The voltage reading on the voltmeter \(V\) in the circuit shown below will be: 
             

If the current in an AC circuit is given as $\mathrm{i}=\left(5\mathrm{A}\right)\sin \left(\mathrm{\omega t}-\frac{\mathrm{\pi }}{3}\right)$ when an emf $\mathrm{e}=\left(200\mathrm{V}\right)\mathrm{sin\omega t}$ is applied across the circuit. The circuit has components-

1.  L and C

2.  L and R

3.  C and R

4.  None of these

A step-down transformer is employed to reduce the main supply of AC from \(220~\text{V}\) to \(11~\text{V}\). If the primary coil draws a current of \(5~\text{A}\) and the secondary supplies a current of \(90~\text{A}\), then the efficiency of the transformer is:
1. \(10\%\)
2. \(95\%\)
3. \(90\%\)
4. \(80\%\)

At resonance, the value of the power factor of the circuit is -

1.  1

2.  Zero

3.  Between zero and 1

4.  More than 1

In a LCR circuit having L = 8.0 henry, C = 0.5 $\mathrm{\mu F}$ and R = 100 $\mathrm{\Omega }$ in series, the resonance frequency (in Hz) is 

1.  $\frac{250}{\mathrm{\pi }}\mathrm{Hz}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

2.  $\frac{600}{\mathrm{\pi }}\mathrm{Hz}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$

3.  $\frac{500}{\mathrm{\pi }}\mathrm{Hz}<mspace\; width="1em"></mspace>$

4.  $500<mspace\; width="1em"></mspace>\mathrm{Hz}$

In a LCR series network, ${\mathrm{V}}_{\mathrm{L}}=40<mspace\; width="1em"></mspace>\mathrm{V},<mspace\; width="1em"></mspace>{\mathrm{V}}_{\mathrm{C}}=20<mspace\; width="1em"></mspace>\mathrm{V}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{V}}_{\mathrm{R}}=15<mspace\; width="1em"></mspace>\mathrm{V}$. The supply voltage is

1.  25V

2.  75V

3.  35V

4.  Zero

The ratio of the number of turns in secondary to primary in an ideal transformer varies from 50 to 550. If the power input is P, then its output power is  -

1.  11P

2.  $\frac{\mathrm{P}}{11}$

3.  $\frac{18P}{111}$

4.  P

Loss in energy in the transformer is due to 

1.  Hysteresis

2.  Eddy current loss

3.  Leakage of flux

4.  All of these

An alternatingcurrentisgivenas \(i = i_1 \cos\omega t+i_2\sin\omega t\). The RMS current is given by;
1. \(\dfrac{i_1+i_2}{\sqrt{2}}\)
2. \(\dfrac{(i_1+i_2)^2}{\sqrt{2}}\)
3. \(\sqrt{\dfrac{i_1^2+i^2_2}{2}}\)
4. \(\dfrac{\sqrt{i_1^2+i^2_2}}{2}\)

In a step-up transformer, the turn ratio is 1:2. A Leclanche cell (e.m.f. 1.5V) is connected across the primary coil. The voltage developed in the secondary coil would be-

1. 3.0 V

2. 0.75 V

3. 1.5 V

4. Zero