In the circuit shown below, the ac source has voltage V = 30cos (ω t) volt with ω = 2000 rad/s. What will be the amplitude of the current? (a) √(5) A (b) 3/√(5) A (c) 3 A (d) 3.3 A

Answer: (c). XL = ω L = 2000 × 5 × 10⁻³ = 10 Ω, XC = 12000 × 50 × 10⁻⁶ = 10 Ω XL = XC: resonance, so Z = total resistance = 6 + 4 = 10 Ω I₀ = V₀/Z = 30/10 = 3 A So the answer is (c) 3 A.

In the circuit shown below, the ac source has voltage V = 30cos (ω t)… | MHT CET 2026 (11 Apr, Shift 1)

Q8MHT CET 2026 · 11 APR · SHIFT 1Medium

Series LCR Circuit

In the circuit shown below, the ac source has voltage $V = 30\cos(\omega t)$ volt with $\omega = 2000$ rad/s. What will be the amplitude of the current?

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Answer: (c)

$X_L = \omega L = 2000 \times 5 \times 10^{-3} = 10\ \Omega$ , $X_C = \frac{1}{2000 \times 50 \times 10^{-6}} = 10\ \Omega$
$X_L = X_C$ : resonance, so $Z =$ total resistance $= 6 + 4 = 10\ \Omega$
$I_0 = \frac{V_0}{Z} = \frac{30}{10} = 3$ A
So the answer is (c) 3 A.

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Series LCR Circuit question from MHT CET 2026 (11 Apr, Shift 1)

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