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 $30 \mathrm{\Omega }$ 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. $2.0 \mathrm{A}$

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

The natural frequency of the circuit shown in the figure is:
    

1. $\frac{1}{2\mathrm{\pi }\sqrt{\mathrm{LC}}}$

2. $\frac{1}{\mathrm{\pi }\sqrt{\mathrm{LC}}}$

3. $\frac{2}{\mathrm{\pi }\sqrt{\mathrm{LC}}}$

4. None of these

The root-mean-square value of an alternating current of 50 Hz frequency is 10 A. The time taken by  the alternating current in reaching from zero to maximum value and peak value will be:

(1) 2×10^–2 s and 14.14 A            (2)  1 × 10^–2 s and 7.07 A

(3) 5×10^–3 s and 7.07 A              (4) 5 × 10^–3 s and 14.14 A