The instantaneous values of alternating

current and voltages in a circuit are given

as 

$\mathrm{i}=\frac{1}{\sqrt{2}}\sin \left(100\mathrm{\pi t}\right) \mathrm{ampere}$
$\mathrm{e}=\frac{1}{\sqrt{2}}\sin (100\mathrm{\pi t}+\mathrm{\pi }/3) \mathrm{volt}$

The average power in Watts consumed in the

circuit is

(a) $\frac{1}{4}$

(b) $\frac{\sqrt{3}}{4}$

(c) $\frac{1}{2}$ 

(d) $\frac{1}{8}$

In an AC circuit an alternating voltage $\mathrm{e}=200 \sqrt{2} \sin 100\mathrm{t}$ volt is connected to a capacitor $1 \mathrm{\mu F}.$ The $\mathrm{rms}$ value of the current in the circuit is

(a) 100 mA

(b) 200 mA

(c) 20 mA

(d) 10 mA

The current i  in a coil varies with time as shown in the figure. The variation of induced emf with time would be

                  (a)                                                                                         (b)

                   (c)                                                                                             (d)

An \(AC\) voltage is applied to a resistance \(R\) and an inductor \(L\) in series. If \(R\) and the inductive reactance are both equal to \(3~ \Omega, \) then the phase difference between the applied voltage and the current in the circuit will be:

The rms value of potential difference V shown in the figure is

(1) ${\mathrm{V}}_{\mathrm{o}}$                                                     

(2) $\frac{{\mathrm{V}}_{\mathrm{o}}}{\sqrt{2}}$

(3)$\frac{{\mathrm{V}}_{\mathrm{o}}}{2}$                                                     

(4) $\frac{{\mathrm{V}}_{\mathrm{o}}}{\sqrt{3}}$   

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. $2.0 \mathrm{A}$

In the given circuit the reading of voltmeter $V_1 and V_2$ are 300V each. The reading to the voltmeter $V_3$ and ammeter A are respectively :

(a) 150V, 2.2A                                             

(b) 220V, 2.2A

(c) 220V, 2.0A                                             

(d) 100V, 2.0A

A 220V input is supplied to a transformer.The output circuit draws a current of 2.0A at 440V. If the efficiency of the transformer is 80%, the current drawn by the primary windings of the transformer is 

1. 3.6A                                        2. 2.8A

3. 2.5A                                        4. 5.0A

A condenser of capacity C is charged to a potential difference of ${\mathrm{V}}_1.$ The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to ${\mathrm{V}}_2$ is

1. ${\left(\frac{\mathrm{C}({\mathrm{V}}_1-{\mathrm{V}}_2{)}^2}{\mathrm{L}}\right)}^{\frac{1}{2}}$                                     

2. $\frac{\mathrm{C}({\mathrm{V}}_1^2-{\mathrm{V}}_2^2)}{\mathrm{L}}$

3. $\frac{\mathrm{C}({\mathrm{V}}_1^2+{\mathrm{V}}_2^2)}{\mathrm{L}}$                                             

4. ${\left(\frac{\mathrm{C}({\mathrm{V}}_1^2-{\mathrm{V}}_2^2)}{\mathrm{L}}\right)}^{\frac{1}{2}}$

Power dissipated in an L-C-R series circuit connected to an AC source of emf $\epsilon$ is 

(a) $\frac{{\epsilon }^{2}R}{\left[{\displaystyle R^2+{\left(L\omega -\frac{1}{C\omega }\right)}^2}\right]}$

(b) $\frac{{\epsilon }^2{\sqrt{R^2+\left(L\omega -{\displaystyle \frac{1}{C\omega }}\right)}}^2}{R}$

(c) $\frac{{\epsilon }^2\left[R^2+{\left(L\omega -{\displaystyle \frac{1}{C\omega }}\right)}^2\right]}{R}$

(d) $\frac{{\epsilon }^{2}R}{\sqrt{R^2+{\left(L\omega -{\displaystyle \frac{1}{C\omega }}\right)}^2}}$