Question

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is then discharged by connecting it directly across a 100-k\Omega100−kΩ resistor. At what point in time after the resistor is connected will the capacitor have 13.5~\mu C13.5 μC of charge remaining on it?

Answer 1

Explanation:

The given data is as follows.

       C = $20 \times 10^{-6} F$

        R = $100 \times 10^{3}$ ohm

        $Q_{o} = 100 \times 10^{-6}$ C

          Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

          $Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

       $13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

               0.135 = $e^{\frac{-t}{2}}$

         $e^{\frac{t}{2}} = \frac{1}{0.135}$

                         = 7.407

           $\frac{t}{2} = ln (7.407)$

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.