Question

A coil has an inductance of 0.7H and is joined in series with a resistance of 220 Ω. When an alternating e.m.f. of 220V at 50 c.p.s. is applied to it, then the watt less component of the current in the circuit is

• A. 5 ampere
• B. 0.5 ampere
• C. 0.7 ampere
• D. 7 ampere

Answer 1

The correct option is B

0.5 ampere

The correct option is (B).

Step 1: Given data:

Inductance = 0.7H

Resistance = 220 Ω

Alternating e.m.f. = 220V

$\mathrm{f}=50\mathrm{cps}$

Step 2: Calculation:

The impedance of the circuit:

Now, $Z=\sqrt{R^2+{\left(2\mathrm{\pi fL}\right)}^2}$

$Z=\sqrt{{220}^2+{(2\mathrm{\pi }\times 50\times 0.7)}^2}$

$Z\approx 220\sqrt{2}\mathrm{\pi }$

The power factor:

$\cos \varphi =\frac{R}{Z}=\frac{\mathrm{\pi }}{4}$

The total current in the circuit:

$$\begin{gathered} I=\frac{V}{Z} \\ I=\frac{220}{220\sqrt{2}} \\ I=\frac{1}{\sqrt{2}}A \end{gathered}$$

Watt less component of current in the circuit:

$$\begin{gathered} I_a=I\sin \left(\varphi \right) \\ {I}_a=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}} \\ {I}_a=0.5A \end{gathered}$$