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

Let the rms voltage of the AC source be $V$. The initial power drawn by the resistor $R$ is:

$P = \dfrac{V^2}{R} \Rightarrow V^2 = P R$

When an inductance is connected in series, the new impedance of the circuit becomes $Z$. The new rms current in the circuit is:

$I' = \dfrac{V}{Z}$

The new power drawn by the circuit is the power dissipated across the resistor (since an ideal inductor does not dissipate real power):

$P' = (I')^2 R = \left(\dfrac{V}{Z}\right)^2 R$

Substituting $V^2 = P R$ into the equation:

$P' = \dfrac{P R}{Z^2} R = P \left(\dfrac{R}{Z}\right)^2$

Answer: $P\left(\dfrac{R}{Z}\right)^2$