Sequential Parallel Resistors

By Ohm's Law, this circuit's behavior is governed by:

$V = I(t) \cdot [r + R(t)]$ (i)

where $R(t)$ is the aggregate load resistance, $r$ the internal resistance, and $I(t)$ is the current at time $t$ . By the parallel combination of load resistors described above, we compute $\frac{1}{R(t) } = t \cdot 1 \Rightarrow R(t) = \frac{1}{t}$ ohms. Solving for $I(t)$ in (i) now gives:

$I(t) = \frac{V}{r+R(t)} = \frac{10}{1 + \frac{1}{t}} = \frac{10t}{t+1}$ (ii)

and the power dissipated in the aggregate load at time $t$ computes to $P_{t} = [I(t)]^2 \cdot R(t) = [\frac{10t}{t+1}]^2 \cdot \frac{1}{t} = \frac{100t}{(t+1)^2}$ . Finally, the total power dissipated in the aggregate load for the first 100 seconds of operation calculates to:

$P_{100} = \Sigma_{t=1}^{100} P_{t} = \Sigma_{t=1}^{100} \frac{100t}{(t+1)^2} = \boxed{356.22}$ joules.