| Initially the capacitor was uncharged. Current in the capacitor just after switching on will be? |
 |
A) \[\frac{\varepsilon }{R}\]
B) \[\frac{\varepsilon }{2R}\]
C) \[\frac{\varepsilon }{5R}\]
D) \[\frac{\varepsilon }{4R}\]
Correct Answer: C
Solution :
| Just after switching, the capacitor will act like a conducting wire. So effective circuit will be |
 |
| By applying KVL at point \[P\frac{V-\varepsilon }{R}+\frac{V-\varepsilon }{R}+\frac{V}{2R}=0\] |
| \[V=\frac{4\varepsilon }{5}\] |
| So, current in the branch containing capacitor\[=\frac{\varepsilon }{5R}.\] |
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