question_answer

In the circuit shown, switch \[{{S}_{2}}\]is closed first and is kept   closed for a long time. Now \[{{S}_{1}}\] is closed. Just after that instant the current through \[{{S}_{1}}\]is

A) \[\frac{\varepsilon }{{{R}_{1}}}\]towards right

B) \[\frac{\varepsilon }{{{R}_{1}}}\]towards left

C)  zero                 

D)  \[\frac{2\varepsilon }{{{R}_{1}}}\]towards left

Correct Answer: B

Solution :

[b]: Just before \[{{S}_{1}}\]is closed the potential across capacitor 2 is \[2\varepsilon \].                          Just after \[{{S}_{1}}\]is closed the potential difference across capacitors 1 and 2 are 0 and \[2\varepsilon \] respectively. So current \[I=\frac{2\varepsilon -\varepsilon }{{{R}_{1}}}=\frac{\varepsilon }{{{R}_{1}}}\] will flow from Q to P.