To solve this problem, we need to understand how power is affected when an inductance is added in series with a resistance in an AC circuit.$\\$Given:$\\$• Resistance $R$ draws power $P$ when connected to an AC source.$\\$• Impedance of the circuit becomes $Z$ when an inductance is added.$\\$The power drawn by a purely resistive AC circuit is given by:$\\ P = I^2 R \\$Where $I$ is the current through the circuit.$\\$When an inductance is added, the impedance $Z$ of the circuit is given by:$\\ Z = \sqrt{R^2 + (X_L)^2} \\$Where $X_L$ is the inductive reactance.$\\$The current in the circuit with impedance $Z$ is:$\\ I' = \frac{V}{Z} \\$The power factor of the circuit is $\frac{R}{Z},$ which is the ratio of resistance to impedance.$\\$The power drawn by the circuit with impedance $Z$ is:$\\ P' = I'^2 R = \left(\frac{V}{Z}\right)^2 R \\$Since $P = \left(\frac{V}{R}\right)^2 R,$ we can express $V^2$ as $P \cdot R. \\$Substitute $V^2 = P \cdot R$ into the expression for $P': \\ P' = \left(\frac{P \cdot R}{Z^2}\right) R = P \left(\frac{R}{Z}\right)^2 \\$Therefore, the power drawn when an inductance is added is $P \left(\frac{R}{Z}\right)^2. \\$This corresponds to Option 4.