Question

27.27 A 10.0-uF capacitor is charged by a 10.0-V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V in a time interval of 3.00 s after charging begins. Find R.

Answer 1

The potential difference on the capacitor

$$V(t)=V_s\left(1-\exp(-t/RC)\right).$$

Hence, the resistance

$$R=-\frac{t}{C\ln(1-V/V_s)}=$$ $$=-\frac{3.00}{10.0\times 10^{-6}\ln(1-4.00/10.0)}=$$ $$=1.96\times 10^6\: \Omega=1.96\:\rm M\Omega$$