KCET · Physics · Alternating Current
In a series resonant R-L-C circuit, the voltage across \(R\) is \(100 \mathrm{~V}\) and the value of \(\mathrm{R}=1000 \Omega\). The capacitance of the capacitor is \(2 \times 10^{-6} \mathrm{~F}\); angular frequency of AC is \(200 \mathrm{rad} \mathrm{s}^{-1}\). Then the potential difference across the inductance coil is
- A \(100 \mathrm{~V}\)
- B \(40 \mathrm{~V}\)
- C \(250 \mathrm{~V}\)
- D \(400 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(250 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The current in the circuit
\(\begin{aligned}\mathrm{i} &=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}} \\&=\frac{100}{1000}=0.1 \mathrm{~A}\end{aligned}\)
At resonance,
\(\begin{aligned}\mathrm{V}_{\mathrm{L}} &=\mathrm{V}_{\mathrm{C}}=\mathrm{iX}_{\mathrm{C}}=\frac{\mathrm{i}}{\omega \mathrm{C}} \\&=\frac{0.1}{200 \times 2 \times 10^{-6}} \\&=250 \mathrm{~V}\end{aligned}\)
\(\begin{aligned}\mathrm{i} &=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{R}} \\&=\frac{100}{1000}=0.1 \mathrm{~A}\end{aligned}\)
At resonance,
\(\begin{aligned}\mathrm{V}_{\mathrm{L}} &=\mathrm{V}_{\mathrm{C}}=\mathrm{iX}_{\mathrm{C}}=\frac{\mathrm{i}}{\omega \mathrm{C}} \\&=\frac{0.1}{200 \times 2 \times 10^{-6}} \\&=250 \mathrm{~V}\end{aligned}\)
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