Question

A 9.0 µF capacitor is charged by a 12.0 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. Find R.

Answer 1

Voltage across a capacitor C IS

   V   =   V₀* ( 1 - e^-t/RC)

   4.00   =   12.0 * (1 - e^{-3.00 / RC})

   e^{-3.00 /RC}   =   1 - 4.00 / 12.0

   e^{3.00 /RC}   = 3/2

   taking log on both sides

   3.00 /RC   = ln (3/2)

   R   =   3.00 /0.3677 * 9 * 10^-6

         = 8.221* 10⁵   Ω

         = 822.1 kΩ

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