Question

A capacitor of capacitance 6 uF is charged to
a potential of 150 V. Its potential falls to 90 V, when another
capacitor is connected to it.Find the capacitance of the second
capacitor and the amount ofenergy lost due to the connection.
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Answer 1

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Answer 2

C1 = 6uF

V1 = 150 V

V2 = 0

C2 =?

V = 90 V

We don't know the value of Q in the second combination so we'll use the formula :-

V = (C1V1 + C2V2) /(C1+C2)

90 = (6×150+0)/(6+C2)

C2 = 4 uF

U = (1/2)CV^2

U1 = 0.5×6×150^2

U1 = 67500 uJ

Cp = C1 + C2

Cp = 6+4

Cp = 10 uF

U2 = 0.5×10×90^2

U2 = 40500 uJ

CHANGE IN U = U2 - U1

= 67500 - 40500

= 27000 uJ

= 0.027 J.