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JEE MAIN Physics QUESTION #995
Question 1
A series LCR circuit is connected to an alternating emf source. The current amplitude at resonance is \(I_0\). If resistance \(R\) becomes twice its initial value, the amplitude of current at resonance will be:
  • 2\(I_0\)
  • \(I_0\)
  • \(\dfrac{I_0}{2}\)✔️
  • \(\dfrac{I_0}{\sqrt{2}}\)
Correct Answer Explanation
At resonance, \(I_0 = \frac{E_0}{R}\). If \(R \to 2R\), then new current \(= \frac{E_0}{2R} = \frac{I_0}{2}\).