A generator produces a voltage that is given by V = 240 sin 120 t, where t is in seconds. The frequency and r.m.s. voltage are:

1. 60 Hz and 240 V

2. 19 Hz and 120 V

3. 19 Hz and 170 V

4. 754 Hz and 70 V

An alternating current is given as i = i₁ cos ωt - i₂ sin ωt. The value of rms current is given by:

1.  $\frac{{\mathrm{i}}_1 + {\mathrm{i}}_2}{\sqrt{2}}$

2.  $\frac{\left|{\mathrm{i}}_1 + {\mathrm{i}}_2\right|}{\sqrt{2}}$

3.  $\sqrt{\frac{{\mathrm{i}}_1^2 + {\mathrm{i}}_2^2}{2}}$

4.  $\sqrt{\frac{{\mathrm{i}}_1^2 + {\mathrm{i}}_2^2}{\sqrt{2}}}$

The peak value of an alternating e.m.f. $\mathrm{E}={\mathrm{E}}_0 \sin \mathrm{\omega t}$ is 10 volt and its frequency is 50 Hz. At a time \(t=\frac{1}{600}~s,\) the instantaneous value of the e.m.f. will be:

1. 1 volt

2. $5\sqrt{3}$ volt

3. 5 volt

4. 10 volt

The variation of the instantaneous current (I) and the instantaneous emf (E) in a circuit are shown in the figure. Which of the following statements is correct?

1. The voltage lags behind the current by π/2.

2. The voltage leads the current by π/2.

3. The voltage and the current are in phase.

4. The voltage leads the current by π.

When an alternating voltage is given as \(E = (6 sin\omega t - 2 cos \omega t)\) volt, what is its RMS value?
1. \(4 \sqrt 2 \) V
2. \(2 \sqrt 5\) V
3. \(2 \sqrt 3\) V
4. \(4\) V

The time required for a 50 Hz sinusoidal alternating current to change its value from zero to the r.m.s. value will be:

1. $1.5\times {10}^{-2} \mathrm{s}$

2. $2.5\times {10}^{-3} \mathrm{s}$

3. ${10}^{-1} \mathrm{s}$

4. ${10}^{-6} \mathrm{s}$

The r.m.s. value of the potential difference V shown in the figure is:

1. ${\mathrm{V}}_0/\sqrt{3}$

2. ${\mathrm{V}}_0$

3. ${\mathrm{V}}_0/\sqrt{2}$

4. ${\mathrm{V}}_0/2$

Voltage and current in an ac circuit are given by \(V=5sin\left ( 100\pi t-\frac{\pi }{6}~ \right )~V\) and \(I=4sin\left ( 100\pi t+\frac{\pi }{6}~ \right )~A.\)
Hence:

1. the voltage leads the current by 30$°.$

2. the current leads the voltage by 30$°.$

3. the current leads the voltage by 60$°.$

4. the voltage leads the current by 60$°.$

The output current versus time curve of a rectifier is shown in the figure. The average value of the output current in this case will be:

1. 0
2. $\frac{I_0}{2}$
3. $\frac{2I_0}{\pi }$
4. I₀

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

1. \(\frac{1}{100}~sec\)
2. \(\frac{1}{200}~sec\)
3. \(\frac{1}{300}~sec\)
4. \(\frac{1}{400}~sec\)