A capacitor of reactance 4 √(3) Ω and a resistor of resistance 4 Ω are connected in series with an ac source of peak value 8 √(2) V . The power dissipation in the circuit is W.

Physics · Alternating Current

JEE Main 2024 — 9 April, Shift 2 — Question 48

A capacitor of reactance $4 \sqrt{3} \Omega$ and a resistor of resistance $4 \Omega$ are connected in series with an ac source of peak value $8 \sqrt{2} \mathrm{~V}$ . The power dissipation in the circuit is $\qquad$ W.

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}^{2} \mathrm{~L}}$

$Z=\sqrt{4^{2}+(4 \sqrt{3})^{2}}=8 \Omega$

$\mathrm{V}_{\mathrm{rms}}=\frac{\mathrm{V}}{\sqrt{2}}=\frac{8 \sqrt{2}}{\sqrt{2}}=(8 \mathrm{~V})$

$\mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}}=\frac{8}{8}=1 \mathrm{~A}$

Power dissipated $=\mathrm{I}_{\mathrm{rms}}^{2} \times \mathrm{R}=1 \times 4=(4 \mathrm{~W})$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Alternating Current
Topic: Series L-R, R-C, L-C Circuits with AC Source