Question

An alternating current I = 14sin(100 πt) A passes through a series combination of a resistor of 30 Ω and an inductor of \(\left(\frac{2}{5 \pi}\right)\)H. Taking \(\sqrt{2} = 1.4\), calculate the

(i) rms value of the voltage drops across the resistor and the inductor

(ii) power factor of the circuit.

Answer 1

Given \(I = 14 sin (100 \pi t)A\)

\(R = 30 \ \Omega\)

\(L = \left(\frac{2}{5 \pi}\right) H\)

Thus angular frequency, \(\omega = 100 \pi\)

(i) rms value of voltage drop across resistor = V_R

_{\(\Rightarrow V_R = I_ {rms}R = \left(\frac{14}{\sqrt2}\right) (30) = \frac{14}{1.4} \times 30 = 300 \ V\)}

rms value of voltage drop across inductor = V_L

_{\(\Rightarrow V_L = I _{rms} X_L = \frac{14}{\sqrt2}\times \omega L = 10 \times 100 \pi \times \frac{2}{5 \pi}\)}

= 400 V

(ii) 
\(\text{Power factor } = cos\phi = \frac{R}{\sqrt{R^2+ X_{L}^{2}}} \)

\(= \frac{30}{\sqrt{30^2+ \left(100 \pi \times \frac{2}{5 \pi}\right)^2}} = \frac{30}{\sqrt{30^2+ 40^2}} = \frac{3}5\)

\(\text{Thus power factor } = \frac{3}{5}\)