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

Answer 1

Power of an alternating circuit

1. Power of an alternating circuit carrying current I with resistance R, impedance V is defined by the form, $P={\left(\frac{V_{rms}}{Z}\right)}^{2}R$, where, $V_{rms}$ is the square the root of the time average of the voltage squared.
2. Impedance of an LR circuit is $Z=\sqrt{\left(R^2+{\left(\omega L\right)}^2\right)}$, where, $R$ is the resistance and $\omega L$ is the inductive reactance due to the inductor L.

Diagram

Power of the circuit

In the absence of the inductance, the power of the circuit is given by

$P={\left(\frac{V_{rms}}{R}\right)}^{2}R................\left(1\right)$.

When the inductance is connected with the resistance in series, then power is

$P\text{'}={\left(\frac{V_{rms}}{Z}\right)}^{2}R................\left(2\right)$

Comparing (1) and (2) we get,

$$\begin{gathered} \frac{P\text{'}}{P}=\frac{{\left({\displaystyle \frac{V_{rms}}{Z}}\right)}^{2}R}{{\left(\frac{V_{rms}}{R}\right)}^{2}R}={\left(\frac{R}{Z}\right)}^2 \\ orP\text{'}=P{\left(\frac{R}{Z}\right)}^2 \end{gathered}$$

Therefore, the power will be $P\text{'}=P{\left(\frac{R}{Z}\right)}^2$.