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An ideal choke draws a current of $8 \mathrm{~A}$ when connected to an AC supply of $100 \mathrm{~V}, 50 \mathrm{~Hz}$. A pure resistor draws a current of $10 \mathrm{~A}$ when connected to the same source. The ideal choke and the resistor are connected in series and then connected to the AC source of $150 \mathrm{~V}, 40 \mathrm{~Hz}$. The current in the circuit becomes
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Verified Answer
The correct answer is:
$\frac{15}{\sqrt{2}} \mathrm{~A}$
Resistance, $\mathrm{R}=\frac{100}{10}=10 \Omega$
Inductive reactance, $\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}$
$$
\begin{gathered}
\frac{100}{8}=2 \pi \times 50 \times \mathrm{L} \\
\Rightarrow \quad \mathrm{L}=\frac{1}{8 \pi} \mathrm{H} \\
\mathrm{X}_{\mathrm{L}}^{\prime}=2 \pi \mathrm{f}^{\prime} \mathrm{L}=2 \pi \times 40 \times \frac{1}{8 \pi}=10 \Omega
\end{gathered}
$$
Impedance of the circuit is $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{\prime 2}}$
$$
=\sqrt{(10)^{2}+(10)^{2}}=10 \sqrt{2} \Omega
$$
Current in the circuit is $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{150}{10 \sqrt{2}}=\frac{15}{\sqrt{2}} \mathrm{~A}$
Inductive reactance, $\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}$
$$
\begin{gathered}
\frac{100}{8}=2 \pi \times 50 \times \mathrm{L} \\
\Rightarrow \quad \mathrm{L}=\frac{1}{8 \pi} \mathrm{H} \\
\mathrm{X}_{\mathrm{L}}^{\prime}=2 \pi \mathrm{f}^{\prime} \mathrm{L}=2 \pi \times 40 \times \frac{1}{8 \pi}=10 \Omega
\end{gathered}
$$
Impedance of the circuit is $\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{\prime 2}}$
$$
=\sqrt{(10)^{2}+(10)^{2}}=10 \sqrt{2} \Omega
$$
Current in the circuit is $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{150}{10 \sqrt{2}}=\frac{15}{\sqrt{2}} \mathrm{~A}$
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