Answer to Question #159610 in Electric Circuits for Vanessa

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:

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