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

A \(220~\text{V}, 50~\text{Hz}\) AC source is connected to an inductance of \(0.2~\text{H}\) and a resistance of \(20~\Omega\) in series. What is the current in the circuit:
1. \(10~\text{A}\)
2. \(5~\text{A}\)
3. \(33.3~\text{A}\)
4. \(3.33~\text{A}\)

The potential difference V and the current i flowing through an instrument in an ac circuit of frequency f are given by $V=5\cos \omega t$ volts and I = 2 sin ωt amperes (where ω = 2πf). The power dissipated in the instrument is 

1. Zero

2. 10 W

3. 5 W

4. 2.5 W

The peak value of an alternating e.m.f. E is given by $E=E_0\cos \omega t$ is 10 volts and its frequency is 50 Hz. At time $t=\frac{1}{600}sec$, the instantaneous e.m.f. is

1. 10 V

2. $5\sqrt{3}V$

3. 5 V

4. 1 V

If a current I given by $I_0\sin \left(\omega t-\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)$ flows in an ac circuit across which an ac potential of $E=E_0\sin \omega t$ has been applied, then the power consumption P in the circuit will be 

1. $P=\frac{E_0{I}_0}{\sqrt{2}}$

2. $P=\sqrt{2}{E}_0{I}_0$

3. $P=\frac{E_0{I}_0}{2}$

4. P = 0