Published by Patrick Mutisya · 14 days ago
Describe the action of a variable potential divider.
A variable potential divider is a resistive device that provides an adjustable output voltage from a fixed supply voltage. By changing the resistance ratio, the fraction of the supply voltage appearing across a portion of the device can be varied continuously.
The device consists of a uniform resistive track with a movable contact (wiper). The total resistance of the track is \$R_{\text{total}}\$. When the wiper is at a distance \$x\$ from one end, the resistances on either side are:
\$\$
R1 = R{\text{total}} \frac{x}{L}, \qquad
R2 = R{\text{total}} \left(1-\frac{x}{L}\right)
\$\$
where \$L\$ is the total length of the resistive track.
The output voltage \$V_{\text{out}}\$ taken between the wiper and the nearer end is given by the potential divider formula:
\$\$
V{\text{out}} = V{\text{s}} \frac{R2}{R1 + R2} = V{\text{s}} \frac{R2}{R{\text{total}}}
\$\$
or, using the position ratio,
\$\$
V{\text{out}} = V{\text{s}} \left(1-\frac{x}{L}\right)
\$\$
Thus moving the wiper changes \$V_{\text{out}}\$ proportionally.
| Feature | Potentiometer | Rheostat |
|---|---|---|
| Number of terminals | 3 | 2 |
| Primary use | Variable voltage source | Variable resistance / current control |
| Current through wiper | Very small (ideally zero) | Significant, carries load current |
| Typical applications | Volume controls, sensor calibration, laboratory voltage references | Light dimmers, motor speed control, heater power adjustment |
Given a 10 kΩ potentiometer connected across a 12 V supply. The wiper is positioned at 30 % of the total length from the left end. Find the voltage between the wiper and the left end.
Solution:
\$\$
x = 0.30 L \quad\Rightarrow\quad V{\text{out}} = V{\text{s}} \left(1-\frac{x}{L}\right) = 12\ \text{V} \times (1-0.30) = 12\ \text{V} \times 0.70 = 8.4\ \text{V}
\$\$
The output voltage is 8.4 V.
A variable potential divider allows a fixed supply voltage to be split into a controllable output voltage by varying the ratio of two resistances within a single component. Understanding the relationship \$V{\text{out}} = V{\text{s}} \frac{R2}{R1+R_2}\$ and the effect of load resistance is essential for accurate use in circuits.