Question

Difficulty: HardResistors and Potentiometers

A 100 kΩ100\text{ k}\Omega linear potentiometer is connected across a 12 V12\text{ V} DC power supply, with terminal 1 connected to +12 V+12\text{ V}, terminal 3 connected to ground (0 V0\text{ V}), and terminal 2 serving as the adjustable wiper. If the wiper is set to 40%40\% of its total travel distance away from terminal 3 toward terminal 1, what is the output voltage measured at terminal 2 relative to ground, and what is the resistance between terminal 1 and terminal 2?

  1. A
    7.2 V7.2\text{ V} and 40 kΩ40\text{ k}\Omega
  2. 4.8 V4.8\text{ V} and 60 kΩ60\text{ k}\OmegaAnswer
  3. C
    4.8 V4.8\text{ V} and 40 kΩ40\text{ k}\Omega
  4. D
    4.8 V4.8\text{ V} and 600 Ω600\ \Omega

Answer

4.8 V4.8\text{ V} output voltage and 60 kΩ60\text{ k}\Omega resistance between terminal 1 and terminal 2
A linear potentiometer configured across a DC voltage source acts as a continuous voltage divider. With terminal 3 connected to ground (0 V0\text{ V}) and the wiper at 40%40\% travel from ground, the lower resistance segment R2−3R_{2-3} is 40 kΩ40\text{ k}\Omega, yielding an output voltage of Vout=12 V×(40 kΩ/100 kΩ)=4.8 VV_{\text{out}} = 12\text{ V} \times (40\text{ k}\Omega / 100\text{ k}\Omega) = 4.8\text{ V}. The upper resistance segment between terminal 1 (+12 V+12\text{ V}) and terminal 2 (wiper) is the remaining portion: 100 kΩ−40 kΩ=60 kΩ100\text{ k}\Omega - 40\text{ k}\Omega = 60\text{ k}\Omega.

Step-by-Step Solution

1
Calculate the resistance of the lower track section between terminal 2 (wiper) and terminal 3 (ground).
R2−3=0.40×100 kΩ=40 kΩR_{2-3} = 0.40 \times 100\text{ k}\Omega = 40\text{ k}\Omega
The wiper is positioned at 40%40\% of total travel measured from the ground terminal.
2
Calculate the output voltage VoutV_{\text{out}} at terminal 2 relative to ground using the voltage divider relationship.
Vout=Vin×(R2−3Rtotal)=12 V×(40 kΩ100 kΩ)=4.8 VV_{\text{out}} = V_{\text{in}} \times \left(\frac{R_{2-3}}{R_{\text{total}}}\right) = 12\text{ V} \times \left(\frac{40\text{ k}\Omega}{100\text{ k}\Omega}\right) = 4.8\text{ V}
A potentiometer connected across a voltage source functions as an adjustable voltage divider.
3
Determine the remaining resistance between terminal 1 (+12 V+12\text{ V}) and terminal 2 (wiper).
R1−2=Rtotal−R2−3=100 kΩ−40 kΩ=60 kΩR_{1-2} = R_{\text{total}} - R_{2-3} = 100\text{ k}\Omega - 40\text{ k}\Omega = 60\text{ k}\Omega
The total resistance is divided into two series segments by the wiper contact.

Key Concept

Potentiometer voltage division and wiper segment resistance distribution
Estimated Time:45s
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