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.

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