Potentiometer Voltage Divider Simulation
Example 3.10: Voltage Division Using Potentiometer
This interactive simulation demonstrates how a potentiometer can be used as a voltage divider. The circuit consists of a power supply (V), a potentiometer (R₀), and a load resistor (R). Adjust the parameters below to see how the output voltage (V₁) changes.
\[ R_1 = \frac{R_0 R}{R_0 + 2R} \] (Parallel combination of R and R₀/2)
\[ R_{total} = R_1 + \frac{R_0}{2} \] (Total resistance)
\[ I = \frac{V}{R_{total}} \] (Circuit current)
\[ V_1 = I \times R_1 \] (Output voltage)
Detailed Solution to Example 3.10
1. Resistance between A and B (parallel combination):
\[ \frac{1}{R_1} = \frac{1}{R} + \frac{2}{R_0} \]
\[ R_1 = \frac{R_0 R}{R_0 + 2R} \]
2. Total resistance between A and C:
\[ R_{total} = R_1 + \frac{R_0}{2} \]
3. Current through potentiometer:
\[ I = \frac{V}{R_1 + \frac{R_0}{2}} = \frac{2V}{2R_1 + R_0} \]
4. Output voltage across R:
\[ V_1 = I R_1 = \left( \frac{2V}{2R_1 + R_0} \right) \times \frac{R_0 R}{R_0 + 2R} \]
Simplified to:
\[ V_1 = \frac{2VR}{R_0 + 4R} \]
This shows that the output voltage depends on both the potentiometer resistance and the load resistance. When R is much larger than R₀, the output approaches V/2 (ideal voltage divider). When R is comparable to or smaller than R₀, the load significantly affects the output voltage.
