Answer to Question #164943 in Electric Circuits for John

Question

A capacitor is charged to 12.0V. This charged capacitor is connected to resistor of 12x10^3 ohms.Find the capacitance of the capacitor,when its voltage is dropped to one-third of its original value in 24.6 seconds.

Answer 1

Explanations & Calculations

• Employing the equation used to describe the discharging behavior of a capacitor through a resistor, this question could be answered.
• Potential difference/ voltage between the two terminals of the capacitor changes over time according to,

$\qquad\qquad \begin{aligned} \small V&= \small V_ie^{-\frac{t}{RC}} \end{aligned}$

• One-third of the original voltage is 4V
• Therefore,

$\qquad\qquad \begin{aligned} \small 4&= \small 12e^{-\frac{t}{RC}}\\ \small \frac{1}{3}&= \small \frac{1}{e^{\frac{t}{RC}}}\\ \small e^{\frac{t}{RC}}&= \small 3\\ \small \frac{t}{RC}&=\small ln3\\ \small C &= \small \frac{t}{R\cdot ln3} \\ &= \small \frac{24.6\,s}{12\times10^3\Omega\times1.0986}\\ &= \small 1.866\times10^{-3}\,F\\ &= \small \bold{1.866\,mF} \end{aligned}$