Physics - Capacitor Question with Solution | TestHub

In a complicated R-C circuit, we can find the heat dissipated ($Q$) by using the equation $Q=\Delta U+W_b$, where $W_b$ is the work done by all the sources (batteries, etc.) when charges flow through them, and $\Delta U$ is the change in electrostatic potential energy of the system of capacitors. (Capacitor C₂ is initially uncharged.)

In the given circuit, at time $t=0$, let us close the key $K$. Assume $q_{C_2}=0$, $C=6\times 10^{-6}F$, $R=0.5\times 10^6\Omega$, and $\epsilon =4V$.