Search any question & find its solution
Question:
Answered & Verified by Expert
In series L-C-R circuit $C=2 \mu \mathrm{F}, L=1 \mathrm{mH}$ and $R=10 \Omega$. What is the ratio of energies stored in the inductor and the capacitor, when the maximum current flows in the circuit?
Options:
Solution:
1802 Upvotes
Verified Answer
The correct answer is:
$1: 5$
At maximum current condition, system is at resonance condition.
$\therefore X_C=X_L$
$\Rightarrow \frac{U_C}{U_L}=\frac{\frac{1}{2} C V^2}{\frac{1}{2} L i^2}=\frac{C i^2 R^2}{L i^2}$
$\Rightarrow \frac{U_C}{U_L}=\frac{2 \times 10^{-6}}{10^{-3}} \times 100$
Hence, we get
$\frac{U_C}{U_L}=\frac{2}{10}=\frac{1}{5}$
$\therefore X_C=X_L$
$\Rightarrow \frac{U_C}{U_L}=\frac{\frac{1}{2} C V^2}{\frac{1}{2} L i^2}=\frac{C i^2 R^2}{L i^2}$
$\Rightarrow \frac{U_C}{U_L}=\frac{2 \times 10^{-6}}{10^{-3}} \times 100$
Hence, we get
$\frac{U_C}{U_L}=\frac{2}{10}=\frac{1}{5}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.