A) 100V
B) 200V
C) 400V
D) 800V
Correct Answer: C
Solution :
\[3\mu F and 6\mu F are connected in series\] \[{{\operatorname{q}}_{3}}={{q}_{6}}\] \[{{\operatorname{C}}_{1}}{{V}_{1}}={{C}_{2}}{{V}_{2}}\] \[3\left( 1200-{{V}_{P}} \right)=6\left( {{V}_{P}}-{{V}_{B}} \right)\] \[{{\operatorname{V}}_{B}} =0 \left[ Grounded \right]\] \[1200-{{V}_{p}}=2{{V}_{p}}\] \[3{{V}_{p}}=1200\] \[{{\operatorname{V}}_{p}}=400V\]
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