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50 Day's Plan For Physics : (Alternating Current)
ANSWER KEY
Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A. 2 3 4 3 1 4 4 2 3 3 3 3 4 3 1 4 1 2 4 4 3 4 1 3 1 3 3 3 3 3
Q. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
A. 3 3 3 2 3 1 1 1 1 2 2 3 3 2 4 2 2 3 2 1
HINT – SHEET
1. Ans ( 2 ) 5. Ans ( 1 )
4 2 4
i = 5 + 5 sin (100 ω t) ∫
1
t dt (t3 /3)1
average value irns = √
= √
4 − 1 3
3 3
< i > = 5 + 5 < sin (100 ω t) > = √
4 − 1
= √ 7
In one time period <sin(100 ω t) > = 0 9
<i> = 5A 6. Ans ( 4 )
2. Ans ( 3 ) I = 100 sin (200 π t)
i2 = i21 cos2 ωt + i22 sin2 ωt + 2i1 i2 cos ωt sin ωt ⇒ 100 = 100 sin (200 π t)
i21 i22 ⇒ sin (200 π t) = 1
⟨ i2 ⟩ = +
2 2 π
1/2 200 π t =
(i21 + i22 ) 2
√⟨ i2 ⟩ = 1
2
t= sec
√
400
3. Ans ( 4 ) 7. Ans ( 4 )
V = 120sin(100 π t) cos(100 π t) In pure inductive circuit current is lagging behind
V = 60 × 2sin(100 π t) cos(100 π t) the voltage by 90°.
= 60sin(200 π t) 8. Ans ( 2 )
V0 = 60 V Erms
lrms = = 200 × 100 × 1 × 10−6 A = 20mA
1/ωc
ω = 200 π
2 π f = 200 π
9. Ans ( 3 )
XL = ω L
f = 100 Hz
= 2 π fL
4. Ans ( 3 ) 1
= 2 π × 50 ×
For DC H1 = I 2 Rt = (2)2Rt = 4Rt
dc
π
2
=100 Ω
2
For AC H2 = IRMS Rt = (
2
) Rt = 2Rt 10. Ans ( 3 )
√ 2 1 1 1
H1 4Rt 2 xc = = xc ∝
= = ωc 2πfc f
H2 2Rt 1
ADPLNEPH21558 HS-1/4
11. Ans ( 3 ) 19. Ans ( 4 )
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Use Xc =
1 tan ϕ = XL
ωC R
12. Ans ( 3 ) tan30º = XL
1 XL
As current flows in a single direction, the device =
√ 3 π √3
allows current only during positive half cycle only ⇒ XL = π
⇒ 2 π fL = π
i0 V 150 L = .01 H
∴ irms = = 0 = = 3.75A.
2 2R 2 × 20
13. Ans ( 4 ) 20. Ans ( 4 )
XL 2πfL
Z= √ R2 + XC2 tan ϕ = =
R R
= √ R2 +
1 21. Ans ( 3 )
ω2 C 2 R
1 cos ϕ =
= √ R2 + Z
4π 2 f 2 C 2 10
cos ϕ =
14. Ans ( 3 ) 20
Apply Vrms2
= VR2 + VL2
cos ϕ = 1
2
obtain Vrms ϕ = 60°
than apply Vrms = 150 22. Ans ( 4 )
where I = 2A at f = 0, Z = ∞ (XC = ∞ )
V 150
Z= = = 75Ω at = f = ∞ , z = ∞ (XL = ∞ )
I 2
15. Ans ( 1 ) so current in both situation will be zero.
V= √ VR2 + VC2
23. Ans ( 1 )
V= √
122 + 52 = 13 V
16. Ans ( 4 )
E = 4cos(1000t) ⇒
Z= √ R2 + XL2 XL = ω L
Z= √
42 + 32 = 1000 × 3 × 10 – 3
Z=5 =3 XC − XL
tan ϕ = =3
V0 4 R 4
i0 (peak) = = 3
Z 5 – 1
ϕ = tan ( )
17. Ans ( 1 ) 4
XC > XL so current lead voltage.
X 3R √
24. Ans ( 3 )
√
tan ϕ = = = 3 ⇒ ϕ = 60o = π/3
R R
18. Ans ( 2 ) Voltage across L and C cancel out.
XC = 1 If ω ↑↑ then XC ↓↓ So, voltage across R is 220 V. Again,
ωC
Z = R 2 + XC 2
√
I0 = 220
A = 2.2A
then Z ↓↓ so i↑↑ 100
HS-2/4
25. Ans ( 1 ) 33. Ans ( 3 )
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Reading of volt meter 1
2
z = √R2 + (ωL − )
Reading of ammeter ωC
V
For f = 0 → Z → ∞
V = VL – VC, i = f = fr → z = R → Minimum
Z
given circuit is at resonance ∵ XL = XC = 25 Ω f→∞→z→∞
So VL = VC & i = V 34. Ans ( 2 )
R R
26. Ans ( 3 ) cos ϕ =
Z
XL ∝ ω , XC ∝ 1 cos30° = R
ω Z
For f > fr, XL > XC ⇒ R = Z cos30°
√ 3
∴ circuit → inductive = 50 × = 25 3 Ω √
2
27. Ans ( 3 ) X = Z 2 − R2 = 25 Ω
√
V = √VR2 + (VL − VC )
2 35. Ans ( 3 )
1√L
2
= √(80) + (60) = 100V
2 Q=
R C
28. Ans ( 3 ) =
1√1
3 9
1
VR2 + (VL – VC)2 = V2 =
9
VR2 + (40)2 = 502 36. Ans ( 1 )
For better tuning, Q-factor must be high.
VR = 30 V
ωo L 1 L 1√L
30 ∴Q= = ( ) =
IR = = 3A R √
LC R R C
10
R and C should be small and L should be high.
29. Ans ( 3 )
R 10 1 37. Ans ( 1 )
cos ϕ = = = ⇒ ϕ = 60° L 1
Z 20 2 Q-factor is given by
30. Ans ( 3 )
√
R C
So, for large quality factor the inductance should be
From phasor diagram it is clear that current is large and resistance and capacitance must be small.
lagging with respect to Erms. This may be happen in
38. Ans ( 1 )
LCR or LR circuit. V
(a) At frequency fr (resonance), Im =
31. Ans ( 3 ) R ↓ ⇒ Im ↑
R
R is independent of ω ∴ graphC (b) Sharpness increases
1
2 ∴ Im ↑ and bandwidth decreases
2
Z= √
R + (ωL − ) ∴ graph A 1√L
ωC (c) Q = R↓ ⇒Q↑
R C
XL = ωL ∴
1
graph B
39. Ans ( 1 )
XC = ∴ graph D 1
ωC f=
32. Ans ( 3 ) 2π√LC
f
For series R-L-C circuit, C′ = 4C ∴ f′ =
2
HS-3/4
40. Ans ( 2 ) 45. Ans ( 4 )
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1√L P = Vi cos ϕ
Q=
R C π
Q∝ 1 Phase difference ϕ = ⇒ P = zero
2
R
R3 > R2 > R1 46. Ans ( 2 )
41. Ans ( 2 ) Pinst = Vinst.Iinst
R R π
cos ϕ = = = V0sin ω t i0sin ( ωt + )
Z √R2 + ω 2 L 2
2
12 V0 i 0
=2× sin ω t cos ω t
2
122 + 4 × π 2 × (60)2 (0.1)2
Pinst = V0 i0 sin(2 ω t)
√
⇒ cos ϕ = 0.30 2
42. Ans ( 3 ) Frequency of power will be double of source.
P = Vrms Irms cos ϕ 47. Ans ( 2 )
= (
100
)(
100
) × 10−3 cos
π At resonance, LCR circuit behaves as purely
2 2 3
√ √
resistive circuit. For purely resistive circuit power
= 2.5 W
43. Ans ( 3 ) factor = 1
Wattless current = Irmssin ϕ 48. Ans ( 3 )
3 = 2sin ϕ
√ Concept
⇒ ϕ = 60° 49. Ans ( 2 )
1 In a choke coil R << XL to avoid power dissipation
P.F. = cos ϕ =
2
44. Ans ( 2 ) 50. Ans ( 1 )
2 2 R
Vrms (30) cos ϕ = ≅0
P= = = 90W Z
R 10
HS-4/4
