﻿ A coil having resistance of 50 ohm and an inductance of 10 ohm, is connected to a 100 volt battery. The energy stored in the coil is | Alternating Current

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# Alternating Current

#### Multiple Choice Questions

1.

A  capacitor of capacitance  2 μF is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA, the voltage across the capacitor will be

• 0.16 V

• 0.32 V

• 79.5 V

• 159 V

A.

0.16 V

XC$\frac{1}{\mathrm{\omega C}}$

=

=

=

Voltage V = i × XC

=  2 × 10-3 ×

V  = 0.16 V

2.

The peak value of alternating current is $5\sqrt{2}$ ampere. The mean square value of current will be

• 5 A

• 2.5 A

• $5\sqrt{2}$ A

• None of the above

A.

5 A

Peak value of alternating current is known as "equivalent" or DC "equivalent" value of AC voltage or current. For sine wave, the RMS value is approximately 0.707 of it's peak value. The crest factor of an AC waveform is the ratio of its peak (crest) to its RMS value.

# 3.A coil having resistance of 50 ohm and an inductance of 10 ohm, is connected to a 100 volt battery. The energy stored in the coil is10 J 20 J 30 J 40 J

B.

20 J

Given:

R = 50 $\mathrm{\Omega }$, L = 10 $\mathrm{\Omega }$, V = 100 V

Energy stored in coil

U =

=

=

=

= 20 J

4.

Assertion: The quantity L/R  possesses dimension of time.

Reason: To reduce the rate of increase of current through a solenoid, we should increase the time constant  ( L/R ).

• If both the assertion and reason are true and reason is a correct explanation of assertion

• If both assertion and reason are true but assertion is not correct explanation of the assertion

• If the assertion is true but the reason is false.

• If both assertion and reason are false.

C.

If the assertion is true but the reason is false.

Induced  e.m.f (ε ) =

current (i) = $\frac{\mathrm{\epsilon }}{\mathrm{R}}$

i =

=

$\frac{\mathrm{L}}{\mathrm{R}}$ = time

In order to increase  the rate of increase of current through a solenoid, we increase the time constant  (L/R). Since L of the given solenoid is constant. Therefore we use a high resistance in series with it.

5.

If in a circuit current lags behind EMF by  $\frac{\mathrm{\pi }}{2}$. Then it is a/an

• resistor circuit

• capacitor

• inductor circuit

• CR circuit

C.

inductor circuit

Figure shows a purely inductive circuit, in which a sinusoidal emf is applied. The induced emf across the inductor is   so that from Kirchoff's law.

εo sinωt  $-$ L$\frac{\mathrm{di}}{\mathrm{dt}}$  = 0

⇒        $\frac{\mathrm{di}}{\mathrm{dt}}$ =  sinωt

⇒         i =  + C

The constant C can be shown to be zero from the fact that, the emf being sinusoidal, the current should also be so, and the average current over one time period has to be zero.

∴        i =  sin

which shows that the current lags behind the emf by a phase of $\frac{\mathrm{\pi }}{2}$.

6.

An AC supply gives 30Vrms which is fed on a pure resistance of 10 ohm. the power dissipated in this is

• 90watt

• 45watt

• 90.2watt

• 180watt

A.

90watt

$\mathrm{P}=\frac{{\mathrm{V}}^{2}}{\mathrm{R}}=\frac{{\left(30\right)}^{2}}{10}=90\mathrm{watt}$

7.

A transmitting station transmits radiowaves of wavelength 360 m. Calculate the inductance of coil required with a condenser of capacity 1.20 μF in the resonant circuit to receive them

• 3.07×10-8 H

• 2.07×10-8 H

• 4.07×10-8 H

• 6.07×10-8 H

A.

3.07×10-8 H

The frequency of radiowaves (speed c=3×108 m/s) transmitted

8.

The core used in transformer and other electromagnetic instruments is laminated, the reason is

• to increase the electric field

• to increase the magnetic field

• to reduce the loss of energy due to eddy currents

• to decrease the resistance

C.

to reduce the loss of energy due to eddy currents

The core used in transformer and other electro-magnetic instruments is laminated,so that the resistance is increased. Due to this increase, energy loss due to eddy current is reduced.

9.

If an ideal parallel LC circuit, the capacitor is changed by connecting it to a d.c source which is then disconnected. The current in the circuit

• becomes zero instantaneously

• grows monotonically

• decays monotonically

• oscillates instantaneously

D.

oscillates instantaneously

When the capacitor is connected to a d.c  source and then disconneccted, it gets charged then it starts discharging through the inductor. An induced emf id produced in the circuit which opposes the growth of the current in L. When the capacitor is fully discharged, the electric he electric energy stored in the capacitor.  gets converted fully into magnetic field energy $\left(\frac{1}{2}{\mathrm{LI}}^{2}\right)$. As soon as the discharge of the capacitor is complete, current stops and the magnetic flux linked with L  starts collapsing. Therefore, an induced emf is again developed which starts recharging the capacitor in the opposite direction. When the recharging is completed, all magnetic energy stored in  L  appears as the electric energy between the plates of the capacitor. The entire process keeps on repeating and the energy taken from the cell keeps on oscillating between C and L. Hence the current in the circuit starts oscillating as soon as the d.c. source charging the capacitor is disconnected.

10.

A capacitor when charged by a potential difference of 200 volt, stores a charge of 0.1 C. By discharging, energy liberated by the capacitor is

• -30 J

• -15 J

• 10 J

• 20 J

C.

10 J

Given : V = 200 volt, Q = 0.1 coulomb

Energy liberated

U = $\frac{1}{2}\mathrm{QV}$

=

= 10 J