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Alternating Current

NEET Physics · 95 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
A capacitor of capacitance 'C', is connected across an ac source of voltage V, given by V = V 0 sinω\omegat The displacement current between the plates of the capacitor, would then be given by :
A Id=V0ωCsin⁡ωt{I_d} = {V_0}\omega C\sin \omega t
B Id=V0ωCcos⁡ωt{I_d} = {V_0}\omega C\cos \omega t
C Id=V0ωCcos⁡ωt{I_d} = {{{V_0}} \over {\omega C}}\cos \omega t
D Id=V0ωCsin⁡ωt{I_d} = {{{V_0}} \over {\omega C}}\sin \omega t
Correct Answer
Option B
Solution

Given, V = V 0 sinω\omegat We know,

q=CVq = CV

Now displacement current I d is given by, I d =

dqdt=CdVdt{{dq} \over {dt}} = {{CdV} \over {dt}}
Id=C(V0ωcos⁡ωt){I_d} = C({V_0}\omega \cos \omega t)
=V0ωCcos⁡ωt= {V_0}\omega C\cos \omega t
Q22
A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ?
A 4A
B 0.2A
C 0.4A
D 2A
Correct Answer
Option B
Solution

Power loss = 0 η\eta = 100% P in = P o/p V P I p = V S I s 220 ×\times I p = 44 I p =

44220=15{{44} \over {220}} = {1 \over 5}

A = .2A

Q23
A series LCR circuit containing 5.0 H inductor, 80μ\muF capacitor and 40Ω\Omega resistor is connected to 230V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be :
A 42 rad/s and 58 rad/s
B 25 rad/s and 75 rad/s
C 50 rad/s and 25 rad/s
D 46 rad/s and 54 rad/s
Correct Answer
Option D
Solution
Q=ωΔω=ωLR⇒Δω=R/L=504=8Q = {\omega \over {\Delta \omega }} = {{\omega L} \over R} \Rightarrow \Delta \omega = R/L = {{50} \over 4} = 8

rad/sec

ω0=1LC=15×80×10−6=50{\omega _0} = {1 \over {\sqrt {LC} }} = {1 \over {\sqrt {5 \times 80 \times {{10}^{ - 6}}} }} = 50

rad/sec

ωmin⁡=ω0−Δω2=46{\omega _{\min }} = {\omega _0} - {{\Delta \omega } \over 2} = 46

rad/sec

ωmax⁡=ω0−Δω2=54{\omega _{\max }} = {\omega _0} - {{\Delta \omega } \over 2} = 54

rad/sec

Q24
A 40 μ\mu F capacitor is connected to a 200 V. 50 Hz ac supply. The rms value of the current on the circuit is, nearly :
A 2.05 A
B 2.5 A
C 25.1 A
D 1.7 A
Correct Answer
Option B
Solution

i rms = Cω\omega

ε\varepsilon

rms C = 40 ×\times 10 -6 F ω\omega = 2

πf\pi f

= 100π\pi

ε\varepsilon

rms = 200 V i rms = 200 ×\times 40 ×\times 10 -6 ×\times 2π\pi ×\times 50 = 2.5 A

Q25
A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is π3{\pi \over 3}. If instead C is removed from the circuit, the phase difference is again π3{\pi \over 3} between current and voltage. The power factor of the circuit is :
A 0.5
B 1.0
C − - 1.0
D zero
Correct Answer
Option B
Solution

The power factor for the LCR circuit will be : When L is removed,

tan⁡ϕ=∣XC∣R\tan \phi = {{\left| {{X_C}} \right|} \over R}
⇒tan⁡π3=XCR\Rightarrow \tan {\pi \over 3} = {{{X_C}} \over R}

...(i) When C is removed,

tan⁡ϕ=∣XL∣R\tan \phi = {{\left| {{X_L}} \right|} \over R}
⇒tan⁡π3=XLR\Rightarrow \tan {\pi \over 3} = {{{X_L}} \over R}

...(ii) Since, X L = X C , the circuit is in resonance, Z = R Power factor = cos ϕ\phi =

RZ{R \over Z}

= 1

Q26
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance
A 0.138 H
B 138.88 H
C 1.389 H
D 13.89 H
Correct Answer
Option D
Solution

From question energy stored in inductor, U = 25 × 10 –3 J Current, I = 60 mA Also, energy stored in inductor, U =

12LI2{1 \over 2}L{I^2}

25 × 10 –3 =

12×L×(60×10−3)2{1 \over 2} \times L \times {\left( {60 \times {{10}^{ - 3}}} \right)^2}

⇒\Rightarrow L =

25×2×106×10−33600{{25 \times 2 \times {{10}^6} \times {{10}^{ - 3}}} \over {3600}}

= 13.89 H

Q27
An inductor 20 mH, a capacitor 100 μ\mu F and a resistor 50 Ω\Omega are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is
A 0.79 W
B 0.43 W
C 2.74 W
D 1.13 W
Correct Answer
Option A
Solution

Average power in impedance Z =

R2+(XC−XL)2\sqrt {{R^2} + {{\left( {{X_C} - {X_L}} \right)}^2}}

where X C = capacitive reactance and X L = inductive reactance. Also X C =

1ωC{1 \over {\omega C}}

and X L =

ωL\omega L

∴\therefore Z =

(50)2+(1314×100×10−6−314×20×10−3)2\sqrt {{{\left( {50} \right)}^2} + {{\left( {{1 \over {314 \times 100 \times {{10}^{ - 6}}}} - 314 \times 20 \times {{10}^{ - 3}}} \right)}^2}}

⇒\Rightarrow Z = 56.15

Ω\Omega

I rms =

VrmsZ{{{V_{rms}}} \over Z}

=

Vm2×Z=102×56{{{V_m}} \over {\sqrt 2 \times Z}} = {{10} \over {\sqrt 2 \times 56}}

Hence power loss in the circuit =

(102×56)2×50{\left( {{{10} \over {\sqrt 2 \times 56}}} \right)^2} \times 50

= 0.79 W

Q28
Figure shows a circuit that contains three identical resistors with resistance R = 9.0 Ω\Omega each, two identical inductors with inductance LL = 2.0 mH each, and an ideal battery with emf ε=18V\varepsilon = 18V. The current ii through the battery just after the switch closed is
A 0.2 A
B 4 A
C 0 ampere
D 2 mA
Correct Answer
Option B
Solution

At t = 0, there will be no current flowing through an inductor as it resists the change in current, whereas current will slowly build in it.

Also, as switch is closed, inductor will give an infinite resistance where capacitor shows zero resistance, so, current through battery, will be I =

E(R/2)=2ER=2×189{E \over {\left( {R/2} \right)}} = {{2E} \over R} = {{2 \times 18} \over 9}

= 4 A

Q29
The potential differences across the resistance, capacitance and inductance are 80 V, 40 V and 100 V respectively in an L-C-R circuit. The power factor of this circuit is
A 0.4
B 0.5
C 0.8
D 1.0
Correct Answer
Option C
Solution

Power factor, cos ϕ\phi =

RZ{R \over Z}

=

VRV{{{V_R}} \over V}

=

VRVR2+(VL−VC)2{{{V_R}} \over {\sqrt {V_R^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} }}

=

80802+(100−40)2{{80} \over {\sqrt {{{80}^2} + {{\left( {100 - 40} \right)}^2}} }}

= 0.8

Q30
A 100 Ω\Omega resistance and a capacitor of 100 Ω\Omega reactance are connected in series across a 220 V source. When the capacitor is 50% charged, the peak value of the displacement current is
A 2.2 A
B 11 A
C 4.4 A
D 112A\sqrt 2 A
Correct Answer
Option A
Solution

Net impedance, Z =

R2+XL2\sqrt {{R^2} + X_L^2}

=

1002 Ω100\sqrt 2 \,\Omega

Peak value of displacement current = Maximum conduction current in the circuit =

ε0Z{{{\varepsilon _0}} \over Z}

=

22021002{{220\sqrt 2 } \over {100\sqrt 2 }}

= 2.2 A

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