A bulb and a capacitor are in series with an A.C. source. On increasing frequency, how will the glow of the bulb change?

1. The glow decreases

2. The glow increases

3. The glow remains the same

4. The glow first decreases then increases

In the circuit shown, X is joining to Y for a long time, and then X is joined to Z. The total heat produced in $R_2$ is ( symbols have their usual meaning)

 $1. \frac{L{E}^2}{2{R_1}^2}$
$2. \frac{L{E}^2}{2{R_2}^2}$
$3.\frac{L{E}^2}{2R_1{R}_2}$
$4. \frac{L{E}^2{R}_2}{2{R_1}^3}$

A transformer has a turn ratio of 1: 5. A DC source of EMF 10 V is applied across its primary. EMF across the secondary will be

1. 50 V

2. 2 V

3. zero

4. 100 V

A series LCR circuit has resistance 2$\Omega$ and self inductance 20 mH. Bandwidth of the circuit is (in radian/s)

1. 1000

2. 100

3. 10

4. $\frac{1}{1000}$

Voltage across the resistor, inductor and capacitor in series LCR circuit are 20 V, 50 V and 50 V respectively. Peak value of the applied source is

1. 20 V

2. 20$\sqrt{2}$ V

3. 120$\sqrt{2}$ V

4. 120 V

The r.m.s value of current for

$l= 3\sin \omega t + 4 \cos \omega t is$

1. 5 A

2. 5$\sqrt{2}$ A

3. $\frac{5}{\sqrt{2}}A$

4. $\frac{7}{\sqrt{2}}A$

If ${\omega }_1$ and ${\omega }_2$ are half power frequencies of a series LCR circuit, then resonant frequency ${\omega }_r$ can be expressed as

$1. {\omega }_r=\frac{{\omega }_1+{\omega }_2}{2}$
$2. {\omega }_r=\sqrt{{\omega }_1{\omega }_2}$
$3. {\omega }_r=\sqrt{\frac{{{\omega }_1}^2+{{\omega }_2}^2}{2}}$
$4. {\omega }_r=\frac{{\omega }_1{\omega }_2}{{\omega }_1+{\omega }_2}$

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:

The r.m.s. voltage of the waveform shown is:

1. 10 V

2. 6.37 V

3. 7 V

4. 10.5 v

If potential differences measured in ac voltmeters across the inductor, capacitor, and resistor in series LCR ac circuit are 100 V, 100 V, and 50 v respectively, then the peak value of the applied ac voltage to the circuit is

1. 50 V

$2. 50\sqrt{2} V$
$3. \frac{50}{\sqrt{2}} V$
$4. \frac{100}{\pi } V$