Question

65. Two insulated metallic spheres of 3 uF
and 5 uF capacitance are charged to 300
V and 500 V respectively. When they are
connected by a wire, the loss of energy is
(1) 3.75 J
(ii) 0.0375 J
(iii) 2.5 J
(iv) none of the above
​

Answer 1

Hey buddy,

◆ Answer -

(ii) ∆U = 0.0375 J

● Explaintion -

# Given -

C1 = 3 uF = 3×10^-6 F

C2 = 5 uF = 5×10^-6 F

V1 = 300 V

V2 = 500 V

# Solution -

Initially, total energy of two isolated capacitors -

Ui = ½C1.V1² + ½C2.V2²

Ui = 1/2 × 3×10^-6 × 300² + 1/2 × 5×10^-6 × 500²

Ui = 0.76 J

Later, capacitance will be -

C = C1 + C2

C = 3×10^-6 + 5×10^-6

C = 8×10^-6 F

Resultant potential will be -

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

V = (3×10^-6×300 + 5×10^-6×500) / (8×10^-6)

V = 425 V

Finally, total energy of two connected capacitors -

Uf = ½C.V²

Uf = 1/2 × 8×10^-6 × 425²

Uf = 0.7225 J

Therefore, loss of energy will be -

∆U = Ui - Uf

∆U = 0.76 - 0.7225

∆U = 0.0375 J

Thanks for asking..