Irodov 3.196: Where the Power Curve Is Flat – Amit & Deepali Sinha Foundation

Problem Irodov 3.196

A constant voltage is kept between the terminals $A$ and $B$ of a circuit in which a resistance $R_x$ is connected in parallel with $R_2 = 30$ $\Omega$ , and this pair is connected in series with $R_1 = 20$ $\Omega$ (as drawn). For what value of $R_x$ does the heat power generated in $R_x$ stay practically unchanged when $R_x$ is varied slightly?

The circuit: R_1 in series with R_2 and R_x in parallel, fed between the terminals A and B. — The circuit: $R_1$ in series with $R_2$ and $R_x$ in parallel, fed between the terminals $A$ and $B$ .

Given

$R_1 = 20$ $\Omega$
$R_2 = 30$ $\Omega$
constant voltage between $A$ and $B$

Find

R_x

for which the power in

R_x

is insensitive to small changes of

R_x

Solution

Step 1The voltage across

R_x

Call $V$ the constant voltage between $A$ and $B$ . The parallel pair $R_2$ , $R_x$ has the resistance $R_2R_x/(R_2 + R_x)$ and is in series with $R_1$ , so it takes the share of $V$ in proportion to its resistance:

U_x = V\,\frac{R_2R_x/(R_2 + R_x)}{R_1 + R_2R_x/(R_2 + R_x)}

Multiplying top and bottom by $R_2 + R_x$ :

U_x = \frac{V R_2R_x}{R_1R_2 + (R_1 + R_2)R_x} \tag{1}

Step 2The power in

R_x

The heat power in $R_x$ is $U_x^2/R_x$ . With (1):

P = \frac{V^2R_2^2\,R_x}{\left[R_1R_2 + (R_1 + R_2)R_x\right]^2}

(2)

It vanishes for $R_x \to 0$ (no voltage) and for $R_x \to \infty$ (no current), so it has a maximum in between, and there a small change of $R_x$ changes $P$ only to second order.

Step 3The stationary point

Write $a = R_1R_2$ and $b = R_1 + R_2$ , so that $P \propto R_x/(a + bR_x)^2$ . Setting the derivative of (2) to zero:

\begin{aligned} \frac{dP}{dR_x} &\propto \frac{(a + bR_x)^2 - 2bR_x(a + bR_x)}{(a + bR_x)^4} = \frac{a - bR_x}{(a + bR_x)^3}\\[2pt] a - bR_x &= 0 \end{aligned}

Numerically:

R_x = \frac{20\cdot 30}{20 + 30}\ \Omega

R_x = \frac{R_1R_2}{R_1 + R_2} = 12\ \Omega

(3)

Check

Dimensions: $R_1R_2/(R_1 + R_2)$ is a resistance. The derivative changes sign from positive ( $R_x < a/b$ ) to negative ( $R_x > a/b$ ), so the stationary point is the maximum of $P$ . Numerically, with $V = 1$ V in (2), at $12$ and at $13$ $\Omega$ :

\begin{aligned} P(12\ \Omega) &= \frac{900\cdot 12}{1200^2}\ \text{W} = 7.50\ \text{mW}\\[2pt] P(13\ \Omega) &= \frac{900\cdot 13}{1250^2}\ \text{W} = 7.49\ \text{mW} \end{aligned}

an $8\%$ change of $R_x$ moves the power by less than $0.2\%$ .

Remark

Seen from the terminals of $R_x$ , the rest of the circuit is a source of emf $VR_2/(R_1 + R_2)$ with internal resistance $R_1R_2/(R_1 + R_2)$ , the two resistors in parallel with $V$ shorted. The answer is the matching rule: the power in a load is greatest, and so flat against small changes, when the load equals the internal resistance of what drives it.