Question

A 12.0 µF capacitor is charged by a 13.0 V battery through a resistance R. The capacitor reaches a potential difference of 4.00 V at a time 3.00 s after charging begins. What is the value of R.

Answer 1

   Voltage across a capacitor C chanrgingthrough a resistance R is

   V   =   V₀* ( 1 - e^-t/RC)

   4.00   =   13.0 * (1 - e^{3.00 / RC})

   e^{-3.00 /RC}   =   1 - 4.00 / 13.0

   e^{3.00 /RC}   =   13.0 / 9.00

   taking log on both sides

   3.00 /RC   =   log_e(1.444)

   R   =   3.00 /0.3677 * 12.0 * 10^-6

         =   6.799* 10⁵   Ω

          =   679.9   kΩ