A) \[4\,\mu C\]
B) \[6\,\mu C\]
C) \[10\,\mu C\]
D) \[16\,\mu C\]
Correct Answer: D
Solution :
Let the potential of the junction be V. then, \[\frac{6-V}{2}+\frac{4-V}{4}+\frac{8-V}{4}=0\] \[\Rightarrow \] \[12-2V+4-V+8-V=0\] \[24=4V\] V = 6 volt \[\therefore \] Potential drop across capacitor \[=6-\,(-10)\] \[=16\,V\] \[\Rightarrow \] Charge on capacitor\[=16\,\mu C\]You need to login to perform this action.
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