To solve for the output voltage $V_o$ in the given circuit, we can follow these steps:

Given:

• $V_1 = 12 \text{ V}$
• $R_{\text{POT}} = 20 \text{ k}\Omega$ (total resistance of the potentiometer)
• $R_{\text{procent}} = 55\%$ (percentage of $R_{\text{POT}}$ between pins 1 and 2)
• $R_1 = 56 \text{ k}\Omega$

Step 1: Calculate the resistance between pins 1 and 2 of the potentiometer

The resistance between pins 1 and 2, denoted $R_2$, is given by:

$R_2 = R_{\text{POT}} \times \frac{R_{\text{procent}}}{100} = 20 \text{ k}\Omega \times \frac{55}{100} = 11 \text{ k}\Omega$

Step 2: Calculate the total resistance seen by the source $V_1$

The total resistance $R_T$ in the circuit is the series combination of $R_2$ and $R_1$:

$R_T = R_2 + R_1 = 11 \text{ k}\Omega + 56 \text{ k}\Omega = 67 \text{ k}\Omega$

Step 3: Calculate the current through the circuit

Using Ohm's Law, the current $I$ through the circuit is:

$I = \frac{V_1}{R_T} = \frac{12 \text{ V}}{67 \text{ k}\Omega} \approx 0.179 \text{ mA}$

Step 4: Calculate the voltage drop across $R_1$ (which is $V_o$)

Since $V_o$ is the voltage across $R_1$, we use Ohm's Law again:

$V_o = I \times R_1 = 0.179 \text{ mA} \times 56 \text{ k}\Omega \approx 10.024 \text{ V}$

Thus, the output voltage $V_o$ is approximately 10.024 V.

Do you have any questions, or would you like a more detailed explanation? Here are some related questions you might be interested in:

1. How would the output voltage change if $R_{\text{procent}}$ were different?
2. What if $R_1$ was reduced to 10 kΩ—how would that affect $V_o$?
3. Can you explain how the potentiometer setting affects the circuit?
4. How would the output voltage change if $V_1$ was increased?
5. What is the power dissipated by $R_1$ in this configuration?

Tip: Remember that the position of the potentiometer's wiper (indicated by $R_{\text{procent}}$) directly influences the division of voltage in the circuit.