Question

Figure shows a LCR circuit connected with a dc battery of emf $\epsilon$ and internal resistance $R$. Initially the capacitor was uncharged. After a long time:

• A. current through the inductor is $\frac{\epsilon }{8R}$
• B. charge stored in the capacitor is $C\frac{\epsilon }{4}$
• C. charge stored in the capacitor is $C\frac{\epsilon }{2}$
• D. potential difference across the terminals of battery is $\frac{\epsilon }{4}$

Answer 1

Answer: (A), (C)

The correct options are
A current through the inductor is $\frac{\epsilon }{8R}$
C charge stored in the capacitor is $C\frac{\epsilon }{2}$

Long after closing the switch the p.d. across the inductor will zero.

total resistance of circuit including internal resistnce = $2R$

current through battery = $\frac{\epsilon }{2R}$

current thorough inductor = $\frac{\epsilon }{2R}\times \frac{1}{2}\times \frac{1}{2}$

charge stored in capacitor = $Q=C\times (\frac{\epsilon }{2R}\times R)$

p.d across battery terminals= $\epsilon -(\frac{\epsilon }{2R}\times R)$