Question

A 2uF capacitor is charged to 8V. It is then connected to a 12V supply and given additional charge

through a 1OMohms resistor. How long will it take to reach a voltage of 10V?

Answer 1

A capacitor connected to a tension through a resistor satisfies the following equation :

$q/C+\dot q R=V$ which is basically juste the expression of tension, using that the tension on the capacitor is $q/C$ and the tension on the resistance is $IR=\dot q R$. Solving this equation yields :

$q/(RC)+\dot q = V/R$

$q=q_0 e^{-t/(RC)}+CV$

By dividing both sides by $C$ and using $U$ for tension on the capacitor, we find

$U=U_0 e^{-t/(RC)}+V$

We know that $U_{t=0}=8V$. Therefore,

$8=U_0 +12$, so $U_0 = -4V$

Now we will find the time when $U$ reaches 10$V$ :

$10=-4\cdot e^{-t/20}+12$

$e^{-t/20}=1/2$

$t=20\ln 2 \approx 13.86 \: s$