A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is I₀. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be (a) I₀/2 (b) (I₀)/(√(2)) (c) 2I₀ (d) I₀

Answer: (a). At resonance Z = R, so I₀ = E₀/R. R → 2R ⇒ I₀' = E₀/2R = I₀/2 So the answer is (a) I₀/2.

A series LCR circuit is connected to an alternating source of emf E… | JEE Main 2025 (22 Jan, Shift 2)

Q28JEE Main 2025 · 22 JAN · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

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

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Answer: (a)

At resonance $Z = R$ , so $I_0 = \dfrac{E_0}{R}$ .
$R \to 2R \Rightarrow I_0' = \dfrac{E_0}{2R} = \dfrac{I_0}{2}$
So the answer is (a) $\frac{I_0}{2}$ .

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