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.You need to login to perform this action.
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