Problem — ranking the power of three light bulbs

Problem

Ranking. Three ohmic light bulbs are connected to identical ideal batteries of EMF $\mathcal{E}$. (A) Resistance $10\ \Omega$. (B) Resistance $20\ \Omega$. (C) Resistance $40\ \Omega$. Rank the dissipated power from smallest to largest (remember: $P = \mathcal{E}^2/R$).

Solution

Reasoning. Each bulb is connected to the same EMF $\mathcal{E}$, so the same voltage is across it. The power dissipated at fixed voltage is $P = \frac{\mathcal{E}^2}{R}$ i.e. inversely proportional to the resistance: for the same $\mathcal{E}$, the one with the smaller resistance dissipates more (it lets more current through).

Comparison (with common $\mathcal{E}$): $P_A = \frac{\mathcal{E}^2}{10}, \qquad P_B = \frac{\mathcal{E}^2}{20}, \qquad P_C = \frac{\mathcal{E}^2}{40}$ Since $10 < 20 < 40$, we have $P_A > P_B > P_C$.

Ranking from smallest to largest: $\ev{P_C < P_B < P_A}$

Links

Topics: Electric circuits Concepts: Power Skills: Ranking reasoning