Question

A coil has an inductance of $0.7H$ and is joined in series with a resistance of $220\Omega$. When the $A.C$ e..m.f of $220V$ is applied to it, find the wattless component of current in the circuit.

• A. $0.2A$
• B. $0.4A$
• C. $0.5A$
• D. $0.7A$

Answer 1

The correct option is C $0.5A$

Wattless component of current is

$i=i_{v}Sin\theta =\frac{E_v}{Z}Sin\theta$

Where, $Z=$ impedence of $L-R$ circuit $=\sqrt{R^2+L^2{\omega }^2}$

So $i=\frac{220}{\sqrt{R^2+L^2{\omega }^2}}Sin\theta$

From impedence triangle,

$Sin\theta =\frac{L\omega }{\sqrt{R^2+L^2{\omega }^2}}$

That is,

$i=\frac{220}{\sqrt{R^2+L^2{\omega }^2}}\frac{L\omega }{\sqrt{R^2+L^2{\omega }^2}}$

$=\frac{220}{R^2+L^2{\omega }^2}L\omega$

$=\frac{220\times 0.7\times 2\pi \times 50}{(220{)}^2+(0.7\times 2\pi \times 50{)}^2}$

$=\frac{220\times 220}{\left(220\right)+(220{)}^2}=0.5A$