To find the Thévenin voltage ($V_{TH}$) using Thévenin's Theorem, we follow these steps:

Step 1: Identify the portion of the circuit

From the diagram, we are asked to find the Thévenin equivalent with respect to the portion on the right of the circuit, beyond the two black dots. This typically involves removing the load (if there is any), which in this case corresponds to the resistors $R_4$ and $R_5$, as we're finding the Thévenin equivalent at their terminals.

Step 2: Deactivate independent sources

There is a 5V voltage source, $V_s$, in the circuit. However, we only need to focus on calculating the Thévenin voltage, which is the open-circuit voltage across the points where $R_4$ and $R_5$ are connected. Therefore, the source remains active.

Step 3: Find the open-circuit voltage across the terminals

We will use the principle of voltage division to calculate the open-circuit voltage across the terminals.

a) Simplify the circuit for voltage calculation

The voltage source $V_s = 5V$ is connected in series with $R_1 = 1 \, k\Omega$ and $R_2 = 1.5 \, k\Omega$. The potential drop across $R_1$ and $R_2$ needs to be calculated. Since $R_3$ forms part of the load side, we are interested in the voltage at the point between $R_2$ and $R_3$.

b) Calculate the total resistance in the voltage division path

$R_{total} = R_1 + R_2 = 1 \, k\Omega + 1.5 \, k\Omega = 2.5 \, k\Omega$

c) Apply voltage division

The Thévenin voltage is the voltage across $R_2$, so using the voltage divider rule: $V_{TH} = V_s \times \frac{R_2}{R_1 + R_2}$ Substituting values: $V_{TH} = 5V \times \frac{1.5 \, k\Omega}{2.5 \, k\Omega}$ $V_{TH} = 5V \times 0.6 = 3V$

Thus, the Thévenin voltage $V_{TH}$ is 3V.

Would you like further explanation or have any questions on these calculations?

Here are some related questions:

1. How do we find the Thévenin resistance for this circuit?
2. What happens if we change the values of $R_4$ and $R_5$?
3. Can you explain how the voltage divider rule works?
4. How do we compute the Thévenin equivalent if the source was a current source?
5. What if we replaced $R_1$ and $R_2$ with different values?

Tip: When simplifying circuits, always identify independent sources first, as they play a key role in Thévenin and Norton theorems.