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Difficulty: EasyResistors and Potentiometers

A linear potentiometer has a total resistance of 20 kΩ20\text{ k}\Omega between its two outer terminals (Terminal 1 and Terminal 3). If the center wiper (Terminal 2) is adjusted so that the resistance between Terminal 1 and Terminal 2 is 6.5 kΩ6.5\text{ k}\Omega, what is the resistance in kilohms (kΩ\text{k}\Omega) between Terminal 2 and Terminal 3?

Answer: 13.5 k\Omega

Answer

The resistance between Terminal 2 and Terminal 3 is 13.5 kΩ13.5\text{ k}\Omega.
In a three-terminal potentiometer, the resistive element connects between the two outer terminals (1 and 3), providing a fixed total resistance of 20 kΩ20\text{ k}\Omega. The adjustable wiper (terminal 2) contacts this element and divides it into two series portions such that R1−2+R2−3=R1−3R_{1-2} + R_{2-3} = R_{1-3}. Subtracting the known section (6.5 kΩ6.5\text{ k}\Omega) from the total (20 kΩ20\text{ k}\Omega) gives 13.5 kΩ13.5\text{ k}\Omega for the remaining section between Terminal 2 and Terminal 3.

Step-by-Step Solution

1
Set up the resistance relationship for the three terminals of a potentiometer.
Rtotal=R1−2+R2−3R_{\text{total}} = R_{1-2} + R_{2-3}
The wiper divides the continuous internal resistive strip into two series sections whose resistances add up to the total end-to-end resistance.
2
Subtract the resistance of the first section from the total end-to-end resistance.
R2−3=20 kΩ−6.5 kΩ=13.5 kΩR_{2-3} = 20\text{ k}\Omega - 6.5\text{ k}\Omega = 13.5\text{ k}\Omega
Isolating the unknown segment gives the remaining resistance along the track to the opposite outer terminal.

Key Concept

Potentiometer Terminal Resistance Division
Estimated Time:30s
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Topics

Resistors and Potentiometers

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