Question

Find the potential difference between the points A and B and between the points B and C of the figure (32−E33) in steady state.

Answer 1

Equivalent capacitance of the circuit can be calculatesd as,
C_eq:
$$\begin{gathered} 1\mathrm{\mu F}\mathrm{is}\mathrm{in}\mathrm{parallel}\mathrm{with},C=\frac{1}{{\displaystyle \frac{1}{(3+3)}}+{\displaystyle \frac{1}{(1+1)}}}=\frac{3}{2}\mathrm{\mu F} \\ \mathrm{So},C_{\mathrm{eq}}=\frac{3}{2}+1=\frac{5}{2}\mathrm{\mu F} \end{gathered}$$
V = 100 V

Total charge in the circuits is,
$Q=C_{\mathrm{eq}}V=\frac{5}{2}\times 100=250\mathrm{\mu }c$
As the volatge across 1μf is 100 V, therfore charge stored on 1 μf capacitors = 100 μC

Charge flowing from A to B = (250 $-$ 100) = 150 μC
C_eq between A and B is 6 μf.
Potential drop across AB,
$$\begin{gathered} V=\frac{Q}{C} \\ \Rightarrow V_{\mathrm{AB}}=\frac{150}{6}=25\mathrm{V} \end{gathered}$$

Potential drop across BC = (100 $-$ 25) = 75 V