JEE MainPhysicsAlternating Current
A series LCR circuit with resistance R is connected to an alternating source of emf and is operating at its resonant frequency. The current amplitude in the circuit is I₀ . An unknown resistance R_x is then connected in series with the circuit. If the frequency of the AC source is kept constant and the current amplitude drops to I₀ 3 , the value of the added resistance R_x is
Options
- A3R
- BR 3
- C2R
- DR 2
Correct answer
C. 2R
Step-by-step solution
Initially, the circuit is at resonance, so the impedance is purely resistive ( Z₁ = R ). The initial current amplitude is I₀ = E₀ R . When R_x is added in series, the total resistance becomes R + R_x . Since the frequency is unchanged, the circuit remains at resonance and the net reactance is still zero. The new current amplitude is I₁ = E₀ R + R_x . Given that I₁ = I₀ 3 , we have: E₀ R + R_x = E₀ 3R R + R_x = 3R R_x = 2R Answer: 2R