Question

A capacitor is discharged through a 30.0 Ω resistor. The discharge current decreases to 26.0% of its initial value in 3.00 ms . What is the value of the capacitor?

Answer 1

Answer:

$74.23\mu F$

Explanation:

We have given the value of resistance R = 30 ohm

Time in which current will decreases to 26% is 3 ms

The discharging equation of the capacitor is given by

$i=i_0e^\frac{-t}{\tau }$ here $\tau$ is time constant which is given as $\tau =RC$

So $0.26i_0=i_0e^\frac{-3}{\tau }$  

$0.26=e^\frac{-3}{\tau }$

$\frac{-3}{\tau }=ln0.26$

$\frac{-3}{\tau }=-1.347$

${\tau }=\frac{3}{1.347}=2.2271ms$

WE know that $\tau =RC=30C$

$C=\frac{2.227}{30}=0.074mF=74.23\mu F$