Skip to content
Home
Courses
Test Series
Books
Free Resources
Result
Gallery
Home
Courses
Test Series
Books
Free Resources
Result
Gallery
Contact us
Back
Report
Report Issue
×
Wrong Answer
Typo Error
Image Issue
Not Clear
Other
JEE MAIN 2024
05-04-24 S1
Question
An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of 20μF is ..............V.
Write Your Answer
Submit Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
View Solution
Solution
$$ \begin{aligned} & X_L=\omega L=100 \times 1=100 \Omega \\ & X_C=\frac{1}{\omega C}=\frac{1}{100 \times 20 \times 10^{-6}}=500 \Omega \\ & Z=\sqrt{\left(X_L-X_C\right)^2+R^2} \\ & \sqrt{(100-500)^2+300^2} \\ & Z=500 \Omega \\ & i_{\text {rms }}=\frac{V_{\text {rms }}}{Z}=\frac{50}{500}=0.1 \mathrm{~A} \end{aligned} $$ rms voltage across capacitor $$ \begin{aligned} & V_{\mathrm{rms}}=X_C i_{\mathrm{rms}} \\ & =500 \times 0.1=50 \mathrm{~V} \end{aligned} $$
Question Tags
JEE Main
Physics
Medium
Start Preparing for JEE with Competishun
YouTube
Books
App
Tests
Scan the code
Powered by
Joinchat
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation.
Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
Open Chat