Assume that the brightness of an ohmic light bulb with resistance $R$ is directly proportional to the power it dissipates. If we connect it to the plates of a charged capacitor with capacitance $C$ at time $t=0$, by what factor is it dimmer at time $t=2 R C$ than its initial brightness?

Question

Quizplus · Accepted Answer

The power dissipated by the bulb is proportional to the square of the voltage across it. The voltage across the capacitor decreases exponentially as $V(t) = V_0 e^{-t/(RC)}$. At $t = 2RC$, the voltage is $V_0 e^{-2}$, so the power is $(V_0 e^{-2})^2 = V_0^2 e^{-4}$. Thus, the brightness is reduced by a factor of $1/e^4$.

Assume That the Brightness of an Ohmic Light Bulb with Resistance