A) 5.5 V
B) 6.0 V
C) 6.5 V
D) 7.0 V
Correct Answer: C
Solution :
At steady state there will no current through capacitor. The current will only pass through resistance part of the circuit as shown in the figure. Hence, \[I=\frac{9}{1+6+\frac{(1+3)\times 4}{(1+3)+4}}=1A\] Pot. Drop across \[3\mu \,F\]capacitor \[=({{V}_{A}}-{{V}_{B}})+({{V}_{B}}-{{V}_{C}})\] \[=6I+1\times I/2=13I/2=13\times 1/2=6.5\,V\] Hence, the correction option is (c).You need to login to perform this action.
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