A) 4 kV
B) 6 kV
C) 8 kV
D) 10 kV
Correct Answer: C
Solution :
We know \[Q=CV\] Hence \[{{({{Q}_{1}})}_{max}}=6\,mC\,\,while\,\,{{({{Q}_{2}})}_{max}}\,=\,12\,mC\] However in series charge is same so maximum charge on \[{{C}_{2}}\] will also be 6 mC (and not 12 mC) and hence potential difference across \[{{C}_{2}}\] will be \[{{V}_{2}}\,=\frac{6mC}{3\mu F}=\,\,2\,kV\] and as in series \[V={{V}_{1}}+{{V}_{2}}\] So \[{{V}_{\max }}\,\,=\,\,6\,kV\,\,+\,\,2kV\,\,=\,\,8kV\]You need to login to perform this action.
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