how did you calculate v thevenin, please elaborate

To calculate the Thevenin voltage \(V_{th}\), we need to find the open-circuit voltage across terminals **a** and **b**. Here's a detailed breakdown of the steps involved:

### 1. Understand the Circuit Configuration:
The circuit has the following elements:
- A \(20 \, V\) voltage source.
- A \(2 \, A\) current source.
- Resistors with values \(10 \, \Omega\), \(6 \, \Omega\), \(5 \, \Omega\), and another \(10 \, \Omega\) across the circuit.

The goal is to find the voltage across the open terminals **a** and **b**, where there is no load connected. This voltage will be the Thevenin voltage \(V_{th}\).

### 2. Simplify the Circuit:
To calculate \(V_{th}\), we simplify the circuit by:
- **Voltage Division**: We calculate the voltage drop across the resistors in the path of the voltage source.

Since there are resistors in series and parallel, we combine them step by step. The key part here is finding the equivalent resistance seen from the perspective of the voltage source, which lets us determine how much of the \(20 \, V\) voltage will appear across the open terminals **a** and **b**.

1. **Parallel Combination of Resistors \(R_1\) and \(R_2\)**:
   - \(R_1 = 10 \, \Omega\)
   - \(R_2 = 10 \, \Omega\)
   
   These two resistors are in parallel, so their combined resistance is:
   \[
   R_{\text{parallel}} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{10 \times 10}{10 + 10} = 5 \, \Omega
   \]

2. **Add the \(6 \, \Omega\) Resistor in Series**:
   - Now, add the \(6 \, \Omega\) resistor in series with the parallel combination:
   \[
   R_{\text{combined}} = R_{\text{parallel}} + 6 \, \Omega = 5 \, \Omega + 6 \, \Omega = 11 \, \Omega
   \]

3. **Find Equivalent Resistance with the \(5 \, \Omega\) Resistor in Parallel**:
   - The combined \(11 \, \Omega\) resistor is in parallel with the \(5 \, \Omega\) resistor, so the total equivalent resistance is:
   \[
   R_{\text{total}} = \frac{R_{\text{combined}} \cdot R_3}{R_{\text{combined}} + R_3} = \frac{11 \times 5}{11 + 5} = 3.4375 \, \Omega
   \]

### 3. Voltage Division to Find \(V_{th}\):
To find the Thevenin voltage \(V_{th}\), we apply the voltage division principle:

- The voltage drop across the combined resistance (\(R_{\text{combined}} = 11 \, \Omega\)) is a fraction of the total \(20 \, V\) voltage. This fraction is determined by the ratio of the combined resistance to the total resistance in the circuit.

Using voltage division:
\[
V_{th} = 20 \, V \times \frac{R_{\text{combined}}}{R_{\text{combined}} + R_3} = 20 \times \frac{11}{11 + 5} = 13.75 \, V
\]

Thus, the open-circuit voltage across terminals **a** and **b** is **13.75 V**, which is the Thevenin voltage \(V_{th}\).

### Conclusion:
- **Thevenin Voltage \(V_{th} = 13.75 \, V\)** is found using the voltage division principle after simplifying the resistors in the circuit.

Let me know if you need further clarification or any additional questions!

Learn how to calculate Thevenin voltage in electric circuits with a step-by-step guide. This problem involves analyzing a circuit with resistors and voltage sources to find the open-circuit voltage using Thevenin's theorem and voltage division. Suitable for advanced-level learners and engineers.

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