A) \[4\,kV\]
B) \[6\,kV\]
C) \[8\,kV\]
D) \[10\,kV\]
Correct Answer: C
Solution :
As \[Q=CV,\text{ }{{\left( {{Q}_{\text{1}}} \right)}_{\text{max}}}=\text{1}{{0}^{\text{6}}}\times \text{6}\times \text{1}{{0}^{\text{3}}}=\text{ 6}mC\] While \[{{\left( {{Q}_{\text{2}}} \right)}_{\text{max}}}=\text{3}\times \text{1}{{0}^{\text{6}}}\times \text{4}\times \text{1}{{0}^{\text{3}}}=\text{12}\mu C\] However in series charge is same so maximum charge on C2 will also be 6 mC (and not 12 mC) and potential difference across it \[{{V}_{\text{2}}}=\text{ 6}\mu C/\text{3}\mu F=\text{2}KV\] and as in series \[V={{V}_{\text{1}}}+{{V}_{\text{2}}}\] so \[{{V}_{\text{max}}}=\text{ 6}KV+\text{ 2}KV=\text{ 8}KV\]
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