Question:

A series LCR circuit is connected to an alternating source of emf \( E \). The current amplitude at resonance frequency is \( I_0 \). If the value of resistance \( R \) becomes twice of its initial value, then amplitude of current at resonance will be:

Show Hint

In a series LCR circuit, the current at resonance is inversely proportional to the resistance. Doubling the resistance reduces the current by half.
Updated On: July 22, 2025
  • \( \frac{I_0}{2} \)
  • \( 2I_0 \)
  • \( I_0 \)
  • \( \frac{I_0}{\sqrt{2}} \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

In an LCR circuit, the current amplitude \( I \) at resonance is given by: \[ I = \frac{E}{R} \] where \( R \) is the resistance, \( E \) is the emf, and \( R \) is the total resistance in the circuit. If the resistance \( R \) is doubled, the current will decrease as the current is inversely proportional to the resistance. Therefore, if the resistance becomes twice the initial value, the current will be half of its initial value: \[ I_{\text{new}} = \frac{I_0}{2}. \] Thus, the correct answer is \( \boxed{\frac{I_0}{2}} \).
Was this answer helpful?
0
0

Top Questions on KCL/ KVL in AC Circuits

View More Questions

Questions Asked in JEE Main exam

View More Questions

JEE Main Notification