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 $30<mspace\; width="1em"></mspace>\mathrm{\Omega }$ and inductive reactance $20<mspace\; width="1em"></mspace>\mathrm{\Omega }$ 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. $4.0<mspace\; width="1em"></mspace>\mathrm{A}<mspace\; width="1em"></mspace>$                                          

 2. $8.0<mspace\; width="1em"></mspace>\mathrm{A}$

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

 4. $2.0<mspace\; width="1em"></mspace>\mathrm{A}$

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

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

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

In $i_1=3\sin <mspace\; width="1em"></mspace>\omega t<mspace\; width="1em"></mspace>\mathrm{an}d<mspace\; width="1em"></mspace>i_2=4\cos <mspace\; width="1em"></mspace>\omega t<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>i_3=i_1+i_2<mspace\; width="1em"></mspace>then<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>i_3$ is –

1. $5<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>(\omega t<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>53°)$                              2. $5<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>(\omega t<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>37°)$

3. $5<mspace\; width="1em"></mspace>\sin <mspace\; width="1em"></mspace>(\omega t<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>45°)$                            4. $5<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>(\omega t<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>53°)$

The voltage time (V-t) graph for triangular wave having peak value $V_0$ is as shown in figure.

The rms value of V in time interval from t = 0 to T/4 is –

1. $\frac{V_0}{\sqrt{3}}$                                             2. $\frac{V_0}{2}$

3. $\frac{V_0}{\sqrt{2}}$                                             4. None of these

When an AC source of emf e = $E_0$ sin (100 t) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be $\mathrm{\pi }/4$ as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C series, find the relationship between the two elements.

1. $R<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1k<mspace\; width="1em"></mspace>\Omega ,<mspace\; width="1em"></mspace>C<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>10\mu F<mspace\; width="1em"></mspace>$                               2. $R<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1k<mspace\; width="1em"></mspace>\Omega ,<mspace\; width="1em"></mspace>C<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1\mu F$

3. $R<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1k<mspace\; width="1em"></mspace>\Omega ,<mspace\; width="1em"></mspace>L<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>10H<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$                                4. $R<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1k<mspace\; width="1em"></mspace>\Omega ,<mspace\; width="1em"></mspace>C<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>1H$