An inductor coil takes current \(8 \mathrm{~A}\) when connected to a \(100 \mathrm{~V}\) and \(50 \mathrm{~Hz} \mathrm{AC}\) source. A pure resistor under the same condition takes current of \(10 \mathrm{~A}\). If inductor coil and resistor are connected in series to a \(100 \mathrm{~V}\) and \(40 \mathrm{~Hz}\) AC supply, then the current in the series combination of above resistor and inductor is

Inductive reactance when \(100 \mathrm{~V}, 50 \mathrm{~Hz}\) source is connected is given by
\(
\mathrm{X}_{\mathrm{L}}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{100}{8}=12.5 \Omega
\)
Resistance \(\mathrm{R}=\frac{\mathrm{V}}{1}=\frac{100}{10}=10 \Omega\)
Inductive reactance at new frequency of \(40 \mathrm{~Hz}\)
\(
\begin{aligned}
& \frac{\mathrm{X}_{\mathrm{L}}^{\prime}}{\mathrm{X}_{\mathrm{L}}}=\frac{2 \pi \mathrm{f}^{\prime} \mathrm{L}}{2 \pi \mathrm{fL}}=\frac{40}{50}=\frac{4}{5} \\
& \mathrm{X}_{\mathrm{L}}^{\prime}=\frac{4}{5} \mathrm{X}_{\mathrm{L}}=\frac{4}{5} \times 12.5=10 \Omega
\end{aligned}
\)
When they are connected in series the impedance will be given by
\(
\begin{aligned}
& Z=\sqrt{R^2+X_L^{\prime 2}}=\sqrt{(10)^2+(10)^2}=10 \sqrt{2} A \\
& I=\frac{V}{Z}=\frac{100}{10 \sqrt{2}}=5 \sqrt{2} \mathrm{~A}
\end{aligned}
\)