Question

A capacitor is charged until it holds 5.0 J of energy. It is then connected across a 10-kΩ resistor. In 11.6 ms , the resistor dissipates 2.0 J.

What is the capacitance?

Answer 1

here,

energy , E = 5 J

resistance , R = 10 000 ohm

time , t = 0.0116 s

Capacitor voltage V decays exponentially

Vt = Vo e^-t/CR

According to E = 0.5 * V^2 C ..V is proportional to sqrtE

so E decays

sqrtEt = sqrtEo e^-t/CR

→ sqrt(Et / Eo) = e^-t/CR ◄Eq1

Et = 3.0J (remaining after 8.5ms)

Eo = 5.0J (initial) .. ►sqrt(Et / Eo) = sqrt(3.0 /5.0) = 0.775

t = 11.6 * 10^-3 s
C = ? F

R = 1*10^4 ohm

(-t/CR) = - (11.6*10^-3)/(C*10^4) = - (8.5*10^-7*1/C)

Sub into Eq1

0.775 = e^-(8.5*10^-7*1/C)

Ln(0.775) = - (8.5*10^-7*1/C) * Ln(e)

Ln(e) = 1

> -0.2549 = - 11.6 * 10^-7*1/C

C = 11.6 * 10^-7 / 0.2549

C = 4.55 * 10^-6 F
(4.55 uF)

the capacitance is 4.55 uF