Two equal capacitors, each of C = 100 µF, and two equal resistors, each of R = 10 kΩ, make a…

Step 1. The start (a). Both capacitors are uncharged, so A is at $0\ \text{V}$ and the whole battery voltage is across the first resistor: $i = \mathcal{E}/R = 1.20\ \text{mA}$.

Step 2. Decay rates (b). At A: $C\dot V_A = \dfrac{\mathcal{E} - V_A}{R} - \dfrac{V_A - V_B}{R}$. At B: $C\dot V_B = \dfrac{V_A - V_B}{R}$. Without the battery term, $e^{-\lambda t}$ works when $(RC\lambda)^2 - 3RC\lambda + 1 = 0$, so $RC\lambda = \dfrac{3 \mp \sqrt5}{2}$. With $RC = 1.0\ \text{s}$ the time constants are $\dfrac{2RC}{3 - \sqrt5} = 2.618\ \text{s}$ and $\dfrac{2RC}{3 + \sqrt5} = 0.382\ \text{s}$.

Step 3. The middle current (c). $V_A - V_B$ is zero at the start and at the end, so it is $A\left(e^{-\lambda_1t} - e^{-\lambda_2t}\right)$. Its derivative is zero when $\lambda_1e^{-\lambda_1t} = \lambda_2e^{-\lambda_2t}$:

Step 4. Its size (d). At $t = 0$, B is uncharged and $\dot V_A = \mathcal{E}/RC$, so $A(\lambda_2 - \lambda_1) = \mathcal{E}/RC$ and $A = \dfrac{\mathcal{E}}{\sqrt5} = 5.37\ \text{V}$. Then $i_{\max} = \dfrac{A}{R}\left(e^{-\lambda_1t^*} - e^{-\lambda_2t^*}\right) = 0.330\ \text{mA}$.

Step 5. Heat (e). At the end both capacitors are at $\mathcal{E}$, so a charge $2C\mathcal{E}$ has passed through the battery. Its work is $2C\mathcal{E}^2 = 28.8\ \text{mJ}$, and the capacitors store $2 \times \tfrac12C\mathcal{E}^2 = 14.4\ \text{mJ}$. The heat is the difference, $C\mathcal{E}^2 = 14.4\ \text{mJ}$.