An inductor coil takes current $8 \mathrm{~A}$ when connected to a $100 \mathrm{~V}$ and $50 \mathrm{~Hz}…
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
$\frac{10}{\sqrt{3}} \mathrm{~A}$
$\frac{5}{\sqrt{2}} \mathrm{~A}$
$10 \sqrt{2} \mathrm{~A}$
$5 \sqrt{2} \mathrm{~A}$
Solution
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
$\begin{aligned}
& Z=\sqrt{R^2+X_L^{\prime 2}}=\sqrt{(10)^2+(10)^2}=10 \sqrt{2} A \\
& I=\frac{V}{Z}=\frac{100}{10 \sqrt{2}}=5 \sqrt{2} \mathrm{~A}
\end{aligned}$
.