 = 8 × 120 = 960 volt
960
= = 96 mA.
104

E-2. A transformer is used to light a 140 watt, 24 volt lamp from 240 V AC mains. The current in the main
cable is 0.7 amp. The efficiency of the transformer is :
140 watt, 24 volt ds ySEi dks çdkf'kr djus ds fy, VªkalQkeZj dk mi;ksx djrs gq, 240 V ds eq[; çR;korhZ /kkjk
lzksr ls tksM+k tkrk gSA eq[; dscy ¼rkj½ esa /kkjk 0.7 amp gS rks VªkalQkeZj dh n{krk gSA
(A) 48% (B) 63.8% (C*) 83.3% (D) 90%
E2  2 140
Sol. % = × 100 = × 100 = 83.3%.
E11 240  0.7

E-3. In a step-up transformer the voltage in the primary is 220 V and the current is 5A. The secondary
voltage is found to be 22000 V. The current in the secondary (neglect losses) is
mPpk;h VªkalQkeZj dh çkFkfed dq.Myh esa foHko rFkk /kkjk ds ifjek.k Øe'k% 220 V rFkk 5A gSA f}rh;d dq.Myh esa
foHko 22000 V çkIr gksrk gS rks f}rh;d esa /kkjk gSA (gkfu;ksa dks ux.; ekuksa)
(A) 5 A (B) 50 A (C) 500 A (D*) 0.05 A
Sol. I1 E1 = I2 E2
E 5  220
I2 = 1 1 = = .05 A
E2 22000

E-4. The core of a transformer is laminated to reduce

(A*) eddy current loss (B) hysteresis loss (C) copper loss (D) magnetic loss
VªkalQkeZj dh ØksM dks iV~Vfyr djds fdldks de fd;k tkrk gS
(A*) Hkaoj /kkjk gkfu;k¡ (B) 'kSfFkY; gkfu;k¡ (C) rkEcz gkfu;k¡ (D) pqEcdh; gkfu;k¡

Section (F) : Miscellaneous fofo/k

F-1. A capacitor is a perfect insulator for :
(A*) constant direct current (B) alternating current
(C) direct as well as alternating current (D) variable direct current

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Alternating Current
la/kkfj=k iw.kZ vpkyd dk O;ogkj gS &
(A*) vpj fn"V /kkjk ds fy,A (B) çR;korhZ /kkjk ds fy,A
(C) fn"V /kkjk rFkk çR;korhZ /kkjk nksuksa ds fy,A (D) ifjofrZ fn"V /kkjk ds fy,A

F-2. A choke coil sould have :

(A) high inductance and high resistance (B) low inductance and low resistance
(C*) high inductance and low resistance (D) low inductance and high resistance
,d pksd dq.Myh es gksuk pkfg,&
(A) mPp çsjdRo rFkk mPp çfrjks/k (B) fuEu çsjdRo rFkk fuEu çfrjks/k
(C*) mPp çsjdRo rFkk fuEu çfrjks/k (D) fuEu çsjdRo rFkk mPp çfrjks/k
F-3. A choke coil is preferred to a rheostat in AC circuit as :
(A*) it consumes almost zero power (B) it increases current
(C) it increases power (D) it increases voltage
çR;korhZ /kkjk ifjiFk esa /kkjk fu;=kd (Rheostat) dh txg pksd dq.Myh dks çkFkfedrk nh tkrh gS D;ksfd &
(A*) blesa O;; 'kfDr yxHkx 'kwU; gSA (B) ;g /kkjk dks c<+krh gSA
(C) ;g 'kfDr dks c<+krh gSA (D) ;g foHko dks c<+krh gSA

F-4. With increase in frequency of an AC supply, the inductive reactance :

(A) decreases (B*) increases directly proportional to frequency
(C) increases as square of frequency (D) decreases inversely with frequency
çR;korhZ /kkjk L=kksr dh vko`fr c<+kus ij çsjdh; çfr?kkr &
(A) ?kVsxkA (B*) vko`fr ds lekuqikrh c<+sxkA
(C) vko`fr ds oxZ ds lekuqikrh c<+sxkA (D) vko`fr ds O;qRØekuqikrh ?kVsxkA

F-5. With increase in frequency of an AC supply, the capacitive reactance :

(A*) varies inversely with frequency (B) varies directly with frequency
(C) varies directly as square of frequency (D) remains constant
çR;korhZ /kkjk lIykbZ dh vko`fr c<+kus ij] /kkjrh; çfr?kkr &
(A*) vko`fr ds lkFk O;qRØekuqikrh ifjofrZr gksrk gSA (B) vko`fr ds lkFk lekuqikrh ifjofrZr gksrk gSA
(C) vko`fr ds oxZ ds lekuqikrh ifjofrZr gksrk gSA (D) fu;r jgrk gSA

F-6. An AC ammeter is used to measure current in a circuit. When a given direct constant current passes
through the circuit, the AC ammeter reads 3 ampere. When an alternating current passes through the
circuit, the AC ammeter reads 4 ampere. Then the reading of this ammeter if DC and AC flow through
the circuit simultaneously, is :
fdlh ifjiFk esa /kkjk ekius ds fy, çR;korhZ /kkjk vehVj dk mi;ksx fd;k tkrk gSA tc ifjiFk esa vpj eku dh
fn"V /kkjk çokfgr dh tkrh gS rks vehVj dk ikB;kad 3 ,sfEi;j gSA tc ifjiFk ls vU; çR;korhZ /kkjk çokfgr dh
tkrh gS rks bldk ikB;kad 4 ,sfEi;j gSA tc çR;korhZ /kkjk rFkk fn"V /kkjk nksuksa dks ,d lkFk çokfgr fd;k tkrk gS
rks vehVj dk ikB;kad gksxk &
(A) 3 A (B) 4 A (C) 7 A (D*) 5 A

F-7. In an a.c. circuit consisting of resistance R and inductance L, the voltage across R is 60 volt and that
across L is 80 volt. The total voltage across the combination is
,d çR;korhZ /kkjk ifjiFk esa çfrjks/k R rFkk çsjdRo L gS rFkk buds fljksa ij foHko Øe'k% 60 volt rFkk 80 volt gS rks
la;kstu ds fljksa dk dqy foHko gS &
(A) 140 V (B) 20 V (C*) 100 V (D) 70 V

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Alternating Current
PART - III : MATCH THE COLUMN
Hkkx - III : dkWye dks lqesfyr dhft, (MATCH THE COLUMN )

1. Match the Physical quantities given in column-I with the parameters they depend on as given in
column-II.
Column I Column II
(A) Inductance of a coil (p) Depends on resistivity
(B) Capacitance (q) Depends on shape
(C) Impedance of a coil (r) Depends on medium inserted
(D) Reactance of a capacitor (s) Depends on external AC voltage source

LrEHk-I esa nh xbZ HkkSfrd jkf'k;ksa dks LrEHk-II esa fn;s x;s izkpyksa ftl ij os fuHkZj djrh gS] ls lqesfyr dhft,A
LrEHk-I LrEHk-II
(A) dq.Myh dk izsjdRo (p) izfrjks/kdrk ij fuHkZj djrh gSA
(B) /kkfjrk (q) vkdkj (shape) ij fuHkZj djrh gSA
(C) dq.Myh dh izfrck/kk (r) izosf'kr (inserted) ek/;e ij fuHkZj djrh gSA
(D) la/kkfj=k dh izfr?kkr (s) ckº; AC oksYVst lzksr ij fuHkZj djrh gSA
Ans. (A) q,r (B) q,r (C) p,q,r,s (D) q,r, s
Sol. (A) Inductance of a coil depends on its shape and magnetic properties of its core (medium inserted)
(B) Capacitance of capacitor depends on its shape and dielectric properties of medium inserted.
(C) Impedance of coil R 2  2L2 depends on resistivity (due to R), shape (for L), magnetic properties
of core inserted and also depends on angular frequency  of external voltage source.
1
(D) Reactance of capacitor = depends on shape (for C), nature of dielectric medium (for C) and
C
external voltage source (due to ).
(A) ,d dq.Myh dh izsjdRo blds vkdkj rFkk bldh ØksM ¼izosf'kr ek/;e esa½ ds pqEcdh; xq.kksa ij fuHkZj djrh gSA
(B) la/kkfj=k dh /kkfjrk blds vkdkj rFkk izosf'kr ek/;e ds ijkoS|qr xq.kksa ij fuHkZj djrh gSA
(C) dq.Myh dh izfrck/kk R 2  2L2 izfrjks/kdrk (R ds dkj.k), vkdkj (L ds dkj.k), izosf'kr ØksM ds pqEcdh;
xq.kksa rFkk  dks.kh; vko`fÙk ds cká oksYVst L=kksr ij Hkh fuHkZj djrh gSA
1
(D) la/kkfj=k dh izfr?kkr = vkdkj (C ds fy,), ijkoS|qr ek/;e dh izd`fr (C ds fy,) rFkk cká oksYVst L=kksr
C
(ds dkj.k) ij fuHkZj djrh gSA

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Alternating Current
2. A steady current 4 A flows in an inductor coil when connected to a 12 V dc source as shown in figure 1.
If the same coil is connected to an ac source of 12 V, 50 rad/s, a current of 2.4 A flows in the circuit as
1
shown in figure 2. Now after these observations, a capacitor of capacitance F is connected in series
50
with the coil as shown in figure 3 with the same AC source :
tc ,d izsj.k dq.Myh dks 12 oksYV ds fn"V oksYVrk L=kksr ls fp=k 1 ds vuqlkj tksM+k tkrk gS rks izsj.k dq.Myh esa
fp=kkuqlkj 4 A dh LFkk;h /kkjk izokfgr gksrh gSA ;fn ;gh leku dq.Myh 12 V o 50 rad/s ds izR;korhZ (AC) oksYVrk
L=kksr ls fp=k 2 ds vuqlkj tksM+h tkrh gS rks ifjiFk esa 2.4 A dh /kkjk izokfgr gksrh gSA vc bu izs{k.kksa ds i'pkr~
1
F /kkfjrk ds la/kkfj=k dks izsj.k dq.Myh ds lkFk Js.khØe esa fp=k-3 ds vuqlkj mlh izR;korhZ oksYVrk L=kksr ls
50
tksM+k tkrk gSA

Column-I Column-II (in S.I units) (S.I. i}fr esa)

(A) The inductance of the coil (nearly) (p) 24
izsj.k dq.Myh dk izsjdRo ¼yxHkx½ &
(B) The resistance of the coil (q) 3
dq.Myh dk izfrjks/k&
(C) Average power (nearly) (r) 0.08
vkSlr 'kfä ¼yxHkx½
(D) Total reactance dqy izfr?kkr
Ans. (A)  (r), (B)  (q), (C)  (p), (D)  (q)
Sol. 1 to 2 : When connected with the DC source tc fn"V lzksr ls tksM+k tkrk gSA
12
R==3
4
When connected to ac source tc izR;korhZ lzksr ls tksM+k tkrk gSA 
V
  =
Z
12
 2.4 =  L = 0.08 H
3  2L2
2

2 2
Vrms Vrms R
Using iz;ksx djus ls P = rms Vrms cos  = cos  = = 24 W
Z 1 2
R 2  ( L – )
C

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Alternating current

 Marked Questions can be used as Revision Questions.

 fpfUgr iz'u nksgjkus ;ksX; iz'u gSA
PART - I : ONLY ONE OPTION CORRECT TYPE
Hkkx - I : dsoy ,d lgh fodYi izdkj ¼ONLY ONE OPTION CORRECT TYPE½

Section (B) : Power consumed in an ac circuit

[k.M (B) : çR;korhZ /kkjk ifjiFk esa 'kfDr O;;
2.2
1. A coil has an inductance of H and is joined in series with a resistance of 220 . When an

alternating e.m.f. of 220 V at 50 c.p.s. is applied to it, then the wattless component of the rms current in
the circuit is
(A) 5 ampere (B*) 0.5 ampere (C) 0.7 ampere (D) 7 ampere
Sol. Wattless current = rms sin 
L 2fL
Where tan  = = =1
R R
v v rms 1
and Irms = rms = =
z 2
R  (L) 2
2
Hindi
2.2
,d dq.Myh dk çsjdRo H gS rFkk bls 220  ds çfrjks/k ds lkFk Js.kh esa tksM+k tkrk gSA tc bldks 220 V

fo-ok-cy, 50 pDdj çfr lSd.M vko`fr okys çR;korhZ lzksr ls tksM+k tkrk gS rks ifjiFk esa oxZek/; ewy /kkjk dk
'kfDrghu ?kVd gS
(A) 5 ,sfEi;j (B*) 0.5 ,sfEi;j (C) 0.7 ,sfEi;j (D) 7 ,sfEi;j
Sol. 'kfDrghu /kkjk = rms sin 
L 2fL
tgk¡ tan  = = =1
R R
v v rms 1
rFkk Irms = rms = =
z R2  (L)2 2

Section (C) : AC source with R, L, C connected in series

[k.M (C) : R, L, C Js.khØe ds lkFk çR;korhZ /kkjk lzksr
2. The current in a circuit containing a capacitance C and a resistance R in series leads over the applied

voltage of frequency by. [REE - 91] [REE - 1991]
2

,d ifjiFk esa la/kkfj=k C rFkk çfrjks/k R Js.khØe esa tqM+s gS rks blesa çokfgr /kkjk] vko`fr okys vkjksfir foHko ls
2
fdruh vkxs gSA
 1   1
(A*) tan–1   (B) tan–1 (CR) (C) tan–1    (D) cos–1 (CR)
  CR   R
xc 1/ c 1
Sol. tan = =  = tan–1
R R CR

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Alternating current
 1   1 
3. An inductor  L  H  , a capacitor  C  F and a resistance (3) is connected in series with
 100   500 
an AC voltage source as shown in the figure. The voltage of the AC source is given as V = 10 cos(100
t) volt. What will be the potential difference between A and B ?
 1   1 
,d çsjdRo  L  H  , ,d la/kkfj=k  C  F  rFkk ,d çfrjks/k (3) fp=kkuqlkj AC foHko lzksr ls
 100   500 
Js.khØe esa tqM+s gSA AC lzksr dk foHko V = 10 cos(100 t) oksYV gSA A rFkk B ds e/; foHkokUrj gksxk ?

(A) 8 cos(100 t – 127º) volt (B) 8 cos(100 t – 53º) volt

(C*) 8 cos(100 t – 37º) volt (D) 8 cos(100 t + 37º) volt
Sol. z= 32  (5  1)2 = 5
i = 2 cos(100t + 53º)
vAB = 8 cos(100t + 53º – 90º) volt
vAB = 8 cos(100t – 37º) volt Ans.

4. An ac voltage source V = V0 sin t is connected across resistance R and capacitance C as shown in

1
figure. It is given that R = . The peak current is 0 . If the angular frequency of the voltage source is
C

changed to keeping R and C fixed, then the new peak current in the circuit is :
3
fp=kkuqlkj izfrjks/k R rFkk /kkfjrk C ds fljksa ij ,d izR;korhZ oksYVst V = V0 sin t dk lzksr tksM+k tkrk gSA ;g
1 
fn;k x;k gS fd R = gSA f'k[kj /kkjk 0 gSA ;fn oksYVst lzksr dh dks.kh; vko`fÙk cnydj dj nh tkrh gS
C 3
(R o C dks fu;r j[krs gq;s) rks ifjiFk esa ubZ f'k[kj /kkjk gksxh &

0 0 0 0
(A) (B*) (C) (D)
2 2 3 3

Sol. The peak value of the current is

/kkjk dk f'k[kj eku gS &
V0 V0
0 = =
1 2 R
R2 
 C2
2


when the angular frequency is changed to
3

tc dks.kh; vko`fÙk cnydj gks tkrh gSA
3
The new peak value is
/kkjk dk u;k f'k[kj eku gS&
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Alternating current
V0 V0 V0 0
0= = =  0=
3 4R 2 2R 2
R2 
2C2
5. For a LCR series circuit with an A.C. source of angular frequency .
,d  dks.kh; vko`fÙk ds izR;korhZ /kkjk L=kksr ls laca) ,d LCR Js.kh ifjiFk ds fy,
1 1
(A) circuit will be capacitive if  > ifjiFk /kkfjrh; gksxk ;fn  >
LC LC
1 1
(B) circuit will be inductive if  = = gks rks ifjiFk iszjdh; gksxk
LC LC
(C*) power factor of circuit will by unity if capacitive reactance equals inductive reactance
ifjiFk dk 'kfDr xq.kkad ,d gksxkA ;fn /kkfjrh; izfr?kkr] izsjdh; izfr?kkr ds cjkcj gSA
1 1
(D) current will be leading voltage if  > /kkjk foHko ls vxzxkeh gksxh ;fn  >
LC LC
 1  1
Sol. (C) The circuit will have inductive nature if  >  L   .
LC  LC 
Hence A is false. Also if circuit has inductive nature the current will lag behind voltage. Hence D is also
false.
1  1 
If  =  L  C  the circuit will have resistive nature. Hence B is false
LC  
R 1
Power factor cos = = 1 it L = . Hence C is true.
1 
2 C

R 2   L –
 C 
1  1 
Sol. ;fn  >  L   gS rks ifjiFk çsj.khd çd`fÙk dk gksxkA
LC  LC 
vr% A vlR; gSA ;fn ifjiFk çsj.khd çd`fÙk dk gS rks /kkjk oksYVst ls ihNs gksxhA vr% D Hkh vlR; gSA
1  1 
;fn  =  L  C  gS rks ifjiFk çfrjks/k çd`fÙk dk gksxkA vr% B vlR; gSA
LC  
R 1
'kfDr xq.kkad cos = = 1 it L = . vr% C lR; gSA
 1 
2 C
R 2   L 
 C 

6. An LCR series circuit with 100  resistance is connected to an AC source of 200 V and
angular frequency 300 radians per second. When only the capacitance is removed, the current lags the
voltage by 60°. When only the inductance is removed, the current leads the voltage by 60º. Then the
current and power dissipated in LCR circuit are respectively
,d LCR Js.kh ifjiFk 100  çfrjks/k ds lkFk 200 V rFkk 300 jsfM;u çfr lSd.M dh dks.kh; vko`fr okys çR;korhZ
/kkjk lzksr ls tqM+k gSA tc ifjiFk ls dsoy la/kkfj=k gVk;k tkrk gS rks /kkjk foHko ls 60° i'pxkeh gks tkrh gSA tc
dsoy çsjdRo gVk;k tkrk gS rks /kkjk foHko ls 60º vxzxkeh gks tkrh gS rks LCR ifjiFk esa /kkjk rFkk 'kfDr O;;
Øe'k% gS
(A) 1A, 200 watt. (B) 1A, 400 watt. (C) 2A, 200 watt. (D*) 2A, 400 watt.

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Alternating current
Sol. When all (L,C,R) are connected then net phase difference = 60 – 60 = 0. So, there will be resoance.
tc lHkh (L,C,R) tqM+s gksaxs rks V rFkk  esa dykUrj 60 – 60 = 0 gks tk;sxk] vr% vuqukn gksxkA
V
= = 2A
R
& rFkk P = 2 R = 400 watt.

175
7. In an L-R series circuit (L = mH and R = 12), a variable emf source (V = V0 sin t) of
11
Vrms = 130 2 V and frequency 50 Hz is applied. The current amplitude in the circuit and phase of
current with respect to voltage are respectively(Use = 22/7)
175
L-RJs.kh ifjiFk (L = mH rFkk R = 12) ls ifjorhZ fo-ok-cy lzksr (V = V0 sin t) tqM+k gSA bldk
11
Vrms = 130 2 V rFkk vko`fr 50 Hz gSA ifjiFk esa /kkjk vk;ke rFkk /kkjk dh foHko ds lkis{k dyk Øe'k% gksaxs &
(= 22/7 dk mi;ksx djsa)
5 5 5
(A) 14.14A, 30° (B) 10 2 A, tan-1 (C) 10 A, tan-1 (D*) 20 A, tan-1
12 12 12

Sol. 0 = 2 rms = 2 Vrms 0 =

2  130 2
Z R2  (L)2
L  L 
tan =  = tan–1  .
R  R 

Section (D) : Resonance

[k.M (D) : vuqukn
8. In LCR circuit at resonance current in the circuit is 10 2 A. If now frequency of the source is changed
such that now current lags by 45° than applied voltage in the circuit. Which of the following is correct :
vuqukn ij LCR ifjiFk esa /kkjk 10 2 A gSA vc L=kksr dh vko`fr bl izdkj ls cnyrh gS fd ifjiFk esa /kkjk
vkjksfir foHkokUrj ls 45° dykdks.k ihNs pyrh gks] rks dkSulk dFku lgh gSA
(A*) Frequency must be increased and current after the change is 10 A
vko`fr fuf'pr :i ls c<+kbZ tkuh pkfg;s rFkk ifjorZu ds i'pkr ifjiFk esa /kkjk 10 A gksxhA
(B) Frequency must be decreased and current after the change is 10 A
vko`fr fuf'pr :i ls ?kVkbZ tkuh pkfg;s rFkk ifjorZu ds i'pkr ifjiFk esa /kkjk 10 A gksxhA
(C) Frequency must be decreased and current is same as that of initial value
vko`fr fuf'pr :i ls ?kVkbZ tkuh pkfg;s rFkk ifjorZu ds i'pkr /kkjk izkjfEHkd eku ds cjkcj gksxhA
(D) The given information is insufficient to conclude anything
dqN fu"d"kZ ds fy;s nh xbZ tkudkjh v/kqjh gSA
Sol. At resonance vuqukn ij
XL = XC
Vrms
rms = = 10 2 ...(i)
R
 
When current lags by : tc /kkjk ls fiNs pyrh gS
4 4
(R = X)
Vrms Vrms
rms = = 10A
2 2
R X 2R

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Alternating current
Section (E) : Transformer
[k.M (E) : VªkalQkeZj
9. The overall efficiency of a transformer is 90%. The transformer is rated for an output of 9000 watt. The
primary voltage is 1000 volt. The ratio of turns in the primary to the secondary coil is 5 : 1. The iron
losses at full load are 700 watt. The primary coil has a resistance of 1 ohm.
VªkalQkeZj dh dqy izHkkoh n{krk 90% gSA bldh fuxZr 'kfDr 9000 watt vafdr gSA izkFkfed dq.Myh ij foHko
1000 oksYV gSA çkFkfed rFkk f}rh;d dq.Myh esa ?ksjksa dk vuqikr 5 : 1 gSA iw.kZ yksM ij ykSg&gkfu;ka 700 watt gSA
çkFkfed dq.Myh esa çfrjks/k 1 vkse gS
(i) The voltage in secondary coil is :
f}rh;d dq.Myh esa foHko gS &
(A) 1000 volt (B) 5000 volt (C*) 200 volt (D) zero volt
E2 N 1
Sol. = 2 =
E1 N1 5
1000
E2 = = 200 volt.
5
(ii) In the above, the current in the primary coil is :
mijksDr ç'u esa] çkFkfed dq.Myh esa /kkjk gS &
(A) 9 amp (B*) 10 amp (C) 1 amp (D) 4.5 amp
Sol. E22 = E11 × %
90
9000 = 1000 × 1 ×
100
1 = 10 amp.
(iii) In the above, the copper loss in the primary coil is :
mijksDr ç'u esa] çkFkfed dq.Myh esa rkEcz gkfu;ka gS &
(A*) 100 watt (B) 700 watt (C) 200 watt (D) 1000 watt
Sol. çkFkfed dq.Myh esa rkez gkfu;k¡
= 12 R1 = (10)2 × 1= 100.
dqy gkfu;k¡ = E1I1 – E2I2
= 10,000 – 9000
= 1000

(iv) In the above, the copper loss in the secondary coil is :

mijksDr ç'u esa] f}rh;d dq.Myh esa rkEcz gkfu;ka gS &
(A) 100 watt (B) 700 watt (C*) 200 watt (D) 1000 watt
Sol. Cu losses in secondary coil f}rh; dq.Myh esa rkez gkfu
= (1000 – 700) – 100
= 200 watt.
(v) In the above, the current in the secondary coil is :
mijksDr ç'u esa] f}rh;d dq.Myh esa /kkjk gS &
(A) 45 amp (B*) 46 amp (C) 10 amp (D) 50 amp
(A) 45 ,sfEi;j (B*) 46 ,sfEi;j (C) 10 ,sfEi;j (D) 50 ,sfEi;j
Sol. E2 2 = 9000 + 200
9200
2 = = 46 A.
200
(vi) In the above, the resistance of the secondary coil is approximately :
mijksDr ç'u esa] f}rh;d dq.Myh esa çfrjks/k yxHkx gS
(A) 0.01  (B*) 0.1  (C) 0.2  (D) 0.4 
Sol. 22 = R2 = 200
200
R2 = = 0.0945.
(46)2

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Alternating current
10._ In the following circuit the current is in phase with the applied voltage. Therefore, the current in the
circuit and the frequency of the source voltage respectively, are
fuEu ifjiFk esa /kkjk vkjksfir foHko ds lkFk dyk esa gSA blfy, ifjiFk esa /kkjk rFkk L=kksr foHko dh vko`fÙk Øe'k% gS

C
Vi ~
R

[Olympiad (Stage-1) 2017]

v 1 1
(A*) i and rFkk (B) zero and 'wkU; rFkk
R 2 LC LC
C 2 C 2
(C) vi and rFkk (D) 4 and rFkk
L  LC LR2 LC
Ans. (A)
1
Sol. f= the circuit is in resonance with the applied voltage.
2 LC
1
f= ifjiFk vkjksfir foHko ds lkFk vuqukn esa gSA
2 LC
v
i= i
R
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
Hkkx - II : ,dy ,oa f}&iw.kk±d eku izdkj ¼SINGLE AND DOUBLE VALUE INTEGER TYPE½
Section (A) : Average, peak and RMS value
[k.M (A) : vkSlr]'kh"kZ rFkk oxZek/; ewy eku
V0
1. The rms value for the saw-tooth voltage of peak value V0 from t=0 to t=2T as shown in figure is .
x
Find the value of x.
V0
fp=kkuqlkj 'kh"kZ eku V0 okys f=kHkqtkdkj foHko dk oxZek/; ewy t = 0 ls t = 2T ds fy, gks rks x dk eku Kkr
x
djksA

Ans. x=3

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Alternating current
Sol. general equation of V
V dh lkekU; lehdj.k
Vo 2V0
V= t – V0 = t – V0
T/2 T
1 1
2
T
 2  2  T  2V  2
0
 V dt    t – V0 dt 
 0   0 T   V0
Vrms = =   = x = 3 Ans.
 T  T 3
   
   
 
Section (C) : AC source with R, L, C connected in series
[k.M (C) : R, L, C Js.khØe ds lkFk çR;korhZ /kkjk lzksr
2. An inductor (xL = 2) a capacitor (xC = 8) and a resistance (8) is connected in series with an ac
source. The voltage output of A.C source is given by v = 10 cos 100t.
,d izsjd dq.Myh (xL = 2), ,d la/kkfj=k (xC = 8) rFkk ,d izfrjks/k (8) Js.kh Øe esa ,d izR;korhZ /kkjk L=kksr
ls tqM+s gS izR;korhZ /kkjk L=kksr ds }kjk iznÙk foHko v = 10 cos 100t gSA
The instantaneous p.d. between A and B is equal to x × 10–1 volt, when it is half of the voltage output
from source at that instant Find out value of x .
A rFkk B ds e/; rkR{kf.kd foHko iru x × 10–1 volt gks rks x dk eku Kkr djks] ;fn bl {k.k ;g foHko iru L=kksr
}kjk iznÙk foHko dk vk/kk gSA

Ans. 48
Sol. impedance of circuit ifjiFk dh izfrck/kk = R 2  (XC  XL )2

Z= 82  (8  2)2 = 10
The current leads in phase by /kkjk dyk dks.k ls  = 37° vkxs ( XC > XL )
 = 37°
10 cos(100t  37)
i = = cos (100 t + 37°)
Z

The instantaneous potential difference across A B is

AB ds fljksa ij fdlh {k.k izfrck/kk
= m (XC – XL) cos (100t + 37° – 90°)
= 6 cos (100 t – 53°)
The instantaneous potential difference across A B is half of source voltage.
AB ds fljksa ij {k.khd oksYV dk L=kksr oksYVrk dk vk/kk gSA
 6 cos (100  t – 53°) = 5 cos 100 t
1 24
solving we get gy djus ij cos 100 t = =
1  (7 / 24) 2 25
24 24
 instantaneous potential difference {k.khd foHkokUrj = 5 × = volts = 48 × 10–1 V
25 5
x = 48
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Alternating current
3. A 2000 Hz, 20 volt source is connected to a resistance of 20 ohm, an inductance of 0.125/ H and a
capacitance of 500/ nF all in series. Calculate the time (in seconds) in which the resistance (thermal
capacity = 100 joule/ºC) will get heated by 10º C. (Assume no loss of heat)
20  çfrjks/k] 0.125/ H çsjdRo rFkk 500/ nF la/kkfj=k Js.kh Øe esa 2000 Hz, 20 volt ds lzksr ls tqM+s gSA rks
fdrus le; esa çfrjks/k dk rki 10ºC tk;sxk \ (izfrjks/k dh Å"ek /kkfjrk = 100 J/ºC ) (Å"ek gkfu ux.; ekfu,A)
Ans. 50
2
 
 
2  20 
Sol. power 'kfDr (P) = rms R P =  R
2
 2  1  
 R   L – c  
   
where tgk¡  = 2 ×  × 2000
(ms)  100  10
H = (ms) = P (t)   t = = = 50 sec.
P P

4. A series LCR circuit containing a resistance of 120 ohm has angular resonance frequency 4 × 105 rad
s–1. At resonance, the voltage across resistance and inductance are 60V and 40 V respectively. At
frequency the current in the circuit lags the voltage by 45º is equal to x × 105 rad/sec. Find value of x.
[REE - 95]
,d Js.kh LCR ifjiFk esa 120 vkse dk çfrjks/k gS rFkk bldh dks.kh; vuquknh vko`fÙk 4 × 105 rad s–1 gSA vuqukn
ij] çfrjks/k rFkk çsjdRo ij foHkokUrj Øe'k% 60V o 40 V gSA x × 105 rad/sec vko`fÙk ij ifjiFk esa /kkjk]
foHkokUrj ls 45º ihNs gS] rks x dk eku Kkr djks ? [REE - 95]
Ans. 8
1
Sol. = ...(i)
LC
V
 (L) = 40 ...(ii)
R
V
 R = V = 60 ...(iii)
R
40
L = R ...(iv)
60
1 40 R
= ...(v)
C 60
From equation (i), (iv) & (v)
leh0 (i), (iv) rFkk (v) ds fy,
1
L = 2 × 10–4 H; C = µF
32

from phasor diagram, VL – VC = VR

Qstj fp=k ls VL – VC = VR
1
 XL – VC = R  L – =R
 C
1 R  R 2  4L / C
 2L – R – =0  = = 8 × 105 rad/sec.
C 2 L
x=8

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Alternating current
SECTION (D) : RESONANCE
[k.M (D) : vuqukn
5. An LCR circuit has L = 10 mH, R = 150  and C = 1 F connected in series to a source of 150 2 cos
t volt. At a frequency that is 50% of the resonant frequency, calculate the average power (in watt)
dissipated per cycle.
,d LCR ifjiFk esa L = 10 mH, R = 150  rFkk C = 1 F Js.khØe esa 150 2 cos t oksYV ds lzksr ls tqM+s gS
;fn vko`fÙk ] vuquknh vko`fÙk dh 50% gks rks izfr pØ O;f;r vkSlr 'kfDr ¼okWV esa½ Kkr djksA
Ans. 75
50 1 1 1
Sol. f= × fr f=  =
100 2 2  LC 2 LC
1
X = L  = 150
C
V0 150 2 0 V0 R
I0 = = PCW = Irms Vrms cos =  = 75 W
Z R 2
 x 2 2 Z

6. In the figure shown an ideal alternative current (A.C.) source of 10 Volt is connected. Find half of the
total average power (in watts) given by the cell to the circuit.
fp=kkuqlkj 10 oksYV dk ,d vkn'kZ izR;korhZ /kkjk (A.C.) L=kksr ifjiFk esa la;ksftr gSA lSy }kjk ifjiFk dks nh xbZ
dqy vkSlr 'kfDr (okWV esa) dk vk/kk eku Kkr djksA

~
10V
Ans. 9

PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE

Hkkx - III : ,d ;k ,d ls vf/kd lgh fodYi çdkj¼ONE OR MORE THAN ONE OPTION
CORRECT TYPE½

Section (B) : Power consumed in an ac circuit

[k.M (B) : çR;korhZ /kkjk ifjiFk esa 'kfDr O;;
1. Average power consumed in an A.C. series circuit is given by (symbols have their usual meaning) :
çR;korhZ /kkjk Js.kh ifjiFk esa vkSlr 'kfDr O;; fuEu }kjk nh tkrh gS ¼tgk¡ ladsrksa dk vFkZ lkekU; gS½ &
2 2
Emax R max | z | cos 
(A*) Erms rms cos (B*) (rms)2 R (C*) (D*)
2(| z |)2 2
Sol. Pav = (A) = (B) = (C) = (D).
2. An AC source supplies a current of 10 A (rms) to a circuit, rms voltage of source is 100 V. The average
power delivered by the source :
(A) must be 1000 W (B*) may be less than 1000 W
(C) may be greater than 1000 W (D*) may be 1000 W
,d çR;korhZ /kkjk lzksr dk foHko 100 V (o-ek-ewy) gS rFkk bl ifjiFk esa 10 A (o-ek-ewy) eku dh /kkjk çokfgr gks
jgk gS rks lzksr }kjk nh xbZ vkSlr 'kfDr :
(A) 1000 W gksuh pkfg,A (B*) 1000 W ls de gks ldrh gSA
(C) 1000 W ls vf/kd gks ldrh gSA (D*) 1000 W gks ldrh gSA

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Alternating current
Sol. Pavr = Irms Vrms cos 
cos  can not be more than 1 so power can not be more than 1000.
cos  ,d ls vf/kd ugh gks ldrk vr% 'kfDr 1000 ls vf/kd ugh gks ldrh gSA

3. Which of the following quantities have zero average value over a cycle. If an inductor coil having some
resistance is connected to a sinusoidal AC source.
(A*) induced emf in the inductor (B*) current
(C) joule heat (D) magnetic energy stored in the inductor
fuEu esa ls dkSulh jkf'k;ksa dk ,d iw.kZ pØ ds fy, vkSlr eku 'kwU; gSA ;fn dqN çfrjks/k okyh çsj.k dq.Myh
T;koØh; çR;korhZ /kkjk lzksr ls tqM+h gSA
(A*) çsjd esa çsfjr fo-ok-cy (B*) /kkjk
(C) twy Å"ek (D) çsjd esa laxzfgr pqEcdh; ÅtkZ
Sol. Joule heat twy Å"ek  2rms R
1 2
Energy in inducting coil çsj.k dq.Myh dh ÅtkZ = L rms .
2

4. In a series LCR circuit with an AC source (Erms = 50 V and  = 50/ Hz), R = 300 , C = 0.02 mF,
L = 1.0 H, Which of the following is correct
Js.kh LCR ifjiFk esa çR;korhZ /kkjk lzksr (Erms = 50 V rFkk  = 50/ Hz) tqM+k gS rFkk R = 300 , C = 0.02 mF,
L = 1.0 H tqM+k gks rks fuEu esa dkSu lgh gSA
(A*) the rms current in the circuit is 0.1 A
ifjiFk esa oxZek/; ewy /kkjk 0.1 A gSA
(B*) the rms potential difference across the capacitor is 50 V
la/kkfj=k ds fljksa ij oxZek/; ewy foHkokUrj 50 V gSA
(C) the rms potential difference across the capacitor is 14.1 V
la/kkfj=k ds fljksa ij oxZek/; ewy foHkokUrj 14.1 V gSA
(D) the rms current in the circuit is 0.14 A
ifjiFk esa oxZek/; ewy /kkjk 0.14 A gSA

Section (C) : AC source with R, L, C connected in series

[k.M (C) : R, L, C Js.khØe ds lkFk çR;korhZ /kkjk lzksr
5. In an AC series circuit when the instantaneous source voltage is maximum, the instantaneous current is
zero. Connected to the source may be a
(A*) pure capacitor
(B*) pure inductor
(C*) combination of pure an inductor and pure capacitor
(D) pure resistor
çR;korhZ /kkjk Js.kh ifjiFk esa tc rkR{kf.kd foHko vf/kdre gksrk gS rc rkR{kf.kd /kkjk 'kwU; gksrh gS rks lzksr ls tqM+k
gks ldrk gS –
(A*) 'kq) la/kkfj=k (B*) 'kq) çsjdRo
(C*) 'kq) çsjdRo rFkk 'kq) la/kkfj=k dk la;kstu (D) 'kq) çfrjks/k

6. A coil of inductance 5.0 mH and negligible resistance is connected to an oscillator giving an output
voltage E = (10V) sin tWhich of the following is correct
(A*) for  = 100 s–1 peak current is 20 A (B*) for  = 500 s–1 peak current is 4 A
(C*) for  = 1000 s–1 peak current is 2 A(D) for  = 1000 s–1 peak current is 4 A
5.0 mH çsjdRo rFkk ux.; çfrjks/k okyh dq.Myh ,d nksfy=k ls tqM+h gSA bldk fuxZr foHko E = (10V) sin
tgSAlgh dFkuksa dk p;u djksA
(A*)  = 100 s–1 ds fy, 'kh"kZ /kkjk 20 A gSA (B*)  = 500 s–1 ds fy, 'kh"kZ /kkjk 4 A gSA
(C*)  = 1000 s–1 ds fy, 'kh"kZ /kkjk 2 A gSA (D)  = 1000 s–1 ds fy, 'kh"kZ /kkjk 4 A gSA

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Alternating current
V0 10
Sol. I0 = =
L   5  103

7. A pure inductance of 1 henry is connected across a 110 V, 70Hz source. Then correct option are
(Use = 22/7):
(A*) reactance of the circuit is 440   (B*) current of the circuit is 0.25 A
 (C) reactance of the circuit is 880   (D) current of the circuit is 0.5 A
1 gsujh dk 'kq) çsjdRo 110 V, 70Hz lzksr ds lkFk tqM+k gS rks lgh dFku gS & (= 22/7):
(A*) ifjiFk dk çfr?kkr 440 gSA  (B*) ifjiFk esa /kkjk 0.25 A gSA
 (C) ifjiFk dk çfr?kkr 880 gSA  (D) ifjiFk esa /kkjk 0.5 A gSA

8. In the circuit shown in figure, if both the bulbs B1 and B2 are identical :
fp=kkuqlkj ;fn cYc B1 rFkk B2 iw.kZr% leku gS rks :

(A) their brightness will be the same

(B*) B2 will be brighter than B1
(C*) as frequency of supply voltage is increased the brightness of bulb B1 will increase and that of B2
will decrease.
(D) only B2 will glow because the capacitor has infinite impedance
(A) mudh ped leku gksxhA
(B*) B2 dh ped B1 ls vf/kd gksxhA
(C*) ;fn L=kksr foHko dh vko`fr c<+k nh tk;s rks B1 dh ped c<+sxh rFkk B2 dh ?kVsxhA
(D) dsoy B2 pedsxk D;ksafd la/kkfj=k dh çfrck/kk vuUr gSA
220
Sol. IC =
2
 1 
R2   
 C 
Brightness of B1 = IC2 R
B1 dh ped = IC2 R
220
IL =
2
R   L 
2

Brightness of B2 = IL2 R
B2 dh ped = IL2 R
here L > C So, B2 will be brighter.
;gk¡ L > C vr%, B2 vf/kd pedsxkA

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Alternating current
9. A circuit is set up by connecting L = 100 mH, C = 5 F and R =100 in series. An alternating emf of
500
(150 2 ) volt, Hz is applied across this series combination. Which of the following is correct

,d ifjiFk ls L = 100 mH, C = 5 F rFkk R =100 Js.kh esa tqM+s gSA Js.kh la;kstu dks (150 2 ) oksYV ,
500
Hz ds çR;korhZ fo-ok-cy lzksr ls tksM+k x;k gS rks lgh dFkuksa dk p;u djks &

(A*) the impedance of the circuit is 141.4 
 ifjiFk dh çfrck/kk 141.4 gSA
(B*) the average power dissipated across resistance 225 W
çfrjks/k ds fljksa ij O;; vkSlr 'kfDr 225 W gSA
(C*) the average power dissipated across inductor is zero.
çsjd ds fljksa ij vkSlr 'kfDr O;; 0 W gSA
(D*) the average power dissipated across capacitor is zero.
la/kkfj=k ds fljksa ij O;; vkSlr 'kfDr 'kwU; gSA
2
 1 
Sol. Z= R 2   L   = (100)2  (100  200)2 = 100 2
 C 
V
Irms = rms
Z
PR = rms2 R
PL = 0
PC = 0

10. In a series RC circuit with an AC source( peak voltage E0 = 50 V and f = 50 /Hz), R = 300 ,C = 25
F. Then :
Js.kh RC ifjiFk çR;korhZ /kkjk lzksr ls tksM+k gSA tgka ( 'kh"kZ oksYVst E0 = 50 V rFkk f = 50 /Hz), R = 300 ,C
= 25 F gS rks :
(A*) the peak current is 0.1 A (B) the peak current is 0.7 A
(C*) the average power dissipated is 1.5 W (D) the average power dissipated is 3 W
(A*) 'kh"kZ /kkjk 0.1 A gSA (B) 'kh"kZ /kkjk 0.7 A gSA
(C*) vkSlr 'kfDr O;; 1.5 W gSA (D) vkSlr 'kfDr O;; 3 W gSA

SECTION (D) : RESONANCE

[k.M (D) : vuqukn
11. In the AC circuit shown below, the supply voltage has constant rms value V but variable frequency f. At
resonance, the circuit :

R 1
F
1
H
 

V,f
~
V
(A*) has a current  given by  =
R
(B*) has a resonance frequency 500 Hz
(C*) has a voltage across the capacitor which is 1800 out of phase with that across the inductor
V
(D) has a current given by  =
2
 1 1
R2    
 

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Alternating current
fn;s x;s çR;korhZ /kkjk ifjiFk esa L=kksr foHko dk oxZek/; ewy eku V fu;r gS rFkk vko`fr f ifjofrZ gS rks ifjiFk esa
vuqukn dh fLFkfr ij &
R 1
F
1
H
 

V,f
~
V
(A*) ifjiFk esa /kkjk  = gSA
R
(B*) ifjiFk dh vuquknh vko`fr 500 Hz gSA
(C*) ifjiFk esa la/kkfj=k ds fljksa ij foHko çsjdRo ij vkjksfir foHko ls 1800 dykUrj ij gSA
V
(D) ifjiFk esa /kkjk  = gSA
2
2  1 1
R   
 
1 1
Sol. Resonance frequency vuqukn vko`fÙk f = = 500 Hz
2 LC
At resonance vuqukn ij
Z=R
V V
& rFkk I = 
z R
L & C are in out of phase. L rFkk C foifjr dyk esa gksrs gSA

Section (E) : Transformer

[k.M (E) : VªkalQkeZj
12. A town situated 20 km away from a power house at 440 V, requires 600 KW of electric power at 220 V.
The resistance of transmission line carrying power is 0.4  per km. The town gets power from the line
through a 3000 V–220 V step-down transformer at a substation in the town. Which of the following
is/are correct
(A*) The loss in the form of heat is 640 kW (B) The loss in the form of heat is 1240 kW
(C*) Plant should supply 1240 kW (D) Plant should supply 640 kW
440 V fo|qr 'kfDr LVs'ku ls ,d 'kgj 20 km nwj fLFkfr gS ftls 220 V ij 600 Kw fo|qr 'kfDr pkfg,A fo|qr
izokg djus okyh rkj ykbZu dk çfrjks/k 0.4  çfr fdeh0 gSA 'kgj dks lapj.k ykbZu }kjk ,d lc&LVs'ku ij
vipk;h VªkalQkeZj (step-down) 3000 V–220 V ls 'kfDr çkIr gksrh gSA fuEu esa ls dkSuls dFku lgh gS@gS\ &
(A*) Å"ek ds :i esa {k; 640 kW gSA (B) Å"ek ds :i esa {k; 1240 kW gSA
(C*) l;a=k dks 1240 kW vkiwfrZ djuh pkfg,A (D) l;a=k dks 640 kW vkiwfrZ djuh pkfg,A

13. 11 kW of electric power can be transmitted to a distant station at (i) 220 V or (ii) 22000 V. Which of the
following is correct
(i) 220 V ;k (ii) 22000 V ij 11 kW dh fo|qr 'kfDr nqjLr LVs'ku ij Hksth tkrh gS rks fuEu esa ls lgh dFkuksa dk
p;u djksA
(A) first mode of transmission consumes less power lapj.k dk çFke rjhdk de 'kfDr O;; djsxkA
(B*) second mode of transmission consumes less power lapj.k dk f}rh; rjhdk de 'kfDr O;; djsxkA
(C) first mode of transmission draws less current lapj.k dk çFke rjhdk de /kkjk çokfgr djsxkA
(D*) second mode of transmission draws less current lapj.k dk f}rh; rjhdk de /kkjk çokfgr djsxkA

14. Power factor may be equal to 1 for :

fuEu eas ls fdldk 'kfDr xq.kkad ,dkad gks ldrk gS &&
(A) pure inductor (B) pure capacitor (C*) pure resistor (D*) An LCR circuit
(A) 'kq) çsjd (B) 'kq) la/kkfj=k (C*) 'kq) çfrjks/k (D*) ,d LCR ifjiFk

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Alternating current
R R
Sol. cos = = 1
z R  (xL  x C )2
2

Because D;ksafd xL = xC

15._ In a series R-C circuit the supply voltage (Vs) is kept constant at 2V and the frequency f of the
sinusoidal voltage is varied from 500 Hz to 2000 Hz. The voltage across the resistance
R = 1000 ohm is measured each time as VR. For the determination of the C a student wants to draw a
linear graph and try to get C from the slope. Then she may draw a graph of [Olympiad (Stage-1) 2017]

1 V2 1 1 VR
(A) f 2 against VR2 (B*) 2
against S2 (C*) against 2 (D*) f against
f VR f2 VR V  VR2
2
s

R-C Js.khØe ifjiFk esa foHko dh vkiwfrZ (Vs) dks 2V ij fu;e j[krs gS rFkk T;koØh; foHko dh vko`fÙk f dks 500
Hz ls 2000 Hz rd cnyrs gSA R = 1000 vkse izfrjks/k ls ikfjr foHko izR;sd le; VR ukik tkrk gSA C ds
fu/kkZj.k ds fy, ,d fo|kFkhZ ,d js[kh; oØ [khapuk pkgrh gS rFkk <ky ls C dks izkIr djus dh dksf'k'k djrh gS rc
og fuEu ds chp oØ [khap ldrh gSA
1 V2 1 1 VR
(A) f 2 rFkk VR2 (B*) 2
rFkk S2 (C*) rFkk 2 (D*) f rFkk
f VR f2 VR V  VR2
2
s

Ans. (BCD)
VR
Sol. VR 
2
 1 
R2   
 2fc 
2
 1  v2
R2     2 R2
 2fc  vR

 1  v 2  vR2
 2fc   R
  vR2
vR
2fc (R) =
v  vR2
2

vR
Graph between f & is a straight line.
v  vR2
2

1 1
2
vs is also straight line
f vR2

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Alternating current

PART - IV : COMPREHENSION

Hkkx - IV : vuqPNsn (COMPREHENSION)

Comprehension - 1
A voltage source V = V0 sin (100 t) is connected to a black box in which there can be either one element
out of L, C, R or any two of them connected in series.
V = V0 sin (100 t) dk foHko lzksr] ,d dkys cDls (black box) ls tqM+k gqvk gSA bl cDls esa L,C,R esa ls dksbZ ,d
vo;o gks ldrk gS ;k nks vo;o Js.kh Øe esa gks ldrs gSaA
Black Box

V=V0sin(100t)

At steady state. the variation of current in the circuit and the source voltage are plotted together with
time, using an oscilloscope, as shown
LFkkbZ voLFkk esa /kkjk rFkk lzksr ds foHko dk le; ds lkFk ifjorZu nksfy=k dh lgk;rk ls iznf'kZr gSA

1. The element(s) present in black box is/are : dkys cDls esa mifLFkr vo;o gS@gksxs %
(A) only C (B) L and C (C) L and R (D*) R and C
(A) dsoy C (B) L rFkk C (C) L rFkk R (D*) R rFkk C
Sol. As current is leading the source voltage, so circuit should be capacitive in nature and as phase

difference is not , it must contain resistor also.
2

gy pwafd /kkjk foHko lzksr ls dyk esa vkxs gSa blfy;s ifjiFk dh izÑfr /kkjrh; ugha gksxh rFkk dykUrj ugha gSA vr%
2
blesa dqN izfrjks/k mifLFkr gSA
2. Values of the parameters of the elements, present in the black box are -
(A*) R = 50 , C = 200 µf (B) R = 50 , L = 2mµ
(C) R = 400  , C = 50 µ f (D) None of these
dkys ckWDl esa mifLFkr vo;o ds izkapkyksa ds eku gksxs &
(A*) R = 50 , C = 200 µf (B) R = 50 , L = 2mµ
(C) R = 400  , C = 50 µ f (D) buesa ls dksbZ ugha
  
Sol. Time delay foyEc le; = =  
 400 4
 1   1
tan–1  RC  = 4  =R
  C

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Alternating current
v0
i0 =
2
 1 
R2   
 C 
2 = 100
 R = 50 
R2  R2
1
and rFkk C = = 200 µF
50  100

3. If AC source is removed, the circuit is shorted for some time so that capacitor is fully discharged and
then a battery of constant EMF is connected across the black box, at t = 0. The current in the circuit will
-
(A) increase exponentially with time constant = 0.02 sec.
(B*) decrease exponentially with time constant = 0.01 sec.
(C) oscillate with angular frequency 20 rad/sec
(D) first increase and then decrease
vc ;fn AC lzksr gVk fn;k tk, rks ifjiFk dqN le; ds fy, y?kqifFkr gks tkrk gS rkfd la/kkfj=k iw.kZr% fujkosf'kr
gks tk, vkSj rRi'pkr ,d fu;r fo|qr okgd cy dh cSVªh dkys ckWDl (black box) ls t = 0 le; ij tksM+h tkrh
gSA rks ifjiFk esa /kkjk &
(A) pj ?kkrkadh; :i ls le; fu;rkad = 0.02 sec. ds lkFk c<+sxhA
(B*) le; fu;rkad = 0.01 sec. ds lkFk pj ?kkrkadh; :i ls ?kVsxhA
(C) dks.kh; vko`fr 20 rad/sec ls nksyu djsxhA
(D) igys c<+sxh rFkk ckn esa ?kVsxhA

Sol. For DC circuit

DC ifjiFk ds fy,
t
–
RC
i = i0 e and rFkk RC = 0.01 sec.
i

1
t

Comprehension-2
An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown.
lek;ksftr nksfy=k vko`fr okys ,d izR;ko`rh tfu=k G dk iz;ksx iznf'kZr ifjiFk esa fd;k tkrk gSA
R=100
L1=1.6mH
C3=2.5F
G S C1=3F
C2=
4.5F
L2=2.4mH

4. Current drawn from the ac source will be maximum if its angular frequency is -
izR;korhZ lzksr ls yh xbZ /kkjk vf/kdre gksxh ;fn bldh dks.kh; vko`fr gS -
(A) 105 rad/s (B) 104 rad/s (C*) 5000 rad/s (D) 500 rad/s

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Alternating current
Sol. Current drawn is maximum at resonant angular frequency. Leq = 4 mH Ceq = 10 F
vf/kdre /kkjk vuquknh vko`fr dh fLFkfr esa feyrh gSA
Leq = 4 mH Ceq = 10 F
1
= = 5000 rad/s
LC

5. To increase resonant frequency of the circuit, some of the changes in the circuit are carried out. Which
change(s) would certainly result in the increase in resonant frequency ?
(A) R is increased.
(B) L1 is increased and C1 is decreased.
(C) L2 is decreased and C2 is increased.
(D*) C3 is removed from the circuit.
ifjiFk dh vuquknh vko`fr c<+kus ds fy, ifjiFk esa dqN ifjorZu djus iM+rs gSaA dkSuls ifjorZu@ifjorZuksa ds
ifj.kkeLo:i vuquknh vko`fr c<+ tk,xhA
(A) R c<+kus ij (B) L1 c<+kus ij rFkk C1 ?kVkus ij
(C) C2 c<+kus ij rFkk L2 ?kVkus ij (D*) ifjiFk ls C3 gVkus ij
Sol. (D) Ceq decreases thereby increasing resonant frequency.
(D) Ceq ?kVus ds dkj.k vuquknh vko`fr c<+ tk,xh

6. If the ac source G is of 100 V rating at resonant frequency of the circuit, then average power supplied
by the source is -
;fn izR;ko`fr lzksr G dk vuquknh vko`fr ij ikB~;kad 100 V gks rks lzksr }kjk nh xbZ vkSlr 'kfDr gksxh -
(A) 50 W (B*) 100 W (C) 500 W (D) 1000 W
100
Sol. At resonance vuqukn ij irms = = 1A
100
Power supplied nh xbZ 'kfDr = Vrms Irms cos  (= 0 at resonance vuqukn ij = 0 gSA) P = 100 W

7. Average energy stored by the inductor L2 (Source is at resonance frequency) is equal to

izsjdRo L2 }kjk lafpr vkSlr ÅtkZ gS (lzksr vuquknh vko`fr ij gS)
(A) zero (B*) 1.2 mJ (C) 2.4 mJ (D) 4 mJ
1 2
Sol. Average energy stored lafpr vkSlr mtkZ = Lirms
2
1
= (2.4 × 10–3 H) . (1 A)2 = 1.2 mJ
2

8. Thermal energy produced by the resistance R in time duration 1 s, using the source at resonant
condition, is
(A) 0 J (B) 1 J
(C) 100  J (D*) not possible to calculate from the given
information
1 s, le;kUrjky esa izfrjks/k R }kjk mRiUu m"eh; ÅtkZ gksxh] ¼lzksr dh vuquknh fLFkfr ij½ -
(A) 0 J (B) 1 J
(C) 100  J (D*) nh xbZ lwpuk ls x.kuk laHko ugha gSA
Sol. As 1µs time duration is very less than time period T at resonance, thermal energy produced is not
possible to calculate without information about start of the given time duration.
1µs le;kUrjky] vuqukn dh fLFkfr esa vkorZ dky T ls cgqr de gS] vr% bl le;kUrjky ds izkjEHk gksus dh lwpuk
ds vHkko esa mRiUu m"eh; ÅtkZ dh x.kuk laHko ugha gSA

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Alternating current
Comprehension-3
vuqPNsn-3 
In the LCR circuit shown in figure unknown resistance and alternating voltage source are connected.

When switch 'S' is closed then there is a phase difference of between current and applied voltage
4
100
and voltage accross resister is V. When switch is open current and applied voltage are in same
2
phase. Neglecting resistance of connecting wire answer the following questions :
fp=k esa n'kkZ, LCR ifjiFk esa vKkr izfrjks/k rFkk izR;kofrZ oksYVst L=kksr tqM+s gq, gSA tc dqath 'S' dks can fd;k
 100
tkrk gS] rks /kkjk o vkjksfir foHko esa dykUrj gS] rFkk izfrjks/k ij foHkokUrj V gSA tc dqath [kksyh tkrh gS
4 2
/kkjk rFkk vkjksfir foHko leku dyk esa gSSA lEidZ rkjksa dk izfrjks/k ux.; ekurs gq, fuEu iz'uksa ds mÙkj nhft,

9. Peak voltage of applied voltage sources is :

vkjksfir foHko L=kksr dh 'kh"kZ oksYVrk gS %
100
(A) 200 2 V (B) 100 V (C*) 100 2 V (D) V
2
10. Resonance frequency of circuit is :
ifjiFk dh vuqukn vko`fÙk gS %
(A) 50 Hz (B*) 25 Hz
(C) 75 Hz (D) Data insufficient for caculation x.kuk ds fy, vkadM+s vi;kZIr
gS

11. Average power consumption in the circuit when 'S' is open :

tc 'S' [kqyh gS]rc ifjiFk esa vkSlr 'kfDr miHkksx gS :
(A) 2500 W (B) 3000 W (C*) 5000 W (D) 1250 W
Sol. When switch is closed tc dqath can gS

vrms (applied) = 100 volts

vpeak (applied) = 100 2
When switch is open tc daqth [kqyh gS

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Alternating current
1 50
f=  = 25 Hz
1 1 2
2 
25  100
Resistance izfrjks/k R = XL = XC = 2fL = 2
2
 100  10000
Average power consumption vkSlr 'kfDr miHkksx =   2 = 5000 W.
 2  2

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Alternating current

 Marked Questions can be used as Revision Questions.

 fpfUgr iz'u nksgjkus ;ksX; iz'u gSA
* Marked Questions may have more than one correct option.
* fpfUgr iz'u ,d ls vf/kd lgh fodYi okys iz'u gS -
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
Hkkx - I : JEE (ADVANCED) / IIT-JEE ¼fiNys o"kksZ½ ds iz'u
1. An AC voltage source of variable angular frequency  and fixed amplitude V connected in series with a
capacitance C and an electric bulb of resistance R (inductance zero). When  is increased :
[JEE 2010; 3/163, –1]
(A) the bulb glows dimmer (B*) the bulb glows brighter
(C) total impedence of the circuit is unchanged (D) total impedence of the circuit increases
,d C /kkfjrk okys la/kkfj=k rFkk R çfrjks/k okys ,d fo|qr cYc ¼ftldk çsjdRo 'kwU; gS½ dks ,d ifjorhZ dks.kh;
vko`fÙk  rFkk fLFkj vk;ke V0 okys AC oksYVrk L=kksr ls Js.khØe esa tksM+k x;k gSA  dk eku c<+kus ij %
[ JEE 2010; 3/163, –1]
(A) cYc dh nhfIr eUn gks tkrh gS (B*) cYc dh nhfIr rhoz gks tkrh gS
(C) ifjiFk dh dqy çfrck/kk ugha cnyrh gS (D) ifjiFk dh dqy çfrck/kk c<+ tkrh gS
Vrms
Sol. irms =
1
R2  2 2
c
when  increases, irms increases so the bulb glows brighter
Vrms
Sol. irms =
1
R2  2 2
c
tc  c<sxh] irms c<sxh blfy;s cYc dh nhfIr rhoz gks tkrh gSA

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Alternating current
2. You are given many resistances, capacitors and inductors. These are connected to a variable DC
voltage source (the first two circuits) or an AC voltage source of 50 Hz frequency (the next three
circuits) in different ways as shown in Column . When a current  (steady state for DC or rms for AC)
flows through the circuit, the corresponding voltage V1 and V2. (indicated in circuits) are related as
shown in Column . Match the two column.
[ JEE 2010; 8/163 ]
vki dks dbZ çfrjks/k] la/kkfj=k ,oa çsjdRo fn;s gSaA budks ,d ifjorhZ DC oksYVrk Jksr ¼igys nks ifjiFk½ ;k 50 Hz
dk AC oksYVrk Jksr ¼ckn ds rhu ifjiFk½ ls vyx&vyx rjhds ls dkWye  esa fn[kk;s x;s fp=kksa ds vuqlkj tksM+k
x;k gSA ifjiFk esa  /kkjk (fLFkj voLFkk DC ds fy;s ;k o-ek-ew- AC ds fy;s) çokfgr gksus ij oksYVrk V1 rFkk V2
(ifjiFkksa esa n'kkZ;h xbZ) dk vkil esa lacU/k dkWye  esa fn[kk;k x;k gSA nksukas dk esy dhft;sA
[ JEE 2010; 8/163 ]
Column  Column 

(A)   0,V1 is proportional to  (p) V

(B)   0,V2 > V1 (q)

(C) V1= 0, V2 = V (r)

(D)   0,V2 is proportional to  (s)

(t)

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Alternating current
dkWye  dkWye 

(A)   0,V1 lh/ks ds lekuqikrh gS (p) V

(B)   0,V2 > V1 (q)

(C) V1= 0, V2 = V (r)

(D)   0, V2 ds lekuqikrh gSA (s)

(t)

Ans. (A) – r,s,t ; (B) – q,r,s,t ; (C) – p,q ; (D) – q,r,s,t

As per given conditions, there will be no steady state in circuit ‘p’, so it should not be
considered in options of ‘c’.
nh xbZ fLFkfr esa ‘p’ esa LFkkbZ voLFkk ugha gSA vr% ;g fodYi ‘c’ esa ugha vkuk pkfg,A

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Alternating current
Sol. (p)

V
As  is steady state current
V1 = 0 ; =0
Hence, V2 = V
So , answer of P  C
(q)

In the steady state ;

d
V1 = 0 as =0
dt
 V2 = V = R
or V2 
and V2 > V1
So , answer of q  B, C, D

(r)

Inductive reactance XL = L
XL = 6 × 10–1 
and resistance = R = 2
So, V1 = XL
and V2 = R
Hence, V2 > V1
So, Answer of r  A,B,D
(s)

Here, V1 = XL, where, XL = 6 × 10–1 

4
10
Also, V2 = XC, where,XC =
3
So, V2 > V1
V1 
V2 
So, answer of s  A,B,D
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Alternating current
(t)

104
Here, V1 = R, where, R = 1000  , XC = 
3
104
V2 = XC , where, XC = 
3
So, V2 > V1
and V1 
V2 
So, answer of t  A,B,D
Ans. (A) – r,s,t ; (B) – q,r,s,t ; (C) – p,q ; (D) – q,r,s,t
Note : For circuit ‘p’ :
Ldi q di d2i dq d2i 1 dq
V–  = 0 or CV = CL + q or 0 = LC 2  or 2

dt C dt dt dt dt LC dt
 1 
So, i = i0 sin  t  0 
 LC 
As per given conditions, there will be no steady state in circuit ‘p’. So it should not be considered in
options
Sol. (p)

V
 LFkkbZ voLFkk /kkjk gS
V1 = 0 ; =0
vr%V2 = V
vr% P dk lgh mÙkj C gSA
(q)

LFkkbZ voLFkk esa

d
V1 = 0 as =0
dt
 V2 = V = R
;k V2 
rFkk V2 > V1
vr% q dk mÙkj  B, C, D
(r)

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Alternating current

çsjdh; çfr?kkr XL = L
XL = 6 × 10–1 
rFkk çfrjks/k = R = 2
vr% V1 = XL
rFkk V2 = R
vr%, V2 > V1
vr% r dk mÙkj  A,B,D gSA
(s)

vr%, V1 = XL, tgk¡, XL = 6 × 10–1 

104
vr%, V2 = XC, tgk¡, XC =
3
vr%, V2 > V1
V1 
V2 
vr% s dk mÙkj  A,B,D
(t)

104
vr%, V1 = R, tgk¡, R = 1000  , XC = 
3
104
V2 = XC , tgk¡, XC = 
3
vr% V2 > V1
rFkk V1 
V2 
vr% t dk mÙkj  A,B,D
Ans. (A) – r,s,t ; (B) – q,r,s,t ; (C) – p,q ; (D) – q,r,s,t
Note : ifjiFk 'p' ds fy,
Ldi q di d2i dq d2i 1 dq
V–  = 0 ;k CV = CL + q ;k 0 = LC  ;k 
dt C dt dt 2 dt dt 2 LC dt
 1 
vr%, i = i0 sin  t  0 
 LC 
nh xbZ fLFkfr esa ‘p’ esa LFkkbZ voLFkk ugha gSA vr% ;g fodYi ‘c’ esa ugha vkuk pkfg,A
Ans. (A) – r,s,t ; (B) – q,r,s,t ; (C) – q ; (D) – q,r,s,t
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Alternating current
3. A series R-C circuit is connected to AC voltage source. Consider two cases; (A) when C is without a
dielectric medium and (B) when C is filled with dielectric of constant 4. The current IR through the
resistor and voltage VC across the capacitor are compared in the two cases. Which of the following
is/are true?
,d lhjht R-C ifjiFk AC oksYVst lzksr ls tqMk gSA nks fLFkfr;ksa ij fopkj dhft;s% (A) tc C esa ijkoS|qr ugh gS
vkSj (B) tc C esa 4 ijk oS|qrkad dk inkFkZ Hkjk gSA izfrjks/k R ls izokfgr /kkjk IR rFkk la/kkfj=k C ij foHkokarj VC
dh rqyuk bu nks fLFkfr;ksa esa dh xbZ gSA rc fuEu esa ls dkSu lgh gS (gSSa) ? [ JEE 2011; 4/160]
(A) RA  RB (B*) RA  BR (C*) VCA  VCB (D) VCA  VCB
A.C. Ans. (BC)

2
 1 
Sol. Case I Z= R2   
 C 

Case II
V
RA  Z´ < Z
Z
V
BR  RA  BR
Z´
VRA  VRB
So. VCA  VCB  VR2  VC2  V02 

2
 1 
Sol. fLFkfr I Z= R2   
 C 

fLFkfr II

V
RA  Z´ < Z
Z
V
BR  RA  BR
Z´
VRA  VRB
vr% VCA  VCB  VR2  VC2  V02 

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Alternating current
4. A series R-C combination is connected to an AC voltage of angular frequency  = 500 radian/s. If the
impedance of the R-C circuit is R 1.25 , the time constant (in millisecond) of the circuit is
Js.kh&c) R rFkk C dks = 500 radian/s dks.kh; vko`fRr ds AC lzksr ls tksM+k x;k gS ;fn R-C ifjiFk dh çfrck/kk
R 1.25 gks rc mldk dkykad (time constant) ¼feyh lsad.M esa½ gksxkA [ JEE 2011; 4/160 ]
Ans. 4
Sol.

W = 500 rad/s
2
 1  2
Z    R = R 1.25
 C 
2 2
 1  2 2  1  2 2 R2
  + R = R (1.25)   +R =R +
 C   C  4
1 R 2 2
   CR = = sec.
C 2  500
2 2  1000
= × 103 ms = ms = 4 ms
500 500

5. In the given circuit, the AC source has  = 100 rad/s. considering the inductor and capacitor to be
ideal, the correct choice (s) is(are) [IIT-JEE-2012, Paper-2; 4/66]
fn;s x;s ifjiFk esa AC L=kksr dh vko`fr  = 100 rad/s gSA iszjdRo rFkk la/kkfj=k dks vkn'kZ ekudj lgh fodYi
(fodYiksa) dk pquko djsaA

(A*) The current through the circuit,  is approximately 0.3 A

(B) The current through the circuit,  is 0.3 2 A.
(C*) The voltage across 100 resistor = 10 2 V
(D) The voltage across 50 resistor = 10V
(A *) ifjiFk esa /kkjk yxHkx I = 0.3 A gSA
(B ) ifjiFk esa /kkjk I = 0.3 2 A gSA
(C *) 100  izfrjks/k ds fljks ij foHko 10 2 V gSA
(D ) 50  izfrjks/k ds fljks ij foHko 10 V gSA
Ans. (A,C or C)
1
Since rms =  0.3 A so A may or may not be correct.
10
1
D;ksafd rms =  0.3 A gS blfy, A dks lgh Hkh ekuk tk ldrk vkSj xyr Hkh ekuk tk ldrk gSA
10

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Alternating current
1 1
Sol. C = 100 F, =
C (100) (100  10 6 )
XC = 100 , XL = L = (100) (.5) = 50 
Z1 = x  100 = 100 2
2
C
2

Z2 = xL2  502 = 502  502 = 50 2

 = 20 2 sin t
20 2
i1 = sin (t + /4)
100 2
1
i1 = sin (t + /4)
5
20 2
I2 = sin (t – /4)
50 2

I= (.2)2  (.4)2

= (.2) 1  4
1 1
= 5=
5 5
1 1 10
(I)rms = = =
2 5 10 10
 0.3A
 0.2  20
(V100)rms = (I1)rms) × 100 =   × 100 = = 10 2 V
 2 2
 0.4  20
V50)rms =   × 50 = = 10 2 V
 2 2
1
Since rms   0.3 A so A may or may not be correct.
10
1
D;ksafd rms=   0.3 A gS blfy, A dks lgh Hkh ekuk tk ldrk vkSj xyr Hkh ekuk tk ldrk gSA
10

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Alternating current
Paragraph for Questions 6 and 7
iz'u 6 vkSj 7 ds fy, vuqPNsn
A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a
place 20 km away from the power plant for consumers' usage. It can be transported either directly
with a cable of large current carrying capacity or by using a combination of step-up and step-down
transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In
the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is
used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-
down transformer is used to supply power to the consumers at the specified lower voltage. It is
reasonable to assume that the power cable is purely resistive and the transformers are ideal with a
power factor unity. All the currents and voltages mentioned are rms values.
_[JEE(Advanced)-2013; 3/60]]
,d rkih; fo|qr la;a=k 600 kW dh 'kfDr 4000 V ij mRikfnr djrk gS] tks 20 km dh nwjh ij miHkksDrkvksa ds
mi;ksx ds fy, ys tk;h tkrh gSA bldks ;k rks mPp /kkjk ogu&{kerk okys dsfcy ls Hkstk tk ldrk gS ;k nksauks
fljksa ij mPpk;h o vipk;h VªkalQkWeZj dk ç;ksx djds fd;k tk ldrk gSA izR;{k izs"k.k dk nks"k ;g gS fd bles
ÅtkZ dk {k; cgqr vf/kd gksrk gS tcfd VªkWUlQkWeZj ds mi;ksx ds rjhds esa {k; cgqr de gksrk gSA bl rjhds esa ,d
mPpk;h VªkWUlQkWeZj la;a=k dh vksj yxk;k tkrk gS ftlls /kkjk dk eku de gks tk,A miHkksDrk ds fljs esa vipk;h
VªkWUlQkWeZj dk iz;ksx fd;k tkrk gS ftlls miHkksdrkvksa dks ,d fo'ks"k de oksYV ij fo|qr 'kfDr nh tk ldsA ;g
ekuk tk ldrk gS fd dsfcy 'kq) izfrjksf/kr gS rFkk VªkWUlQkWeZj vkn'kZ gSa] o mudk 'kfDr xq.kkad ,d gSA mfYyf[kr
leLr /kkjkvksa o oksYVrkvksa dk eki rms gSA

6. If the direct transmission method with a cable of resistance 0.4  km–1 is used, the power dissipation (in
%) during transmission is :
;fn ,sls dsfcy dk mi;skx fd;k tk, ftldk izfrjks/k 0.4  km–1 gS rc izR;{k isz"k.k dh fLFkfr esa 'kfDr {k;
( % esa) gS :
(A) 20 (B*) 30 (C) 40 (D) 50
Ans. (B)
Sol. P = 600 × 1000 = 4000 × I  I = 150 A
Power loss 'kfDr {k; = 2r = (150)2 × 0.4 ×20 = 180kW
Power loss 180
Power loss percentage izfr'kr esa 'kfDr {k; = ×100 =  100
Power input 600
 30 %

7. In the method using the transformers, assume that the ratio of the number of turns in the primary to that
in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied
at 200V, the ratio of the number of turns in the primary to that in the secondary in the step-down
transformer is :
VªkWUlQkWeZj ds iz;ksx djus okyh fof/k esa] ;g ekusa mPpk;h VªkWUlQkeZj ds izkFkfed o f}rh;d esa yisVksa dh la[;k dk
vuqikr 1 : 10 gSA ;fn fo|qr 'kfDr] miHkksDrkvksa dks 200V ij nh tkrh gS rks vipk;h VªkWUlQkWeZj esa izkFkfed o
f}rh;d ds yisaVksa dh la[;k dk vuqikr gS :
(A) 200 : 1 (B) 150 : 1 (C) 100 : 1 (D) 50 : 1
Ans. (A)
Np 40,000 200
Sol.  =
Ns 200 1

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Alternating current
8. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and alternating
current 
(t) = 0cos(t,), with 0 = 1A and  = 500 rad s–1 starts flowing in it with the initial direction shown in the
7
figure. At t = , the key is switched from B to D. Now onwards only A and D are connected. A total
6
charge Q flows from the battery to charge the capacitor fully. If C = 20, R = 10 and the battery is
ideal with emf of 50V, identify the correct statement (s) [JEE (Advanced)-2014,P-1, 3/60]

7
(A) Magnitude of the maximum charge on the capacitor before t = is 1 × 10–3 C.
6
7
(B) The current in the left part of the circuit just before t = is clockwise
6
(C*) Immediately after A is connected to D. the current in R is 10A.
(D*) Q = 2 × 10–3 C.
fp=k esa n'kkZ, x;s ifjiFk esa le; t = 0 ij fcUnq A dks fLop }kjk fcUnq B ls tksM+k tkrk gSA blls ifjiFk esa ,d
izR;korhZ /kkjk (t) = 0cos(t) fp=k esa fn[kkbZ xbZ fn'kk esa izokfgr gksus yxrh gS] tgk¡ 0 = 1A rFkk  = 500 rad s–1A
7
le; t = ij fLop dks fcUnq B ls gVkdj fcUnq D ls tksM+k tkrk gSA blds i'pkr~ flQZ A rFkk D tqM+s gq, gSA
6
la/kkfj=k dks iwjh rjg vkosf'kr djus ds fy, cSVjh ls dqy vkos'k Q izokfgr gksrk gSA ;fn C = 20, R = 10 rFkk
cSVjh 50V fo|qr okgd cy okyh vkn'kZ cSVjh gks rc lgh fodYi@fodYiksa dks pqfu,A

7
(A) la/kkfj=k ij le; t = ls igys vf/kdre vkos'k dk ifjek.k 1 × 10–3 C gSA
6
7
(B) ck¡, ifjiFk esa le; t = ls Bhd igys fo|qr /kkjk nf{k.kkorhZ (clockwise) gSA
6
(C*) fcUnq A dks fcUnq D ls tksM+us ds rqjUr i'pkr~ izfrjks/k R esa fo|qr /kkjk dk eku 10A gSA
(D*) Q = 2 × 10–3 C.
Ans. (C), (D)

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Alternating current

Sol. Charge on capacitor will be maximum at t =
2
Qmax = 2 × 10–3 C
7
(A) charge supplied by source from t = 0 to t =
6
7 7 7
6 sin
 sin500t  6  6 = –1mC
Q=  cos(500t)dt =   =
0  500  0 500
Just after switching In steady state

Apply KVL just after switching

Q
50 + 1  R  0   = 10 A
C
In steady state Q2 = 1mC
net charge flown from battery = 2mC

Hindi la/kkfj=k ij vkos'k t = ij vf/kdre gksxk
2
Qmax = 2 × 10–3 C
7
(A) t = 0 ls t = rd L=kksr }kjk lIykbZ vkos'k
6
7 7 7
6 sin
 sin500t  6  6
Q =  cos(500t)dt =   = 500 = –1mC
0  500  0

dqath can djus ds Bhd ckn LFkk;h voLFkk esa

Q1
dqath can djus ds Bhd ckn KVL yxkus ij 50 +  R  0   = 10 A
C
LFkk;h voLFkk esa Q2 = 1mC cSVjh ls izokfgr dqy vkos'k = 2mC

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Alternating current
9. In the circuit shown, L = 1 H, C = 1 F and R = 1 k. They are connected in series with an a.c. source
V = V0 sint as shown. Which of the following options is/are correct ?
[JEE (Advanced) 2017, P-1, 4/61, –2]
L = 1H C = 1F R = 1k

~ V0sint

(A) The current will be in phase with the voltage if  = 104 rad.s–1
(B) At  >> 106 rad.s–1, the circuit behaves like a capacitor
(C*) The frequency at which the current will be in phase with the voltage is independent of R
(D*) At  ~ 0 the current flowing through the circuit becomes nearly zero
fp=k esa fn[kk;s x, ifjiFk L = 1 H, C = 1 F, R = 1 k gSA ,d ifjorhZ oksYVrk ( V = V0 sint) L=kksr ls Js.kh
lac/k gSA fuEu esa dkSu lk (ls) dFku lgh gS@gSaA?
L = 1H C = 1F R = 1k

~ V0sint

(A) tc  = 104 rad.s–1 gksxh rc fo|qr /kkjk (electric current) oksYVrk dh ledyk esa gksxhA
(B) tc  >> 106 rad.s–1, ifjiFk la/kkfj=k (capacitor) dh rjg O;ogkj djrk gSA
(C*) tc fo|qr /kkjk oksYVrk dh ledyk esa gksxh rks og vkofÙkZ R ij fuHkZj ugh djsxhA
(D*) tc  ~ 0 gksxh rc ifjiFk esa cgrh /kkjk 'kwU; ds fudV gksxhA

Ans. (CD)
Sol. Current will be in phase with voltage at resonant frequency.
vuquknh vko`fÙk ij /kkjk rFkk oksYVrk leku dyk esa gksxsaA
1 1
L =  0 = = 106 sec–1
C LC

If ;fn  > 0 Circuit behaves like inductive. ifjiFk izsjdh; gksxkA

If ;fn ~0 Z  I0

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Alternating current
10. The instantaneous voltages at three terminals marked X, Y and Z are given by
rhu VfeZuyksa ds fcUnqvksa X, Y ,oa Z ds fy, rkR{kf.kd oksYVrk (instantaneous voltage) nh xbZ gSA
VX = V0 sint,

 2   4 
VY = V0 sin  t  and VZ = V0 sin  t  3 
 3   
An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is
connected between points X and Y and then between Y and Z. The reading (s) of the voltmeter will be
[JEE(Advanced) 2017 ; P-2, 4/61, –2]
,d vkn'kZ oksYVekih (ideal voltmeter) nks fcUnqvksa ds foHkokUrj dk vkj ,e ,l (root mean square, Vrms) eku
nsrk gSA ;g oksYVekih fcUnq X ,oa Y ls tksMk tkrk gS fQj Y ,oa Z ls tksMk tkrk gSA bl oksYVekih dk ekiu
gksxk@gksxsaA

rms 3
(A*) VXY  V0
2

rms 1
(B) VYZ  V0
2
(C*) independent of the choice of the two terminals (fdlh Hkh nks fcUnqvksa ds p;u ij fuHkZj ugha djrk)
rms
(D) VXY  V0
Ans. (AC)
Sol. Vxy = Vx – Vy = (Vxy)0 sin (t + 1)
2
(Vxy)0 = V02  V02  2V02 cos  3V0
3
(Vxy )0 3
(Vxy)rms =  V0
2 2
Vyz = Vy – Vz = (Vyz)0 sin (t + 2)
2
(Vyz)0 = V02  V02  2V02 cos  3V0
3
(Vyz )0 3
(Vyz)rms =  V0
2 2
Vxz = Vx – Vz = (Vxz)0 sin (t + 3)
4
(Vxz)0 = V02  V02  2V02 cos  3V0
3
(Vyz )0 3
(Vxz)rms =  V0
2 2

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Alternating current
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
Hkkx - II : JEE (MAIN) / AIEEE ¼fiNys o"kksZ½ ds iz'u
1. A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be
:
fdlh ifjiFk dk çfrjks/k 12 ohm rFkk çfrck/kk 15 ohm gSA ifjiFk dk 'kfDr xq.kkad gksxk &
[AIEEE 2005; 4/300]
(1*) 0.8 (2) 0.4 (3) 1.25 (4) 0.125
Sol. Power factor
R
= cos  =
Z
12 4
= 
15 5
= 0.8

2. The phase difference between the alternating current and emf is /2. Which of the following cannot be
the constituent of the circuit?
çR;korhZ /kkjk rFkk fo|qrokgd cy ds chp dykUrj /2 gSA fuEufyf[kr esa ls dkSu bl ifjiFk dk vo;o ugha gks
ldrk gS ? [AIEEE 2005; 4/300]
(1) C alone (2*) R, L (3) L, C (4) L alone
(1) dsoy C (2*) R, L (3) L, C (4) dsoy L
X 1
Sol. tan  = =   R=0
R 0

3. In a series LCR circuit R = 200  and the voltage and the frequency of the main supply is 220 V and 50
Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by
30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power
dissipated in the LCR circuit is
,d Js.kh LCR ifjiFk esa R = 200 vkSj eq[; iznk;h L=kksr dh oksYVrk ,oa vko`fÙk Øe'k% 220 V ,oa 50 Hz gSA
ifjiFk esa ls la/kkfj=k fudky ysus ij /kkjk oksYVrk ls 30º i'p gks tkrh gSA ifjiFk esa ls izsjd fudky ysus ij /kkjk
oksYVrk ls 30º vxz gks tkrh gSA LCR ifjiFk esa [kir 'kfä gS [AIEEE 2010; 4/144, –1]
(1) 305 W (2) 210 W (3) zero 'kwU; W (4*) 242 W
XL R 200
Sol. tan 30º =   XL = =
R 3 3
XC 200
tan 30º =  Xc =
R 3
2
Z= R  (XL  XC ) = 200 
220
irms = = 1.1
200
P = (irms )2 × R = (1.1)2 × 200
P = 242 W

4. An arc lamp requires a direct current of 10 A at 80 V to function. if it is connected to a 220 V(rms), 50

Hz AC supply, the series inductor needed for it to work is close to : [JEE (Main) 2016, 4/120, –1]
,d vkdZ ySEi dks izdkf'kr djus ds fy;s 80 V ij 10 A dh fn"V /kkjk (DC) dh vko';drk gksrh gSA mlh vkdZ
dks 220 V(rms), 50 Hz izR;korhZ /kkjk (AC) ls pykus ds fy;s Js.kh esa yxus okys izsjdRo dk eku gSA
(1) 0.08 H (2) 0.044 H (3) 0.065 H (4) 80 H
Ans. (3)

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Alternating current
80
Sol. R= = 8
10
80V

VL2  80 2  220 2
VL2 = (220 + 80) (220 – 80)

= 300 × 140  VL = 204.9

Irns XL = 204.9
220
xL = 2.5
64  x L2

1
5. For an RLC circuit driven with voltage of amplitude m and frequency 0  the current exhibits
LC
resonance. The quality factor, Q is given by : [JEE (Main) 2018; 4/120, –1]
1
m vk;ke rFkk 0  vko`fÙk ds foHko }kjk pfyr ,d RLC ifjiFk vuqukfnr gksrk gSA xq.krk dkjd Q dk
LC
eku gksxk :
R CR 0L 0R
(1) (2) (3*) (4)
(0C) 0 R L
Ans. (3)

v0

Sol. vm

vm
2

w1 w0 w
w2
R
Band width csUM pkSM+kbZ 2 – 1 =
L
0 L
Quality factor fo'ks"krk xq.kkad Q =  0
2  1 R

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Alternating current
6. In an a.c circuit, the instantaneous e.m.f and current are given by
,d a.c ifjiFk ds fo|qr okgy cy rFkk /kkjk dk rkR{kf.kd eku fuEufyf[kr lehdj.kksa ls fn;k x;k gSA
e = 100 sin 30t
 
i = 20 sin  30 t – 
 4
In one cycle of a.c the average power consumed by the circuit and the wattless current are,
respectively: [JEE (Main) 2018; 4/120, –1]
a.c ds ,d iw.kZ pØ es aifjiFk }kjk vkSlr 'kfDRk O;; rFkk okVghu /kkjk ds eku] Øe'k% gS %
50 1000
(1) ,0 (2) 50,0 (3) 50,10 (4*) ,10
2 2
Ans. (4)
Sol. e = 100 sin30t
 
i = 20sin  30t  
 4
100 20 1 1000
Pav = erms irms cos = . .  W
2 2 2 2
I sin  20 1
wattless current 'kfDrghu /kkjk = 0 = . = 10A
2 2 2

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Alternating current

SUBJECTIVE QUESTIONS
fo"k;kRed iz'u ¼SUBJECTIVE QUESTIONS½

1. A current of 4 A flows in a coil when connected to a 12 V d.c. source. If the same coil is connected to a
12 V, 50 rad/s, AC source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil.
Also, find the power developed in the circuit if a 2500 µF condenser is connected in series with coil.
[REE - 1993]
tc fdlh dq.Myh dks 12 V ds fn"V /kkjk lzksr ls tksM+k tkrk gS rks mlesa 4 A dh /kkjk çokfgr gksrh gSA ;fn blh
dq.Myh dks 12 V, 50 rad/s, ds çR;korhZ /kkjk lzksr ls tksM+rs gS rks 2.4 A dh /kkjk çokfgr gksrh gS rks dq.Myh dk
çsjdRo Kkr djks\ ;fn dq.Myh ds lkFk 2500 µF dk la/kkfj=k Js.khØe esa tksM+ fn;k tk;s rks ifjiFk esa mRiUu
'kfDr Kkr djks \ [REE - 1993]
Ans. .08 H; 17.28 watt
Sol. given that fn;k gSaµ
V 12
R= = = 3
I 4
12
Z= = 5
2.4
Z2 = R2 + (L)2
25 = 9 + (50 × L)2  L = 0.08 H
Now for LCR circuit, Z = R 2  (XL – XC )2
vc LCR ifjiFk ds fy, Z = R 2  (XL – XC )2
2
2  1 
= 3   50  0.08 – –6 
= 5 .
 50  2500  10 
2 2
 12   12 
So Power mRiUu 'kfDr =    R =    3 = 17.28 watt.
 Z   5 

2. A box P and a coil Q are connected in series with an AC source of variable frequency. The EMF. of
source is constant at 10 V. Box P contains a capacitance of 1 µF in series with a resistance of 32 .
Coil Q has a self inductance 4.9 mH and a resistance of 68 . The frequency is adjusted so that the
maximum current flows in P and Q. Find the impedance of P and Q at this frequency. Also find the
voltage across P and Q respectively. [REE - 98]
,d ckWDl P rFkk ,d dq.Myh Q dks Js.khØe esa ,d ifjorhZ vko`fÙk okys çR;korhZ&/kkjk lzksr ls tksM+k tkrk gSA lzksr
dk fo|qrokgd cy 10 V ij fLFkj jgrk gSA cDls P esa 1 µF dk la/kkfj=k rFkk 32  dk ,d çfrjks/k Js.khØe esa tqM+s
gSA dq.Myh Q dk LoçsjdRo 4.9 mH rFkk çfrjks/k 68  gSA lzksr dh vko`fÙk bl çdkj lek;ksftr dh tkrh gS fd P
rFkk Q esa egÙke /kkjk çokfgr gksrh gSA bl vko`fÙk ij fd P rFkk Q dh çfrck/kk;sa Kkr dhft,A P rFkk Q ds fljksa
ij foHko dh Hkh x.kuk dhft,A
[REE - 1998]
Ans. P=76.96 ,Q=97.59  P  7.7 V; Q = 9.8 V, net impedance = 100 
 P=76.96 ,Q=97.59  P  7.7 V; Q = 9.8 V, dqy çfrck/kk = 100 

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Alternating current
Sol.

Here ;gk¡ L = 4.9 mH

C = 1F
R = 68 + 32 = 100 
1 1
= =  105 rad/sec
LC 7
ZP = (32)2  (1/ C)2 = 76.96 

ZQ = (68)2  (L)2 = 97.59 

Total imedence dqy izfrck/kk, Z = R 2  (XL – XC )2

2
2  105 7  10 –5 
= 100    4.9  10 –3 –  = 100
 7 10 –6 
10
Irms = A
100
VP = ZP × Irms = 7.7 V
VQ = ZQ × Irms = 9.8 V

50
3._ In a series LCR circuit with an ac source of 50 V, R = 300, frequency  = Hz. The average electric

field energy, stored in the capacitor and average magnetic energy stored in the coil are 25 mJ and 5 mJ
respectively. The RMS current in the circuit is 0.10 A. Then find:
(a) Capacitance (c) of capacitor
(b) Inductance (L) of inductor.
(c) The sum of rms potential difference across the three elements.
50
LCR Js.khØe ifjiFk esa izR;korhZ L=kksr 50 V, R = 300, vko`fÙk  = Hz gV~Zt gSA la/kkfj=k esa lafpr vkSlr

fo/kqr ÅtkZ rFkk dq.Myh esa lafpr vkSlr pqEcdh; mtkZ Øe'k% 25 mJ rFkk 5mJ gS tc fd ifjiFk esa oxZek/;ewy
/kkjk 0.10 A gSA rc Kkr djs &
(a) la/kkfj=k dh /kkfjrk (c)
(b) dq.Myh dk izsjdRo (L)
(c) rhuksa rRoksa ds fljksa ij oxZek/; ewy foHkokUrj dk ;ksxA
Ans. (a) C = 20 F (b) 1 H (c) 90 V
1 2 
Sol. Av. electric field energy vkSlr fo|qr ÅtkZ =  CVrms  = 25 × 10 J
–3

2 
1 1
 × c.2rms × 2 2 2
= 25 × 10–3 J  C = 20 F
2 4  c
1 
Av. magnetic energy vkSlr pqEcdh; ÅtkZ  L2rms  = 5 × 10–3
2 

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Alternating current
2  5  10 3
 L=  L= 1 henry gsujh
(0.10)2

VR = rms.R VC = rms XC VL = rms × L

1 50
= 0.10 × 300 = (0.10)  = 0.10  2  (1)
 50  6 
2    20  10
  
= VR = 30 V VC = 50 V VL = 10 V
rms voltate of source L=kksr dk oxZ ek/; ewy Erms = 50 V (given in the question) iz'u esa fn;k x;k gSA

4. An inductor 20 × 10–3 Henry, a capacitor 100 µF and a resistor 50 are connected in series across a
source of EMF V = 10 sin 314 t. Find the energy dissipated in the circuit in 20 minutes. If resistance is
removed from the circuit and the value of inductance is doubled, then find the variation of current with
time (t in second) in the new circuit.
20 × 10–3 gsujh izsjdRo] 100 µF dk la/kkfj=k rFkk 50 dk izfrjks/k V = 10 sin 314 t fo-ok-cy L=kksr ls Js.kh Øe
esa tqM+s gSA ifjiFk esa 20 feuV es gksus okyh ÅtkZ gkfu Kkr djks\ ;fn ifjiFk ls çfrjks/k gVk fn;k tk;s rFkk
çsjdRo dk eku nqxuk dj fn;k tk;s rks u;s ifjiFk esa /kkjk dk le; ¼lSd.M½ ds lkFk ifjorZu Kkr djks\
[REE - 99]
Ans. 951.52 J; 0.52 cos 314 t
Solu. Given that fn;k gqvk gS
L = 20 × 10 H
–3

C = 100 × 10–6 F
R = 50   = 314
V = 10 sin 314 t t = 20 × 60 second
2
2
 V0 
(i) H =  R (t) =  R × t
rms  2 Z 
 
2
 1  2
here z = R 2   L –  = (50)2   6.28 – 31.8  = 56.15 
 c 
V0
(ii) I = I0 sin (314 t + /2) where tgk¡ I0 =
 1 
 (2L) – c 
 

5. The electric current in an AC circuit is given by i = i0 sin t. What is the time taken by the current to
change from its maximum value to the rms value?
izR;korhZ/kkjk ifjiFk esa i = i0 sin t /kkjk izokfgr gks jgh gSA /kkjk dks blds vf/kdre eku ls o-ek- ewy eku rd
ifjofrZr gksus esa fdruk le; yxsxk\ [REE - 99]

Ans. T/8 or
4
Solu. I1 = I0 sin  t1
I0 = I0 sin  t1
/2 T
t1 = =
2 / T 4

& I2 = 0 = I0 sin ( t2)
2
/ 4 T
t2 = = 
2 / T 8
T T T
 t = t1 – t2 = – = Ans
4 8 8

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Alternating current
6. A circuit containing a 0.1 H inductor and a 500 F capacitor in series is connected to a 230 volt, 100/
Hz supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element. (c) What is the average power
transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the
total average power absorbed by the circuit? [‘Average’ implies average over one cycle.]
,d ifjiFk esa 0.1 H çsjdRo rFkk 500 F dk la/kkfj=k Js.khØe esa 230 volt, 100/ Hz ds L=kksr ls tqM+s gSA ifjiFk
dk çfrjks/k ux.; gSA (a) /kkjk dk 'kh"kZ eku rFkk o-ek-ewy eku Kkr djks \ (b) çR;sd vo;o ds fljksa ij foHkoikr
dk oxZek/; ewy eku Kkr djks \ (c) çsjdRo dks LFkkukUrfjr vkSlr 'kfDr Kkr djks \ (d) la/kkfj=k dks LFkkukUrfjr
vkSlr 'kfDr Kkr djks \ (e) ifjiFk dh dqy vkSlr 'kfDr Kkr djks \ [‘;gk¡ vkSlr eku] dqy ,d pØ ds fy, Kkr
djuk gS]
Ans. (a) 23 2 A, 23 A (b)460 volt, 230 volt(c) zero 'kwU; (d) zero 'kwU; (e) zero 'kwU;.
Sol. Give that (fn;k x;k gS)
L = 0.1 H
C = 500 × 10–6 F
100
Vrms = 230 volt, f = Hz

V 230
(a) Irms = rms =
Z Z
1 1
where tgk¡ Z = L – = 2f L – = 20 – 10 = 10
c 2fc
& IO = 2 × Irms = 23 2
 1 
(b) VL = Irms (L) VC = Irms  
 c 
(c) < PL> = Irms Vrms cos  here tgk¡  = 90º So <P> = 0
(d) <PC> = 0 (e) < PNet> = 0

7. A series LCR circuit with L = 0.125/ H, C = 500/ nF, R = 23  is connected to a 230 V variable frequency
supply.
(a) What is the source frequency for which current amplitude is maximum? Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain
the value of this maximum power. (c) For what reactance of the circuit , the power transferred to the
circuit is half the power at resonance? What is the current amplitude at this reactance? (d)If  is the
angular frequency at which the power consumed in the circuit is half the power at resonance,write an
expression for  (e) What is the Q-factor (Quality factor) of the given circuit?
Js.kh LCR ifjiFk 230 V ifjofrZr vko`fr okys lzksr ls tqM+k gS] tgka L = 0.125/ H, C = 500/ nF, R = 23 gS
(a) /kkjk vk;ke ds vf/kdre eku ds fy, lzksr dh vko`fr D;k gksxh \ bldk vf/kdre eku Kkr djksA
(b) lzksr dh fdl vko`fr ds fy, ifjiFk }kjk O;; vkSlr 'kfDr vf/kdre gksxh \ vf/kdre 'kfDr dk eku Kkr
djksA (c) ifjiFk ds fdl çfr?kkr ds fy, ifjiFk dks LFkkukUrfjr 'kfDr vuqukn ij izkIr 'kfDr dh vk/kh gS \ bl
çfr?kkr ij /kkjk vk;ke D;k gS \ (d) ;fn dks.kh; vko`fÙk  ij ifjiFk esa O;; 'kfDr vuqukn ij 'kfDr dh vk/kh gS
rks bl ds fy, O;atd Kkr djks\ (e) fn, x, ifjiFk dk Q-xq.kkad ¼fo'ks"krka xq.kkad½ D;k gS ?
Ans. (a) 2000 Hz, 10 2 A (b) 2000 Hz, 2300 watt (c) 23  10 A.
0.125 1 109
(d)     23 (e) 500/23
 500


Sol. Given that (fn;k x;k gS)
500
L = 0.125/ C= × 10–9 F

R = 23 Vrms = 230 volt
1 1 230
(a) f R = = 2000 Hz Irms (max) = = 10 Amp.
2 CL R
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Alternating current
2
(b) at Resonance vuqukn ij fR = 2000 Pmax = rms R = (10)2 × 23 = 2300 watt
2
Pmax 2 rms R 2 
(c) = ´rms R = ´rms R = ´rms = rms
2 2 2
Vrms rms 230 10
= = = x = 23 
2 2 2 2
R x 2 (23)  x 2
´0 =  ´rms × 2 = 10 Amp.
1
(d) L – = + 23    = ?
c
0.125
2    2000 
r L   500 .
(e) Q = =
R 23 23
8. The maximum values of the alternating voltages and current are 400 V and 20 A respectively in a circuit
connected to 50 Hz supply and these quantities are sinusoidal. The instantaneous values of the voltage
and current are 200 2 V and 10 A respectively. At that instant both are increasing positively.
Determine the average power consumed in the circuit.
,d ifjiFk ftls 50Hz vko`fÙk L=kksr ls tksM+k x;k gS] esa izR;korhZ oksYVrk rFkk /kkjk dk vf/kdre eku Øe'k% 400 V
rFkk 20 A gS o ;s T;koØh; :i gSA oksYVrk o /kkjk dk rkR{kf.kd eku Øe'k% 200 2 V rFkk 10 A gSA bl {k.k
nksuksa /kukRed fn'kk esa o`f)eku gSA ifjiFk esa O;f;r vkSlr 'kfDr Kkr dhft,A
Ans. P = 3864 W
Sol.
 = 2f = (100) rad/s
From the above two figures we can write :

 
V = 400 sin(t + 1) = 400 sin 100t  
 4
 
i = 20 sin(t + 2) = 20 sin 100t  
 6
Phase difference between V and i :

 = (/4 – /6) = or 15º
12
 400   20 
P = Vrmsirms cos =     cos15º = 3864 W
 2   2

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Alternating current
gy

 = 2f = (100) rad/s

mi;qZDr nks fp=kksa ls ge fy[k ldrs gS :
 
V = 400 sin(t + 1) = 400 sin 100t  
 4
 
i = 20 sin(t + 2) = 20 sin 100t  
 6
V rFkk i esa dykUrj :

 = (/4 – /6) = or 15º
12
 400   20 
P = Vrmsirms cos =     cos15º = 3864 W
 2   2
9. A 750 Hz, 20 V source is connected to a resistance of 100 , a capacitance of 1.0 F and an
inductance of 0.18 H in series. Calculate the following quantities : (Olympaid 2013-14)
(a) Impedence of the circuit
(b) Draw an impedence diagram with suitable scale
(c) Power factor
(d) The time in which the resistance will get heated by 10ºC, provided that the thermal capacity of
resistance = 2 J/ºC
750 Hz vko`fr, 20 V L=kksr ds lkFk 100 izfrjks/k, 1.0 F la/kkfj=k rFkk 0.18 H dk izsjdRo Js.kh Øe esa tqM+s gq,s
gSA fuEu jkf’'k;ksa dk eku Kkr djksA (Olympaid 2013-14)
(a) ifjiFk dh izfrck/kk
(b) mi;qDr iSekus ds lkFk izfrck/kk vkjs[k iznf'kZr djks
(c) 'kfDr xq.kkad
(d) og le; Kkr djks ftlesa izfrjks/k 10ºC xeZ gks tkrk gS, fn;k x;k gS fd izfrjks/k dh Åf"e; /kkfjrk = 2 J/ºC gS
Sol.

z= (XL  XC )2  R 2
  2n  2(750)  1500
XL = L  (1500) (0.18)  848.5 
1 106
XC = = = 212.12 
C 1500

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Alternating current
z= (636.4)2  (100)2 = 100 (6.364)2  12 = 100 × 6.44  644 
XL  XC 848.5  212.12
tan   = = 6.36
R 100
R 100
(C) cos   = = 0.155
z 644
(B) impedence is constant as n is constant izfrck/kk fu;r gS D;ksfd vko`fr n fu;r gS
 20
(D) irms = rms = A
z 644
2
2  20 
H =  irms  Rt    (100)t
 644 
2
 20 
(ms) () =   (100) t
 644 
2
 20 
(2) (10) =   (100) t
 644 
2
 644  1
t=    × 20
 20  100
t = 207.36 sec.

10. In the given circuit fn;s x;s ifjiFk esa

R1 = 4 XL = 3

R2 = 6 XC = 8

R3 = 10 XL = 10

XC = 10

~
Vrms = 100 volt

Calculate Kkr dhft;sA

(a) Current in each branch. izR;sd 'kk[kk esa fo|qr /kkjk
(b) Power generated in each resistance izR;sd izfrjks/k esa mRiUu 'kfDrA
(c) Total power generated in the circuit. ifjiFk esa mRiUu dqy 'kfDrA
(d) Net current drawn from source. L=kksr ls izokfgr dqy /kkjkA
(e) Net impedance of the circuit. ifjiFk dh dqy izfrck/kk
Sol. Impedence of each branch izR;sd 'kk[kk dh izfrck/kk
Z1 = R12  XL2  42  32  5

Z2 = 6 2  82  10
Z3 = (10)2  (10  10)2  10
(a) Current in each branch izR;sd 'kk[kk esa /kkjk
V 100
 1 =   20 Amp.
Z1 5

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Alternating current
100
2   10 Amp.
10
100
3   10 Amp.
10
(b) Power in each branch. izR;sd 'kk[kk esa 'kfDr
P1 = (1)2 R1 = (20)2 (4) = 1600 watt
P2 = (2)2 R2 = (10)2 (6) = 600 watt
P3 = (3)2 R3 = (10)2 (10) = 1000 watt
(c) Net power of the circuit. ifjiFk dh dqy 'kfDr
P = P1 + P2 + P3 = 3200 watt
phase difference between voltage & current in each branch.
izR;sd 'kk[kk esa /kkjk ,oa foHkokUrj ds e/; dykUrj
XL 3
tan 1    1  37
R1 4
XC 8
tan 2    2  53
R2 6
tan 3  0  3  0

53°
V
37°

(d) Net current drawn from source. L=kksr ls izokfgr /kkjk

(1 cos 37   2 cos 53   3 )2  (1 sin37   2 sin53)2  1040 
V 100
 (e) Net impedance of the circuit. ifjiFk dh izHkkoh izfrck/kk Z  
 1040

11. A metallic coil of N turns of radius a, resistance R, and inductance L is held fixed with its axis along a

spatial uniform magnetic field B whose magnitude is given by B0 sin(t).
f=kT;k a rFkk N Qsjksa okyh tM+or~ /kkfRod dq.Myh (coil) dk çfrjks/k R rFkk çsjdRo L dks le&LFkkfud pqEcdh;

{ks=k B ftldk ifjek.k B0 sin(t) gS] esa bl rjg j[kk gS rkfd dq.Myh dk v{k pqEcdh; {ks=k ds vuqfn'k gSA
(a) Write the emf equation for the current i in the coil.
(b) Assuming that in the steady state i. oscillates with the same frequency  as the magnetic field,
obtain the expression for i.
(c) Obtain the force per unit length. Further obtain its oscillatory part and the time-averaged
compressional part.
(d) Calculate the time-averaged compressional force per unit length given that B0 = 1.00 tesla, N = 10,
a = 10.0 cm,  = 1000.0 rad-s–1, R = 10.0  , L = 100.0 mH.
(a) /kkjk i ds fy, emf dh lehdj.k fyf[k,A
(b) LFkkbZ voLFkk esa i, pqEcdh; {ks=k dh vko`fÙk  ls gh nksfyr ekurs gq,] i ds fy, O;atd çkIr dhft,A
(c) çfr bdkbZ yEckbZ ds fy, cy çkIr djsA bldk nksfyr eku rFkk lEihM+u Hkkx dk le;&vkSlr eku Kkr djsaA
(d) çfr bdkbZ yEckbZ lEihM+u cy dk le; vkSlr eku x.kuk djsaA fn;k x;k gS B0 = 1.00 Vslyk, N = 10, a =
10.0 cm,  = 1000.0 rad-s–1, R = 10.0  , L = 100.0 mH

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Alternating current
di
Sol. (a) i R + L = – Na2 B0  cos t
dt
N a2 B0  R cos t  L sin t 
(b) i =
R  2L2
2

dF NB2 a2 
(c) = – 20 2 2 (R sin t cos t +  L sin2 t)
d R  L
dF NB02 a2 2L
= =–
d av 2(R 2  2L2 )

dF NB02 a 2
= =– (R sin 2 t – L cos 2 t)
d osc 2(R2  2L2 )
dF
(d) = 1.55 N.m–1
d av

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