TS EAMCET · Physics · Alternating Current
A series \(L C R\) circuit with \(L=05 \mathrm{H}\) and \(R=10 \Omega\) is connected to an AC supply with \(\mathrm{rms}\) voltage and frequency equal to \(200 \mathrm{~V}\) and \(\frac{150}{\pi} \mathrm{Hz}\), respectively. The magnitude of the capacitance is varied so that current amplitude in the circuit becomes maximum. The rms voltage difference across the inductor is
- A \(3000 \mathrm{~V}\)
- B \(2500 \mathrm{~V}\)
- C \(2000 \mathrm{~V}\)
- D \(2600 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(3000 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Key Idea Current amplitude in series LCR circuit is maximum, if there is resonance i.e, \(\omega L=\frac{1}{\omega C}\). Given, \(L=0.5 \mathrm{H}, R=10 \Omega, V_{r m s}=200 \mathrm{~V}\) and \(f=\frac{150}{\pi} \mathrm{Hz}\) Angular frequency,…
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