MHT CET · Physics · Alternating Current
A series \(L-C-R\) circuit containing a resistance ' \(R\) ' has angular frequency ' \(\omega\) '. At resonance the voltage across resistance and inductor are ' \(\mathrm{V}_{\mathrm{R}}\) ' and ' \(\mathrm{V}_{\mathrm{L}}\) ' respectively, then value of capacitance will be
- A \(\frac{V_R}{V_L \omega R}\)
- B \(\frac{V_L}{V_R R \omega^2}\)
- C \(\frac{V_R}{V_L R \omega^2}\)
- D \(\frac{V_L R}{V_R \omega}\)
Answer & Solution
Correct Answer
(A) \(\frac{V_R}{V_L \omega R}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { At resonance, } C=\frac{1}{\omega^2 L} \\ & L=\frac{V_L}{I \omega} \text { and } I=\frac{V_R}{R} \\ \therefore \quad & L=\frac{V_L R}{V_R \omega} \\ \therefore \quad & C=\frac{V_R \omega}{V_L R \omega^2}=\frac{V_R}{V_L R \omega}\end{aligned}\)
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