Physics · Alternating Current

JEE Main 2024 — 5 April, Shift 1 — Question 56

An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of $20 \mu \mathrm{~F}$ is $\qquad$ V. $V=50 \sqrt{2} \sin 100 t \text { volt }$

Answer: 50

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=100 \times 1=100 \Omega$

$\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{100 \times 20 \times 10^{-6}}=500 \Omega$

$\mathrm{Z}=\sqrt{\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}+\mathrm{R}^{2}}$

$\sqrt{(100-500)^{2}+300^{2}}$

$\mathrm{Z}=500 \Omega$

$\mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\mathrm{Z}}=\frac{50}{500}=0.1 \mathrm{~A}$

rms voltage across capacitor

$V_{\text {rms }}=X_{C} i_{\text {rms }}$

$=500 \times 0.1=50 \mathrm{~V}$

Answer key and solution verified before publishing.

Practise Alternating Current

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor