A light bulb and an inductor coil are connected to an AC source through a key as shown in the figure below. The key is closed and after some time an iron rod is inserted into the interior of the inductor. The glow of the light bulb:
1. | decreases |
2. | remains unchanged |
3. | will fluctuate |
4. | increases |
The instantaneous values of alternating current and voltages in a circuit are given as,
\(i=\frac{1}{\sqrt{2}}\sin\left ( 100\pi t \right )~\text{Ampere}\)
\(e=\frac{1}{\sqrt{2}}\sin\left ( 100\pi t+\pi /3 \right )~\text{Volt}\)
What is the average power consumed by the circuit in watts?
1. | \( \frac{\sqrt{3}}{4} \) | 2. | \( \frac{1}{2} \) |
3. | \( \frac{1}{8} \) | 4. | \( \frac{1}{4}\) |
The r.m.s. value of the potential difference V shown in the figure is:
1.
2.
3.
4.
A coil has a resistance of \(30~ \mathrm{ohm}\) and inductive reactance of \(20 ~\mathrm{ohm}\) at a \(50~\mathrm{ Hz}\) frequency. If an \(\mathrm{ac}\) source of \(200 ~\mathrm{volts,}\) \(100~\mathrm{ Hz}\) is connected across the coil, the current in the coil will be:
1. | \(2.0~\mathrm{ A}\) | 2. | \(4.0~\mathrm{ A}\) |
3. | \(8.0~\mathrm{ A}\) | 4. | \(20/\sqrt{13}~\mathrm{A}\) |
The value of the quality factor is:
1.
2.
3.
4. L/R
In a circuit \(L\), \(C\), and \(R\) are connected in series with an alternating voltage source of frequency \(f\). The current leads the voltage by \(45^\circ\). The value of \(C\) is:
1. \(\frac{1}{2 \pi f(2 \pi f L-R)}\)
2. \(\frac{1}{2\pi f(2 \pi f L+R)}\)
3. \(\frac{1}{ \pi f(2 \pi f L-R)}\)
4. \(\frac{1}{\pi f(2 \pi f L+R)}\)
Turn ratio of a step-up transformer is \(1: 25\). If current in load coil is \(2~\text{A}\), then the current in primary coil will be:
1. | \(25~\text{A}\) | 2. | \(50~\text{A}\) |
3. | \(0.25~\text{A}\) | 4. | \(0.5~\text{A}\) |
A coil of \(40\) H inductance is connected in series with a resistance of \(8~\Omega\) and the combination is joined to the terminals of a \(2~\text{V}\) battery. The time constant of the circuit is:
1. \(1/5~\text{s}\)
2. \(40~\text{s}\)
3. \(20~\text{s}\)
4. \(5~\text{s}\)
For a series \(\mathrm{LCR}\) circuit, the power loss at resonance is:
1. \(\frac{V^2}{\left[\omega L-\frac{1}{\omega C}\right]}\)
2. \( \mathrm{I}^2 \mathrm{~L} \omega \)
3. \(I^2 R\)
4. \( \frac{\mathrm{V}^2}{\mathrm{C} \omega} \)
A capacitor of capacity C has reactance X. If capacitance and frequency are doubled, then the reactance will be:
1. 4X
2.
3.
4. 2X