## Solution to 1992 Problem 32

 We first find the equivalent resistance seen by the battery. Resistors $R_3$ and $R_4$ combine to give a $20 \;\Omega$ resistor, which combines $R_5$ to give a $50 \;\Omega$ resistor, which combines with $R_2$ to give a $25 \;\Omega$ resistor, which combines with $R_1$ to give a $75 \;\Omega$. The current through this equivalent resistor (which is also the current through $R_1$) is \begin{align*}I_1 = \frac{3 \mbox{ V}}{75 \;\Omega} = 0.04 \mbox{ A}\end{align*}We can find the current through $R_2$ and $R_5$ by the current division formula\begin{align*}I_2 &= I_1 \frac{50 \;\Omega}{50 \;\Omega + 50 \;\Omega} = 0.02 \mbox{ A} \\I_5 &= I_1 \frac{50 \;\Omeg...Similarly, the currents through $R_3$ and $R_4$ are\begin{align*}I_3 &= I_5 \frac{R_4}{R_3 + R_4} = 1/150 \mbox{ A} \\I_4 &= I_5 \frac{R_3}{R_3 + R_4} = 2/150 \mbox{ A...Therefore, the power dissipated by each resistor is\begin{align*}P_1 &= I_1^2 R_1 = \left(0.04 \mbox{ A}\right)^2 \cdot 50 = 0.08 \mbox{ W} \\P_2 &= I_2^2 R_2 = \left(0...Therefore, resistor $\boxed{R_1}$ dissipates the most power. Therefore, answer (A) is correct.