Question

Find the potential drop across the capacitor C in the given circuit

• A. 0.2 V
• B. 0.4 V
• C. 0.8 V
• D. 1.0 V

Answer 1

The correct option is B 0.4 V

When capacitor is fully charged , the capacitor no current draws. So no current will pass in the capacitor branch and we can removed the branch from the circuit which is shown in figure.
The equivalent resistance of the whole circuit is $R_{eq}=0.4+(1+2)||(4+3)=0.4+\frac{3\times 7}{3+7}=0.4+2.1=2.5\Omega$
The current in the circuit is $I=\frac{2}{2.5}=0.8A$
The current through branch PAQ is $I_1=\frac{7}{10}I=\left(7/10\right)0.8=0.56A$ (by current division rule)
The current through branch PAQ is $I_1=\frac{3}{10}I=\left(3/10\right)0.8=0.24A$
Thus, $V_P-V_A=1I_1=0.56V....\left(1\right)$ and
$V_P-V_B=4I_2=4\left(0.24\right)=0.96V....\left(2\right)$
$\left(2\right)-\left(1\right)\Rightarrow V_A-V_B=0.96-0.56=0.4V$ . This is the potential across the capacitance.