Question

A resistor of 10 kilo-ohms is connected in series to a capacitor which are connected to a 10 volt battery. After some time, the charge in the capacitor is 10 micro-coulombs. At that time, the battery is disconnected and 5 milli-seconds after the battery has been disconnected, the charge in the capacitor is 3.7 micro-coulombs. What is the voltage across the resistor 1.9 milli-seconds after the battery has been disconnected in volts?

Answer 1

Given

RC circuit  

R = 10000 ohm

V = 10 V

before disconnecting the battery the charge is Q = 10*10^-6C

after 5 milli seconds from the battery disconnected the charge is q(5ms) =3.7*10^-6 C

that is q(t) = Q*e^(-t/RC)

3.7*10^-6 = 10*10^-6*e^((-5*10^-3)/(10000*C))

solvinf for C

C = 0.50289*10^-6 F

voltage across the resistor 1.9 ms after the battery has been disconnected is  

v(t) = Ve^(-t/RC)

V(1.9ms) = 10e^(-(1.9*10^-3)/(10000*0.50289*10^-6))V

V(1.9 ms) = 6.853564 V