In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An e.m.f. $E=E_0\cos \left(\omega t\right)$ applied to the circuit. The power consumed in the circuit is:

(1) $\frac{E_0^2}{R}$

(2) $\frac{E_0^2}{2R}$

(3) $\frac{E_0^2}{4R}$

(4) $\frac{E_0^2}{8R}$

In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter reading respectively will be:

1. \(0~\text{V}, 3~\text{A}\)
2. \(150~\text{V}, 3~\text{A}\)
3. \(150~\text{V}, 6~\text{A}\)
4. \(0~\text{V}, 8~\text{A}\)

An AC source of variable frequency \(f\) is connected to an \(LCR\) series circuit. Which of the following graphs represents the variation of the current \(I\) in the circuit with frequency \(f\)?

1.		2.
3.		4.

Which of the following plots may represent the reactance of a series LC combination 

(1) a

(2) b

(3) c

(4) d

In an ac circuit, I = 100 sin 200 $\mathrm{\pi t}$. The time required for the current to achieve its peak value will be

1. $\frac{1}{100}\sec$

2. $\frac{1}{200}\sec$

3. $\frac{1}{300}\sec$

4. $\frac{1}{400}\sec$

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 $30<mspace\; width="1em"></mspace>\mathrm{\Omega }$ 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. $2.0<mspace\; width="1em"></mspace>\mathrm{A}$

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