question_answer

Find the potential difference across the capacitor in the circuit shown in the figure if the capacitance of the capacitor is \[3\mu F.\]

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).