An electric bulb rated 100 W, 40 V has to be operated across 50 V, 50 Hz a.c supply. The capacitance of the capacitor which has to be connected in series will the bulb is

$1. 12 F$
$2. 24\pi F$
$3. \frac{6}{\pi }F$
$4. \frac{1}{1200\pi }F$

An electricbulb is connected in series with an inductor across a.c supply. When an iron is inserted in the inductor, the brightness of bulb

1. Increases

2. Decreases

3. First increases the deceares

4. Remains the same

The quality factor of the series resonant circuit is

$1. 2\pi x\frac{Maximum energy stored }{Energy dissipation per cycle }$
$2. \frac{Resonant frequency }{Band width }$
$3. \frac{Potential difference across capacitor }{Potential difference across resistor }$
$4. All of these$

When alternating current flows through a conductor, the rate of flux change

1. Is higher in the inner part of the conductor

2. Is lower in the inner part of the conductor

3. Is uniform throughout the conductor

4. Depends on the resistivity of the conductor

In a step-up transformer, the turn ratio is \(1:20\). The resistance of \(100~\Omega\) connected across the secondary is drawing a current of \(2~\text{A}\). What are the primary voltage and current respectively?
1. \(100~\text{V}, 0.5~\text{A}\)
2. \(200~\text{V},10~\text{A}\)
3. \(10~\text{V}, 40~\text{A}\)
4. \(10~\text{V}, 20~\text{A}\)

An ideal inductor of 20 H is joined in series with a resistance of 5 ohms and a battery of 5V. The current flowing in the circuit after 4s is ( in amperes)

$1. \frac{1}{e}$
$2. e$
$3. 1-\frac{1}{e }$
$4. 1-e$

The RMS value of the current in a series LCR circuit is plotted with angular frequency as shown below. The ohmic resistance is R, inductance is L and capacitance is C in the circuit. The value of the difference (${\omega }_2-{\omega }_1)$is

$1. \frac{R}{L}$
$2. \frac{L}{R}$
$3. \frac{1}{\sqrt{LC}}$
$4. \frac{1}{\sqrt{2LC}}$

In the circuit diagram shown below, the reading of the a.c. voltmeter is

1. 200 V

2. 100 V

3. $50\sqrt{2} V$

4. zero

In a series RLC alternating current circuit, the r.m.s. value of current is $l$, the average power loss in the circuit is

$1. l^2 R$
$2. l^2 \omega L$
$3. l^{2 }\omega C$
$4. l^2 \left(\omega L-\frac{1}{\omega C}\right)$

A step-down transformer transforms 200 V to 100 volts. If the current in primary and secondary coils are 6 A and 9.6 A respectively, the efficiency of the transformer is

1. 60%

2. 70%

3. 80%

4. 90%