22

ALTERNATING CURRENT 22
22.1 (B)
E Sin ot

Ifi=,sin (at- ) then v,=lsin (uf-¢ -a2)

and v (aL) 1, sin (ot-¢ +2).
So
Ve +v,+ v, =, sin ot
0+v, , sin ut Ya E, sin ot
ol-1
Also tan= Q 0 =0, so i=i sin ot
R

Hence answer is (B) z= R

22.2 (D)
When capcitance is removed

lan oL
R
or ol 100tan 60° ...(1)
when inductance is remuved
1 1
tan = 100 tan 60° ...(2)
(uCNR)
From equation (1) 8& (2) oL=ac
So it is condition of resonance.
so z = R= 100 Q
I= v/R= 200/100 =2A
Power P=1R=4 x 100 = 400 W
22.3 (D)
From the rating of the bulb, the resistance of the bulb can be calculated.

200V ,50 Hz
2 A
Rens=100 2
P

R=1009Q
For the bulb to be operated at its rated value the rms current
through it should be 1A
Also, =m

200
.
1=J100 4+(2n50L)?
22.4 (A)
According to given problem,
V
. (1)

.2)
and,

Substituting the value ofI from Equation (1) in (2).

9

112

Ans.
R R

22.5 (C)
j=3 sin ot + 4 cos ot

=5 (sin (ot+8)]

Sid
5 mean value =
rms value =

intervals.
mean value will be different for various time
:. Initial value of time is not given hence the equation (1) indicates that it is ahead of V by 5 where
V, sinat then i given by
If voltage applied is V= C.
0<S<90 which indicates that the circuit contains R &
Hence (C).
22.6 (A)
n/4)
VE V, sin (ot+ x/4) = V,cos (ot- can
Since V Ilags current, an inductor bring it in phase with current.

22.7 (B)
cK,= 100)
200v RE
1002) L K= 20002)
Ve200

=2

IR
V 200
00-2A I= i+r2 =2/7 Amp.
22,8
ol>.
The circuit will have inductive natureifa>naturethe current willlag behind voltage Hence Dis alsota
inductive
Hence Ais faise. Alsoif circuitthas
nature. Hence Bis false
1 1
oGthe circuit will have resistance

1
=.Hence Cis true.
R =1 itol
Power factor cos =

22.9 (C)
X= Xç at resonance
=1. for both circuits

22.10 (D)
4 greater than 1)
Since, cose
R IR 8
(Also cose can never be

Hence (C) is wrong.

Also, Ix
Current will be leading.
In a LCR circuit
= (6-12)2 +82
V 10; which is less than voltage drop across capacitor.
22.11 (C)
If we have all R.Land C then I vs. E will be:
To obtain a leading phase difference of n/4:
if X, < X, and we use all R, L and C in the circuit,
then the resultant graph will be : IX

tant
Resultan

which can give a leading phase difference of n/4:

Similarly, if we have only resistance and capacitor then we can obtain a phase difference of
a/4 (leading) for suitable values of I, X, and R. But we cannot obtain aleading phase difference ofal4
if we use only capacitor (phase difference of n/2), or only (inductor and resistor) (phase differenca of |
2), or only resistor (phase difference of 0).
22.12 (A)
z= 3+4 -5 Ans.
22.32 0

2
2:7
, sinotdt Im1-coso 2res

=0
Sol. 2r
27

area of i-t graph of one cycle is zero.

It can be seen graphically that the
</> in one cycle
 22.33 10

Sol. impedance of circuit=R+(Xc-X

z- +(0-2) =100
22.34 24 i

Sol. The currant leads in phase by (:: Xo>X)

=37°

10 cos (100 nt + 37°) ECos (100 rt+ 37°) K=37

R=8

The instantaneous potential difference across ABis

=I e-X,)cos (100t +37°-90°)
=6 cos (100 xt- 53°)
The instantanoous potential difference across ABis half of source voltage.
6 cos (100 t-53°)=5 cos 100 t
1
24
solving we get cos 100 nt =(7124)2:= 25
24 24
instantaneous potential difference =5 % 25 volts
5

22.35 128

Sol. 128
Vm= 16 +202 =25.6 V=V
22.36 3

Sol. V TI4 T

T 1/2

TI4

22.37 8

C=10002)
Sol. 200V V-200
1002) (X= 2002)

200 2A
Ie100
200
r=-Xe 100
=2A
I= +12 = 242 Amp.
22.38 200

tan 60 = = Voc
R or tan 60 R
oL= 1l oC
Impendance of circuit =R
Current in the circuit

2Ap.
Average power P-vJ,cos (as cos =1)
P= 200 x2 x1= 400 wat.

22.39 x=3
Lremoved

100 W3
cOs 30° =
2
1o0+X

Cremoved
100
Cos 60°=
/ho0°x

u (LC) =3
oC

1 300
=10043

10043
. frequency = Hz
27
= resonant frequency
22.40 1
R Z=
Sol. Given 7 =0.8 4
1
1
Since o = G. resonant frequency LC

Xe=R