A capacitor of capacitance 150.0 μ F is connected to an alternating source of emf given by E = 36 sin (120π t) V. The maximum value of current in the circuit is approximately equal to: (a) 1/√(2) A (b) 2√(2) A (c) √(2) A (d) 2 A

Answer: (d). XC = (1)/(ω C) = 1120π × 150 × 10⁻⁶ ≈ 17.7 Ω I₀ = (E₀)/(XC) = (36)/(17.7) ≈ 2.04 A So the answer is (d) 2 A.

A capacitor of capacitance 150.0 μ F is connected to an alternating… | JEE Main 2023 (6 Apr, Shift 2)

Q41JEE Main 2023 · 6 APR · SHIFT 2Easy

Pure R, L and C Circuits

A capacitor of capacitance $150.0\ \mu F$ is connected to an alternating source of emf given by $E = 36 \sin(120\pi t)\text{ V}$ . The maximum value of current in the circuit is approximately equal to:

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Answer: (d)

$X_C = \frac{1}{\omega C} = \frac{1}{120\pi \times 150 \times 10^{-6}} \approx 17.7\ \Omega$
$I_0 = \frac{E_0}{X_C} = \frac{36}{17.7} \approx 2.04$ A
So the answer is (d) $2$ A.

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