Question

A capacitor of capacitance 5μF is charged to 50 V and another capacitor of capacitance 7μF is charged to 42 V. These two are connected together. On doing so, what is the energy lost by the 5μF capacitor?

(a) 2.35 mJ

(b) 2.45 mJ

(c) 2.55 mJ

(d) 2.25 mJ

I had been asked this question during an online interview.

My question is based upon Energy Stored in a Capacitor in division Electrostatic Potential and Capacitance of Physics – Class 12

Answer 1

Right answer is (d) 2.25 mJ

The explanation is: Common potential, V = \( \frac {(C_1V_1 + C_1V_2)}{(C_1 + C_2)}\)

V = \( \frac {[(5 \times 50) + (7 \times 40)] \times 10^{-6}}{(5 + 7) \times 10^{-6}}\)

V = \( \frac {(250 + 280)}{12}\) = 44.4 V

Energy lost by 5 μF capacitor = \(\frac {1}{2}\)C1V1^2 – \(\frac {1}{2}\)C1V^2 = \(\frac {1}{2}\)C (V1^2 – V^2)

\(\frac {1}{2}\) × (5 × 10^-6) × (50^2 – 40^2) = 0.00225

U = 2.25 × 10^-3 J

U = 2.25 mJ

Therefore, the energy lost is 2.25 mJ