Starets Academy

A CLASS ABOVE. GUARANTEED Time: 60 Mins

Date: 16-06-2024
11-12-2024
ALTERNATING CURRENT - SOLUTIONS

Course: NEET Total Marks: 180

Total Questions: 50
Section: Test
PHYSICS
Series Standard: ATHENA
XII 25
7. [1]
PHYSICS A
Bvl
I= ; P = I 2R.
1. [3] R
The power consumed in an RL circuit- Bvl 0.5 × 2 × 2 1
l= = = A
P = V rms × I rms × cosϕ R 6 3
Power factor, 1 2
R
P= l 2R = ×6= W
9 3
cosϕ =
√R 2 + ( ωL ) 2 8. [3]
cosϕ is nearly zero when R ≈ 0 or when L has a very 200√2
large value. P av = V rmsI rmscosθ =
√2
⋅
2

√2
( )
cos 30 ∘ = 100√6 w
Therefore, real power consumption in a circuit is the
9. [3]

y
least when it contains low R and high L.
P av = E vI vcosϕ = 100 × 10cosϕ = 1000cosϕ Watt
2. [1]

m
So depending upon value of ϕ, P av can be equal to

e
Consider the expression for the average power loss.
or less than 1000 W.

d
P avg = V rms ⋅ I rms ⋅ cosop

a
May be less than 1000 W.
R

c
= V rms ⋅ I rms ⋅ 10. [1]
√R + (x )
A
2
L − XC 2 Ev R
P = E vI vcosϕ; P = E v

ts
Z Z
3. [4]
[1]

e
Conceptual. E 2v R 110 × 110 × 11

r
or P = = = 275 W

ta
4. [4] Z2 22 × 22
< P⟩ = IVcosϕ

S
11. [3]
20 10 I = 3sinωt + 4cosωt
= × × cos60 ∘
√2 √2 I 2 = 9sin 2ωt + 16cos 2ωt + 24sinωtcosωt
= 50 W I 2 = 9 + 7cos 2ωt + 12sinωt
T
5. [3] 2
7
P avg = V mmsI mscos(Δϕ) ∫ I dt = 9T + 2 T + 0
0

=
100
√2
×
100 × 10 − 3
√2
× cos
()
π
3
=
25
2
T

10 4
1 ∫ T0 I 2dt 25
= × × 10 − 3 I 2rms = T
= T
2 2 2
T
dt
10
= = 2.5 W 25 5

6. [1]
4 I rms =
√ 2
=
√2
< P >= IVcosϕ 12. [2]
20 2
= × × cos60 ∘
√2 √2
= 10 W
13. [1] 19. [1]
Average value of 2sin100ωt over one time period is Angle traversed by phasor = 60 ∘
zero. T π
⇒ Δt = 6 = 3 × 100π
i avg = 3 A 1
= 300 s
14. [1]
Conceptual. 20. [2]
V rms
15. [3] i rms =
Z
Conceptual.
16. [1] Z= √R + (X
2
L − XC 2 )
I = I 0 + I 1sinωt
τ
Z= √22 + (4 − 2) 2 = 2√2 Ω
∫ I 2dt
2π 220 / √2 220
2 0
I rms = τ T= ⇒ i rms = = = 55 A
ω 2√2 4
∫ dt
0 21. [4]
T
( )
Lω o 1 L

√
2 2
∫ I 0 + I 1sin 2ωt + 2I 0I 1sinωt dt Q factor is R = R ,
C
0
=

y
T
∫ dt since ω o = 1 / √(LC)

m
O
22. [3]

e
2
I 1T
I 20T + 2 + O Z = X L − X C = 10Ω
2
I rms =
T

a dV 0 20

c
I0 = = = 2A
2 Z 10

√
I1

A
2 2
⇒ I rms = I0 + I rms = = √2 A

ts
2 2
√
17. [2]

re 23. [2]

ta
∫ I 2dt 1 1
I rms =
√ ∫ dt
ω0 =
√LC
=
√5 × 80 × 10 − 6
= 50 rad / s

4
S R

√
∫ 4tdt ω = ω0 ±
2 2L
= 4
20
∫ dt ω 1 = 50 − = 50 − 2 = 48 rad / s
2 2×5
20
ω 2 = 50 + = 50 + 2 = 52 rad / s
2×5
( 2t )
√
2 4
2 24. [4]
= 4 = 2√3 A
( t )2 The given circuit is under resonance as X L = X C,
ϕ = 0 ∘ . Hence power dissipated in the circuit is
V2
18. [2] P = R = 161.33 ≈ 161 W
An inductance and a resistance are connected in
series with an AC circuit. In this circuit the current
and the potential difference across the resistance
lags behind potential difference across the
inductance by π / 2.
25. [2] 28. [2]
V = 100sin(314t) Initially, inductor does not allow flux through it to
ω = 314 change, therefore current passes through the series
V R = IR resisters 10Ω and 20Ω respectively.
80 But, after a long time (i.e. in steady state), inductor
I=
20 acts as a short. Therefore, 10Ω will not be effective
2 1
I = 4A Hence, current through battery, I = 20 = 10 = 0.1 A
V 2R + V 2C = 100
√ 29. [4]
When switch is connected in position a, then
(80) 2 + V 2C = (100) 2
capacitor will be charged.
V 2C = (100) 2 − (80) 2 And charge on capacitor is q = C. E(E = V)
V C = 60 Potential in battery
V C = IX C We know that,
q2
1 Energy store in capacitor (U) = 2C
60 = 4 ×
ωC Again,
4 When switch is position b then circuit behaves as
C= = 212 μF
60 × 314 L.C combination.
26. [1]

R=
V2
=
40 × 40
= 80 Ω y
and amplitude of current in LC circuit is I o

m
Hence, apply energy conservation,

e
P 20
Maximum electrical energy = maximum magnetic

d
40
i = 80 = 0.5A

a
energy

c
V rms 220 q2 1
i rms = 0.5A =
Z
=
Z = LI 2o ( ∵ q = CE)

A
2C 2
Z = 440 Ω

ts
2
C 2E 2 2
√R + = 440
2 = LI o
XC
C

√6400 + X = 440 ⇒ X
2 2

re
= (440) 2 − 6400 I o = E √C / L

ta
C C

X 2C 30. [1]

S
= 187200 ⇒ X C = √187200 R
X C = 432.67 Ω cosϕ = Z
1 R
27. [3] =
√ 2 R 2 + (Lω) 2
√
In a series LCR circuit variation of charge on
capacitor equation is given by 1 R2
=
d 2q
dt 2
R dq
+ L dt +
( )1
LC
q= L
V 2 R 2 + (Lω) 2
R 2 + (Lω) 2 = 2R 2
If you compare this with the spring mass damped R
L=
oscillator 1000

()
d 2x b dx k R
+ m dt + x=0 cosϕ ′ =

√ (
dt 2 m

where b is the damping constant, then R is the

equivalent damping constant in the LCR oscillation
R2 +
R
1000
× 2000 ) 2

R
equation. =
√ R 2 + (2R) 2
1
=
√5
31. [1] 35. [1]
q = q 0cosωt For L − C oscillations, Energy stored in inductor =
dq Energy stored in capacitor
I= = − ωq 0sinωt 1 1
dt 2 2
2
Li m
= 2
CV m
q2 1
∵ = Li 2 Given, V m = 25 V, C = 10μF = 10 − 5 F
2C 4
and L = 100mH = 10 − 1H
1
ω= or
√LC
10 − 5
√ √
C
tanωt = −1 im = Vm = 25
√2, ωt = tan √2 L 10 − 1
−1
√LCtan √2
= 25 × 10 − 2 A = 0.25 A
32. [3]
1q 12 q 20 PHYSICS B
U=
2C
=
2C ( 2
q 0e − t / τ =
2C )
e − 2t / τ
36. [2]
(where τ = CR ) V p = 360 V, V s = 36 V, V sI s = 36 W
U = U ie − 2t / τ V sI s 36
Current in primary coil I p = V = 360 = 0.1 A
p
1
U i = U ie − 2t 1 / τ

y
4 37. [1]
1 εs Ns

m
= e − 2t 1 / τ =

e
4 εp Np

d
t 1 = τln2 εs 10

a
t =

c
−τ 220 100
Now q = q 0e
ε s = 22 V

A
q0 t2
−

ts
= q 0e τ 38. [1]
8

e
t 2 = τln8 = 3τln2 Conceptual.

t1
=
t2 3
1

ta r 39. [2]
PI 3000

S
Solu. : I P = =
VP 200
33. [2]
[3]
90
Conceptual. Po = PI ×
100
34. [2] 90
As it can be easily seen by the direction of I that Q = 3000 × = 2700 W
100
is decreasing thus, energy of capacitor is decreasing P 0 2700
and hence, energy of inductance is increasing or VQ = = = 450 V
IS 6

( )
1
2
LI 2 gives that I is increasing. 40. [1]
ε1 N1 100
= = ⇒ ε 2 = 22 V
ε2 N2 10
22
I= = 1 mA, V 0 = 7 V
22 × 10 3
41. [2] 44. [2]
Given, Given, C = 40μF = 40 × 10 − 6 F, and
Capacity of condenser C = 20μF = 20 × 10 − 6 F L = 16mH = 16 × 10 − 3H Angular frequency of
Inductor L = 10 mH = 10 × 10 − 3 H oscillating circuit,
This is the condition of Resonance. 1 1
ω= =
∴ Resonance frequency or frequency of vibration √LC √(
1 1 )(
16 × 10 − 3 40 × 10 − 6 )
f=
2π √Lc 10 4
= = 1.25 × 10 3 s − 1
1 8
f=
2π 10 × 10 − 3 × 20 × 10 − 6 45. [3]
√ Given, before introducing the dielectric, the resonant
1
f= frequency is
2π × √200 × 10 − 9 f0 =
1
2π√LC
f = 355.88 ≈ 356 cycle / sec
After introducing the dielectric, the resonant
42. [3]
frequency is
8 −1
λ = 300 m, c = 3 × 10 m s 1
c
Frequency f = λ = 300 = 10 6 Hz
3 × 10 8 f= C
2π√L(16C) new
(= ϵ rC = 16C )
Resonance frequency
1 1 f=
f0
4

my
e
f= or L =
2π√LC C4π 2f 2

d
46. [3]

a
1 Conceptual.
∴L= ≡ 10 − 8 H
( )
c
2
4π 2 10 6 × 2.4 × 10 − 6 47. [2]

A
Conceptual.

ts
43. [1]
48. [3]

e
We know that resonant frequency of a tuning circuit

r
1 Conceptual.
is given by f 0 = .

ta
2π√LC

From here we can find C, which is given as

C=
1
2
4π 2f 0L
We are given the range of frequency and we have to
S
find the range of variable capacitor. By putting the
extreme values of frequency, we can find the range
of capacitor.
By putting f 0 = 500 kHz, we get
1
C1 = = 253 pF
4π 2 ( 500 × 10 3 ) 2
× 400 × 10 − 6

By putting f 0 = 1.5 MHz, we get

1
C2 = = 28 pF
4π 2 ( 1.5 × 10 6 ) 2
× 400 × 10 − 6

So the range of variable capacitor comes as:

28 pF − 253 pF.
49. [1] 50. [3]
The rms value of voltage across the source, For series R − L − C circuit,
100√2
V rms = = 100 V
Z= √R + (X
2
L )
− XC 2
√ 2
∴ ω = 1000 rad / s

∴ i rms =
V rms
=
V
= √ 2
(
(300) + 1000 × 0.9 −
10 6
1000 × 2 )
2
= 500Ω

|Z|
√R + (X 2
L − XC 2 )
V rms
=

√ (
R2 + ωL − ωC
1
) 2

100
=

√ (
(1000) 2 + 1000 × 2 −
1
1000 × 1 × 10 − 6 ) 2

= 0.0707 A

y
The current will be same everywhere in the circuit,
therefore,
PD across inductor, V L = i rms X L

e m
d
= 0.0707 × 1000 × 2 = 141.4 V

a
PD across capacitor, V C = i rmsX C

= 0.0707 ×
1
1 × 1000 × 10 − 6
= 70.7 V

Ac
re ts
S ta