Equal Heat from Different External Resistors

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Electric Circuits Intermediate Ohm's Law

Source: High school physics (Chinese)

Problem

A power source with internal resistance $r$ is sequentially connected to two different external resistors, $R_1$ and $R_2$. During the same time interval, $R_1$ and $R_2$ are measured to dissipate equal amounts of heat.

  1. Must $R_1$ and $R_2$ be equal? If not, what relationship must hold among $R_1$, $R_2$, and $r$?
  2. Assuming $R_1 > R_2$, which resistor is more reasonable to use?
$R_1$ and $R_2$ need not be equal; they must satisfy $r = \sqrt{R_1 R_2}$. Use $R_1$ for higher efficiency.

Power dissipated in an external resistor $R$ is $P = I^2 R = \dfrac{\varepsilon^2 R}{(R+r)^2}$, and heat over equal time $t$ is $Q = P t$.

Setting $Q_1 = Q_2$ gives $\dfrac{R_1}{(R_1+r)^2} = \dfrac{R_2}{(R_2+r)^2}$.

Cross-multiplying: $R_1(R_2+r)^2 = R_2(R_1+r)^2$. Expanding and simplifying yields $(R_2 - R_1)(R_1 R_2 - r^2) = 0$.

If $R_1 e R_2$, then $R_1 R_2 = r^2$, i.e., $r = \sqrt{R_1 R_2}$ (geometric mean).

The efficiency $\eta = \dfrac{R}{R+r}$ increases monotonically with $R$. Since $R_1 > R_2$, using $R_1$ delivers higher efficiency.