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

The voltage across the capacitor after time t is

$V=V_0(1-e^{- \frac{t}{RC}})$

Given $V=\frac{V_0}{3}$

$\frac{V_0}{3}=V_0(1-e^{- \frac{t}{RC}})$

$e^{- \frac{t}{RC}}= \frac{2}{3}$

$e^{- \frac{24.6 }{(12\times 10^3)C}}= \frac{2}{3}$

On solving this equation ,we get

$C=5.0559\times 10^{-3} F= 5.056 mF$