Question

4. Find the voltage across the capacitor, Vc, 0.15 second after theswitch is closed.

Answer 1

This is and RC circuit. We know that in this kind of circuits the potencial acroos the capacitor during the charging process is:

`V_{}=E(1-e^{-(t)/(RC)})`

where E is the voltage on the battery, t is the time, R is the total resistance and C is the capacitance.

In this case we have two resistance in series, then the total resistance is the sum of each of them. Therefor the equivalent resistance in this case is 1100 Ohms.

Plugging the values given we have:

`\begin{gathered} V=35(1-e^{-(0.15)/((75*10^(-6))(1100))}) \\ V=29.32 \end{gathered}`

Therefore after 0.15 seconds the voltage in the capacitor is 29.32 V.