Irodov 3.179: The Loaded Potentiometer Sags – Amit & Deepali Sinha Foundation

Problem Irodov 3.179

A potentiometer is a uniform resistive bar of length $l$ and resistance $R_0$ , with a voltage $V_0$ applied between its two ends. A device of resistance $R$ is connected between the left end of the bar and a sliding contact that touches the bar at a distance $x$ from that end, so that the voltage $V$ fed to the device can be varied. Find $V$ as a function of $x$ , and examine separately the case $R \gg R_0$ .

The potentiometer. V_0 is applied across the whole bar R_0; the device R is joined to the bar's left end and to the slider, a distance x along it. — The potentiometer. $V_0$ is applied across the whole bar $R_0$ ; the device $R$ is joined to the bar’s left end and to the slider, a distance $x$ along it.

Given

$V_0$
bar of length $l$ , resistance $R_0$
device $R$
slider at $x$ from the end joined to the device

Find

V(x)

· the case

R \gg R_0

Solution

Step 1The bar cut in two by the slider

The bar is uniform, so its resistance is shared in proportion to length. The part of length $x$ , between the left end and the slider, and the remaining part of length $l - x$ have

R_x = \frac{R_0x}{l}, \qquad R_0 - R_x = \frac{R_0(l - x)}{l} \tag{1}

The device is connected across the part $R_x$ only, in parallel with it; the other part is in series with that pair.

Step 2The parallel pair and the divider

The device and the part $R_x$ in parallel act as one resistance

R_\parallel = \frac{RR_x}{R + R_x}

The same current passes $R_\parallel$ and the rest of the bar, so $V_0$ is shared between them in proportion to their resistances, and the device takes the share of $R_\parallel$ :

V = V_0\,\frac{R_\parallel}{R_\parallel + (R_0 - R_x)} \tag{2}

Step 3Clearing the fractions

Multiplying numerator and denominator of (2) by $R + R_x$ , then expanding $(R_0 - R_x)(R + R_x)$ , whose term $-RR_x$ cancels the first $RR_x$ :

\begin{aligned} V &= V_0\,\frac{RR_x}{RR_x + (R_0 - R_x)(R + R_x)}\\[2pt] &= V_0\,\frac{RR_x}{RR_0 + (R_0 - R_x)R_x} \end{aligned}

Substituting (1):

\begin{aligned} V &= V_0\,\frac{RR_0x/l}{RR_0 + R_0^2(l - x)x/l^2}\\[2pt] &= V_0\,\frac{Rx/l}{R + R_0(l - x)x/l^2} \end{aligned}

Multiplying top and bottom by $l$ :

V = \frac{V_0Rx}{Rl + R_0(l - x)x/l}

(3)

V against x for three loads, keyed by colour. Every curve runs from 0 at x = 0 to V_0 at x = l; the heavier the load (smaller R), the further the curve sags below the straight line. — $V$ against $x$ for three loads, keyed by colour. Every curve runs from $0$ at $x = 0$ to $V_0$ at $x = l$ ; the heavier the load (smaller $R$ ), the further the curve sags below the straight line.

Step 4A light load,

R \gg R_0

The product $(l - x)x$ is at most $l^2/4$ , at the middle of the bar, so the second term of the denominator of (3) never exceeds $R_0l/4$ . For $R \gg R_0$ it is negligible against $Rl$ :

V \approx \frac{V_0Rx}{Rl}

V \approx V_0\,\frac{x}{l}

(4)

The voltage then grows in proportion to the slider’s travel: the straight line in the graph.

Check

At $x = 0$ the device is shorted by a bar of zero length, and (3) gives $V = 0$ ; at $x = l$ it sits across the whole supply, and the formula gives $V_0Rl/(Rl) = V_0$ , whatever $R$ is. Units: $R_0(l - x)x/l$ is a resistance times a length, like $Rl$ , so the fraction is dimensionless.

Remark

The load pulls the voltage down by the largest fraction in the middle of the travel; the shortfall in volts, $V_0x/l - V$ , peaks further right, at $x = 2l/3$ for a light load. At $x = l/2$ the exact value is $V = V_0/(2 + R_0/(2R))$ , so the relative shortfall from the straight line is about $R_0/(4R)$ when $R \gg R_0$ . That is why a potentiometer used as a divider is chosen with $R_0$ well below the load it feeds, at the price of the current $V_0/R_0$ it wastes.