Question

Total energy stored in a 900 microcoulombs capacitor of 100 volts is transferred into a 100 microcoulomb capacitor. The potential drop across the new capacitor is (in volts) :-

(1) 900
(2) 200
(3) 100
(4) 300​

Answer 1

Dear Student,

◆ Answer -

∆V = 200 V

● Explanation -

# Given -

V1 = 100 V

C1 = 900 uF

V2 = ?

C2 = 100 uF

# Solution -

Given that energy is transferred from 900 uF capacitor to 100 uF capacitor.

E1 = E2

1/2 C1V1^2 = 1/2 C2V2^2

C1V1^2 = C2V2^2

V2^2 = C1V1^2/C2

V2 = V1 √(C1/C2)

V2 = 100 √(900/100)

V2 = 100 × √9

V2 = 300 V

Potential drop across the new capacitor is ∆V = 300-100 = 200 V.

Thanks dear. Hope this helps you...