From the circuit given, we can see that the three
6 μF capacitors connected in upper branch are in series.
Now simplifying the circuit and redrawing it,
Since the combination of three
6 μF capacitors are connected in parallel with battery. So,
Sum of potential difference across
6 μF capacitors = potential difference of battery
Since, all the three capacitors are identical, hence
V1=V2=V3=V′
⇒V1+V2+V3=100V
⇒3V′=100 V
∴V′=1003 V
From the shown figure, direction of flow of positive charge it is clear that
Vc>Vd
⇒Vc−Vd=+1003 V
Alternative solution:
Let the three
6 μF capacitors be replaced an equivalent capacitor
1Ceq=16+16+16
⇒Ceq=63=2μF
Charge flow through the
2 μF,
q1=100×2=200 μC
The potential difference,
Vc−Vd=(q1)6 μF
⇒Vc−Vd=200 μC6 μF=1003V
Hence, option (c) is correct answer.