In the diagram shown find the potential difference between the points A and B and between the points B and C in the steady state.

Answer

`V_(AB) = 25 V, V_(BC) = 75 V`

Explanation

The two capacitors of `3 muF` each are in parallel. Their equivalent capacity `= 3 + 3 = 6muF`. Again, two capacitanace of `1 muF` each are also in parallel. Their equivalent capacity `= 1 + 1 = 2 muF`

Therefore, the given network reduces tp the eqivalent circuit shown in Fig. In the steady state, when all the capacitors are charged, there is no current in the circuit. Therefore potential drop across both the resistances is Zero.
`:.` Potentail diff, across `1 muF` capacitor = `p.d` of 100 V
As `6 muF` and `2 muF` capacitors are in series, the p.d of 100 V is divided between them in the inverse ratio their capacitance, i.e., in the ratio `2 : 6 or 1 : 3`
`:. V_(AB) = p.d` across `6 muF = (1)/(4) xx 100 = 25 V`
and `V_(BC) = p.d ` across `2 muF = (3)/(4) xx100 = 75 V`