Integration question

Question

When connected to earth, any charged capacitor will discharge, and the voltage across the capacitor will decrease. This is modelled by the differential equation dV/dT=-V/RC, where V = voltage, and R and C (resistance in ohms and capacitance in farads respectively) are constants. If V drops from 10 volts to 1 volt in two seconds, and is C = 3*10^-6 farads, determine the value of R.

Answer 1

dV/dt = -V / RC

 

dV/V = -dt / RC

 

V(t) = V(0)e^{-t / RC}

 

C = 3 x 10^-6 F

V(0) = 10

V(2) = 1

 

V(2) / V(0) = e^{-2 / RC}

 

1/10 = e^{-2 / RC} = 0.1

 

-2 / RC = ln(0.1)

 

R = -2 / [C ln(0.1)] = -2 / [(3 x 10^-6 F) ln(0.1)] ≅ 289,530 Ω