A) 100 V
B) 200 V
C) 400 V
D) 600 V
Correct Answer: C
Solution :
Given circuit can be reduced as follows In series combination charge on each capacitor remain same. So using Q = CV Þ \[{{C}_{1}}{{V}_{1}}={{C}_{2}}{{V}_{2}}\] Þ \[3\,(1200-{{V}_{p}})=6({{V}_{P}}-{{V}_{B}})\] Þ \[1200-{{V}_{p}}=2{{V}_{p}}\] (Q \[{{V}_{B}}=0)\] Þ 3Vp = 1200 Þ \[{{V}_{p}}=\text{ 4}00volt\]You need to login to perform this action.
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