Question

How long would it take to discharge a 0.025 μF capacitor to 1% of its full charge through a resistance of

(a) 10 M$\Omega$, (b) 1 M$\Omega$, (c) 1 k$\Omega$?

Short answer

It would take 1.15s, 0.115s, and 0.000115s to discharge the capacitor to 1% of its full charge with 10M$\Omega$, 1M$\Omega$, and 1k$\Omega$ resistors respectively.

Step-by-step solution

Given information:

Given a capacitor of 0.025 μF, discharging time to 1% of the full charge on the capacitor is to be calculated.

Calaculations:

The charge Q(t) after time t of discharging of the capacitor is given by:

$$\begin{gathered} Q\left(t\right)=Q_0{e}^{-\frac{t}{RC}} \\ where{Q}_{0}istheinitialcharge. \\ Thefinalchargeonthecapacitoris1\%oftheinitial,thatis: \\ {Q}_0\times \frac{1}{100}=Q_0{e}^{-\frac{t}{RC}} \\ \frac{1}{100}=e^{-\frac{t}{RC}} \\ \ln \left(\frac{1}{100}\right)=-\frac{t}{RC} \\ -2\ln \left(10\right)=-\frac{t}{RC} \\ t=2RC\ln \left(10\right) \\ t=4.605RC \\ \left(i\right)R=10M\Omega \\ thenRC=10M\Omega \times 0.025\mu F \\ RC=0.25s \\ t=4.605\times 0.25s \\ t=1.15s \\ \left(ii\right)R=1M\Omega \\ thenRC=1M\Omega \times 0.025\mu F \\ RC=0.025s \\ t=4.605\times 0.025s \\ t=0.115s \\ \left(iii\right)R=1k\Omega \\ thenRC=1k\Omega \times 0.025\mu F \\ RC=0.000025s \\ t=4.605\times 0.000025s \\ t=.000115s \end{gathered}$$