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In the circuit as shown in figure the switch is closed at \( \mathrm{t}=0 \).
\( \mathrm{P} \)
W
A long time after closing the switch
- (A) voltage drop across the capacitor is E.
(B) current through the battery is \( \frac{E}{R_{1}+R_{2}} \)
(C) energy stores in the capacitor is \( \frac{1}{2} C\left(\frac{R_{2} E}{R_{1}+R_{2}}\right)^{2} \)
(D) current through the capacitor becomes zero.
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