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 |
The peak value of an alternating e.m.f. is 10 volts 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. | \(5 \sqrt{3}\) volts |
3. | 5 volts | 4. | 10 volts |
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.
2.
3.
4.
The r.m.s. value of the potential difference V shown in the figure is:
1.
2.
3.
4.
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. | \(I_0 \over 2\) |
3. | \(2I_0 \over \pi\) | 4. | \(I_0\) |
In an ac circuit \(I=100~sin~200~ \pi t.\) 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. | \(\frac{1}{400}~sec\) |