KCET · Physics · Alternating Current
In a series \(L C R\) circuit, \(R=300 \Omega, L=0.9 \mathrm{H}\), \(C=2.0 \mu \mathrm{F}\) and \(\omega=1000 \mathrm{rad} / \mathrm{s}\), then impedance of the circuit is
- A \(900 \Omega\)
- B \(500 \Omega\)
- C \(400 \Omega\)
- D \(1300 \Omega\)
Answer & Solution
Correct Answer
(B) \(500 \Omega\)
Step-by-step Solution
Detailed explanation
Given, \(R=300 \Omega, L=0.9 \mathrm{H}, \omega=1000 \mathrm{rad} / \mathrm{s}\) \(C=2.0 \mu \mathrm{F}=2 \times 10^{-6} \mathrm{~F}\)
Impedance of given \(R-L-C\) circuit,
\(
\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\left[\therefore X_L=\omega L \text { and } X_C=\frac{1}{\omega C}\right] \\
& =\sqrt{(300)^2+\left(1000 \times 0.9-\frac{1}{1000 \times 2 \times 10^{-6}}\right)^2} \\
& =\sqrt{90000+(900-500)^2} \\
& =\sqrt{90000+160000}=\sqrt{250000}=500 \Omega
\end{aligned}
\)
Impedance of given \(R-L-C\) circuit,
\(
\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\left[\therefore X_L=\omega L \text { and } X_C=\frac{1}{\omega C}\right] \\
& =\sqrt{(300)^2+\left(1000 \times 0.9-\frac{1}{1000 \times 2 \times 10^{-6}}\right)^2} \\
& =\sqrt{90000+(900-500)^2} \\
& =\sqrt{90000+160000}=\sqrt{250000}=500 \Omega
\end{aligned}
\)
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