A 0.25μF capacitor is charged to 50 V . It is then connected in series with...

Question

Question

A 0.25μF capacitor is charged to 50 V . It is then connected in series with a 55Ω resistor and a 150 Ω resistor and allowed to discharge completely.

How much energy is dissipated by the 55Ω resistor?

Answer 1

Answer #1

Given,

The capacitance, C = 0.25 F = 0.25 * 10^-6 F

The Voltage, V = 50 V

We know,

Q = C * V

= 0.25 * 10^-6 * 50

= 12.5 * 10^-6

The energy, U = 1/2 * Q² / C

= 0.5 * (12.5 * 10^-6)² / (0.25 * 10^-6)

= 3.125 * 10^-4 J

The energy dissipated by the 55 resistor is,

E = 3.125 * 10^-4 * 55 / (55+ 150)

E = 8.38 * 10^-5 J

Answer 2

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