JEE Mains · Physics · STD 12 -7. Alternating current
A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is \(I_0\). If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be :
- A \(2 \mathrm{I}_0\)
- B \(I_0\)
- C \(\frac{\mathrm{I}_0}{2}\)
- D \(\frac{\mathrm{I}_0}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{I}_0}{2}\)
Step-by-step Solution
Detailed explanation
Initially, \(\mathrm{I}_0=\frac{\varepsilon_{\mathrm{m}}}{\mathrm{R}}\) Finally, \(\mathrm{I}_0^1=\frac{\varepsilon_{\mathrm{m}}}{2 \mathrm{R}}=\frac{\mathrm{I}_0}{2}\)
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