Question

A 10.0 x10^(-6)F capacitor is charged by a 10.0V battery through a resistance R. The capacitor reaches a potential difference of 4.00V in a time interval of 3.00s after charging begins. Find R.

Answer 1

The voltage across the capacitor at any instant in time during the charging period can be found as follows:

$$V_c=V_s(1-e^{-\dfrac{t}{RC}}),$$ $$4=10\cdot(1-e^{-\dfrac{3}{RC}}),$$ $$e^{-\dfrac{3}{RC}}=0.6,$$ $$ln(e^{-\dfrac{3}{RC}})=ln(0.6),$$ $$-\dfrac{3}{RC}=ln(0.6),$$ $$R=\dfrac{-3}{ln(0.6)\cdot10\cdot10^{-6}\ F}=587.28\ k\Omega.$$