A) \[3\mu C\]
B) \[6\mu C\]
C) \[12\mu C\]
D) \[24\mu C\]
Correct Answer: B
Solution :
From figure it is clear that, current will flow only through the branch containing\[{{R}_{2}}\]. \[\therefore \] \[I=\frac{E}{{{R}_{2}}+r}=\frac{5}{4+1}=1A\] So, potential difference across \[{{R}_{2}};\]\[V=I{{R}_{2}}=1\times 4=4V\] Let \[q\] be the charge on each plate of each capacitor, then \[\frac{q}{C}+\frac{q}{C}=4\] \[\Rightarrow \] \[\frac{2q}{C}=4\] \[\Rightarrow \] \[q=2\times C=2\times 3=6\mu C\]You need to login to perform this action.
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