Question

A series combination of two capacitances of value
0.1 uF and 1uF is connected with a source of
voltage 500 volts. The potential difference in volts
across the capacitor of value 0.1 uF will be :
(1) 50
(2) 500
(3) 45.5
(4) 454.5​

Answer 1

Answer: 454.5

Explanation: we know in series combination

Q1 =Q2

So, C1V1 = C2V2

V2 = C1V1/C2

We know that,

V = V1 +V2

V = V1 +(C1V1/C2)

V =V1 (1+ C1/C2)

V= V1 (C2+C1/C2)

V1 = VC2/(C2 + C1)

V1 = 500×1/1+0.1

V1 = 454.5