The capacity of a pure capacitor is 1 farad. In dc circuits, its effective resistance will be 

1. Zero

2. Infinite

3. 1 ohm

4. 1/2 ohm

In the circuit shown in the figure, the ac source gives a voltage $V=20\cos \left(2000\text{\hspace{0.17em}}t\right).$ Neglecting source resistance, the voltmeter and ammeter reading will be: 

1. 0V, 0.47A

2. 1.68V, 0.47A

3. 0V, 1.4 A

4. 5.6V, 1.4 A

A small-signal voltage V(t)=V_o sinωt is applied across an ideal capacitor C 

1. over a full cycle, the capacitor C does not consume any energy from the voltage source

2. current I(t) is in phase with voltage V(t)

3. current I(t) leads voltage V(t) by 180°

4. current I(t) lags voltage V(t) by 90°

An inductor 20 mH, a capacitor 50μF, and a resistor 40Ω are connected in series across a source of emf V=10sin340t. The power loss in the AC circuit is:

1. 0.67 W

2. 0.78W

3. 0.89 W

4. 0.46 W

A coil has resistance $30 \mathrm{\Omega }$ and inductive reactance $20 \mathrm{\Omega }$ at 50 Hz frequency. If an AC source of 200 V, 100 Hz, is connected across the coil, the current in the coil will be:

 1. $4.0 \mathrm{A}$                                          

 2. $8.0 \mathrm{A}$

 3. $\frac{20}{\sqrt{13}}\mathrm{A}$                                         

 4. $2.0 \mathrm{A}$

The impedance of a coil, when DC supply is replaced by AC supply:
1. will remain the same
2. will increase
3. will decrease
4. will be zero

A step-down transformer is connected to 2400 volts line and 80 amperes of current is found to flow in output load. The ratio of the turns in primary and secondary coil is 20 : 1. If transformer efficiency is 100%, then the current flowing in primary coil will be 

1. 1600 A

2. 20 A

3. 4 A

4. 1.5 A

An alternating current is given as \(i = i_1 \cos\omega t+i_2\sin\omega t\). The RMS current is given by;
1. \(\dfrac{i_1+i_2}{\sqrt{2}}\)
2. \(\dfrac{(i_1+i_2)^2}{\sqrt{2}}\)
3. \(\sqrt{\dfrac{i_1^2+i^2_2}{2}}\)
4. \(\dfrac{\sqrt{i_1^2+i^2_2}}{2}\)

Loss in energy in the transformer is due to 

1.  Hysteresis

2.  Eddy current loss

3.  Leakage of flux

4.  All of these

A step-down transformer is employed to reduce the main supply of AC from \(220~\text{V}\) to \(11~\text{V}\). If the primary coil draws a current of \(5~\text{A}\) and the secondary supplies a current of \(90~\text{A}\), then the efficiency of the transformer is:
1. \(10\%\)
2. \(95\%\)
3. \(90\%\)
4. \(80\%\)