The capacitive network shown in the figure is assembled with

The capacitive network shown in the figure is assembled with initially uncharged capacitors. A potential difference, V_ab = +100V, is applied across the network. The switch S in the network is kept open. Assume that all the capacitances shown are accurate to two significant figures. What is potential difference V_cd across the open switch S?

Solution

9 uF and 15 uF are in parallel ,so

9+15 =24 uF

4 uF,8 uF and 12 uF are in parallel ,so

4+8+12=24 uF

Voltage at Point c

V_c = V_ab *(24/24+16) =100*(24/40 ) =60 Volts

Voltage at Point d is

V_d = V_ab*(6/6+24) =100*(6/30) =20 Volts

So Potential across the switch when it is open is

V_cd = V_c-V_d =60-20

V_cd =40 Volts