Question

What is superposition theorem with an example?

Answer 1

The superposition theorem is a fundamental principle in linear circuit analysis that simplifies the process of analyzing circuits with multiple independent sources (voltage or current sources). The theorem states that in any linear circuit with multiple independent sources, the response (voltage or current) at any point in the circuit can be determined by considering the effect of each source individually while turning off the other sources.

### The Superposition Theorem Explained

1. **Turn Off All Sources Except One**: To apply the superposition theorem, first, turn off all but one of the independent sources in the circuit. For voltage sources, turn them off by replacing them with a short circuit (a wire). For current sources, turn them off by replacing them with an open circuit (a break in the circuit).

2. **Analyze the Circuit**: Solve the circuit with only the one active source. Determine the contribution of this source to the desired response (voltage or current) at the point of interest.

3. **Repeat for Each Source**: Repeat the process for each independent source in the circuit, turning off all other sources while analyzing the circuit.

4. **Combine the Results**: Add up all the contributions from each source to find the total response at the point of interest.

### Example of Superposition Theorem

Consider a simple circuit with two voltage sources and two resistors:

- **Circuit Description**:
  - **V1** = 10V
  - **V2** = 5V
  - **R1** = 1kΩ
  - **R2** = 2kΩ

   The resistors are connected in series with the voltage sources arranged as follows: V1 is connected in series with R1 and then in series with the parallel combination of V2 and R2.

```
      +----R1----+----+----R2----+
      |          |    |          |
     V1         V2   (Node)     Ground
      |          |    |
      +----------+----+
```

   **Objective**: Find the voltage across R2.

**Step 1: Consider V1 Only**

- Turn off V2 by replacing it with a short circuit. The circuit then looks like:

```
      +----R1----+----+----R2----+
      |          |    |          |
     V1        Short  (Node)     Ground
      |          |    |
      +----------+----+
```

  - The total resistance in the circuit is R1 + R2.
  - The total current through the circuit is \( I = \frac{V1}{R1 + R2} = \frac{10V}{1kΩ + 2kΩ} = \frac{10V}{3kΩ} \approx 3.33mA \).
  - The voltage across R2 is \( V_{R2} = I \times R2 = 3.33mA \times 2kΩ = 6.67V \).

**Step 2: Consider V2 Only**

- Turn off V1 by replacing it with a short circuit. The circuit then looks like:

```
      +----Short----+----+----R2----+
      |             |    |          |
     V2           (Node)  (Node)     Ground
      |             |
      +-------------+
```

  - The voltage across R2 is directly V2 (since it's a short circuit around R1, so it doesn’t affect the voltage across R2).
  - Thus, the voltage across R2 in this case is simply V2 = 5V.

**Step 3: Combine the Results**

- Add the contributions from V1 and V2:
  - Contribution from V1: 6.67V
  - Contribution from V2: 5V

  - Total voltage across R2: 6.67V + 5V = 11.67V

### Summary

The superposition theorem allows us to break down a complex circuit with multiple sources into simpler parts. By analyzing the effect of each source individually and then combining the results, we can simplify the analysis of the circuit significantly.

Answer 2

The Superposition Theorem is a fundamental principle used in linear circuit analysis. It states that in a linear circuit with multiple independent sources (voltage or current), the voltage or current at any point in the circuit can be determined by summing the contributions of each source individually, while all other sources are turned off.

### Key Points of Superposition Theorem:

1. **Linearity**: The theorem applies to linear circuits where components follow Ohm's Law and the principle of superposition is valid.

2. **Turning Off Sources**:
   - **Voltage Source**: Replaced by a short circuit (0V).
   - **Current Source**: Replaced by an open circuit (0A).

3. **Finding Individual Contributions**:
   - Analyze the circuit with only one independent source active at a time.
   - Determine the voltage or current due to that single source.

4. **Summing Contributions**:
   - After finding the contributions from each source, add them algebraically to get the total voltage or current.

### Example:

Consider a simple circuit with two voltage sources and two resistors:

- **Circuit Components**:
  - Resistor \( R_1 = 2 \Omega \)
  - Resistor \( R_2 = 3 \Omega \)
  - Voltage Source \( V_1 = 10V \)
  - Voltage Source \( V_2 = 5V \)

**Objective**: Find the current through \( R_2 \).

#### Steps Using Superposition Theorem:

1. **Analyze with \( V_1 \) Active and \( V_2 \) Turned Off**:
   - Replace \( V_2 \) with a short circuit.
   - The circuit now has \( V_1 \) and \( R_1 \) in series, with \( R_2 \) in parallel.

   The total resistance seen by \( V_1 \) is:
   \[
   R_{\text{total}} = R_1 + \left(\frac{R_2}{R_1 + R_2}\right) = 2 \Omega + \frac{3 \Omega}{5 \Omega} = 2 \Omega + 0.6 \Omega = 2.6 \Omega
   \]
   The current due to \( V_1 \):
   \[
   I_{V_1} = \frac{V_1}{R_{\text{total}}} = \frac{10V}{2.6 \Omega} \approx 3.85A
   \]
   Current through \( R_2 \) when \( V_1 \) is active:
   \[
   I_{R_2}^{V_1} = I_{V_1} \times \frac{R_1}{R_1 + R_2} = 3.85A \times \frac{2 \Omega}{5 \Omega} \approx 1.54A
   \]

2. **Analyze with \( V_2 \) Active and \( V_1 \) Turned Off**:
   - Replace \( V_1 \) with a short circuit.
   - The circuit now has \( V_2 \) across \( R_2 \) only.

   The current through \( R_2 \) when \( V_2 \) is active:
   \[
   I_{R_2}^{V_2} = \frac{V_2}{R_2} = \frac{5V}{3 \Omega} \approx 1.67A
   \]

3. **Sum the Contributions**:
   \[
   I_{R_2} = I_{R_2}^{V_1} + I_{R_2}^{V_2} = 1.54A + 1.67A \approx 3.21A
   \]

Therefore, the total current through \( R_2 \) is approximately \( 3.21A \).

This example demonstrates how to apply the Superposition Theorem to analyze complex circuits by breaking them down into simpler parts.