Question

Find the potential difference between the points A and B and between the points B and C of the figure in steady state.

Answer 1

Equivalent capacitance of the circuit can be calculatesd as,
C_eq:-

\[1  \mu F\text{ is in parallel with, }C = \frac{1}{\frac{1}{(3 + 3)} + \frac{1}{(1 + 1)}} = \frac{3}{2}\mu F\] 

So, \[C_{eq}  = \frac{3}{2} + 1   =   \frac{5}{2}  \mu F\]

V = 100 V

Total charge in the circuits is,

\[Q =  C_{eq} V = \frac{5}{2} \times 100 = 250  \mu c\]

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,

\[V = \frac{Q}{C}\]

\[ \Rightarrow  V_{AB}  = \frac{150}{6} = 25  V\]

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