Question

A resistance ‘R‘ draws power ‘P‘ when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes ‘Z‘, the power drawn will be :

• A. P
• B. P(R/Z)²
• C.
• D. P(R/Z)

Answer 1

The correct option is B

P(R/Z)²

A resistance R draws power P when connected to an AC source.

The magnitude of voltage of the AC source is

$V^2=RP$

$\therefore V=\sqrt{PR}$

An inductor of inductance L and reactance $\omega L$ is now placed in series with the resistance.

The impedance Z is given by

$Z=\sqrt{R^2+{\omega }^2{L}^2}$

$tan\varnothing =\frac{\omega L}{R}ta{n}^2\varnothing =\frac{{\omega }^2{L}^2}{R^2}1+ta{n}^2\varnothing =\frac{1+{\omega }^2{L}^2}{R^2}=\frac{R^2+{\omega }^2{L}^2}{R^2}=se{c}^2\varphi co{s}^2\varnothing =\frac{R^2}{R^2{\omega }^2{L}^2}cos\varnothing =\frac{R}{(R^2+{\omega }^2{L}^2{)}^{\frac{1}{2}}}=\frac{R}{Z}$
Power drawn is VI' $cos\varnothing =V\left(\frac{V}{Z}\right)\left(\frac{R}{Z}\right)$
$=\frac{V^{2}R}{Z^2}=\frac{V^2}{Z^2}=P{\left(\frac{R}{Z}\right)}^2$