Use Superposition theorem to find the voltage V in a given network.
2 Answers
To use the Superposition Theorem to find the voltage \( V \) in a given network, you need to follow these steps:
### Superposition Theorem Overview
The Superposition Theorem states that in a linear circuit with multiple independent sources (voltage or current sources), the voltage or current at any point in the circuit can be determined by considering the effect of each source independently and then summing the effects.
### Steps to Apply the Superposition Theorem
1. **Identify All Independent Sources**:
- Independent sources include voltage sources and current sources.
- For each source, you will need to consider its effect on the circuit separately.
2. **Turn Off All Sources Except One**:
- To turn off a voltage source, replace it with a short circuit.
- To turn off a current source, replace it with an open circuit.
3. **Analyze the Circuit for the Active Source**:
- With only one source active, solve the circuit to find the voltage or current of interest. This might involve using techniques like Ohm’s Law, Kirchhoff's Voltage and Current Laws, or node and mesh analysis.
4. **Repeat for Each Source**:
- Turn on each source one at a time, turn off the others, and find the voltage or current of interest for each case.
5. **Sum the Results**:
- Add the contributions from each source to find the total voltage or current at the point of interest.
### Example Problem
Let's solve a simple example circuit using the Superposition Theorem. Consider a circuit with two independent voltage sources \( V_1 \) and \( V_2 \) connected in series with resistors \( R_1 \) and \( R_2 \). We want to find the voltage \( V \) across \( R_2 \).
#### Given:
- Voltage Source \( V_1 \)
- Voltage Source \( V_2 \)
- Resistor \( R_1 \)
- Resistor \( R_2 \)
**Step 1: Analyze with \( V_1 \) Active and \( V_2 \) Turned Off**
- Replace \( V_2 \) with a short circuit.
- The circuit now has \( V_1 \) in series with \( R_1 \) and \( R_2 \).
The total resistance in the circuit is \( R_1 + R_2 \). The current \( I_1 \) through the circuit is:
\[ I_1 = \frac{V_1}{R_1 + R_2} \]
The voltage \( V_{R_2} \) across \( R_2 \) is:
\[ V_{R_2}^{(V_1)} = I_1 \cdot R_2 = \frac{V_1 \cdot R_2}{R_1 + R_2} \]
**Step 2: Analyze with \( V_2 \) Active and \( V_1 \) Turned Off**
- Replace \( V_1 \) with a short circuit.
- The circuit now has \( V_2 \) in series with \( R_1 \) and \( R_2 \).
The total resistance in the circuit is \( R_1 + R_2 \). The current \( I_2 \) through the circuit is:
\[ I_2 = \frac{V_2}{R_1 + R_2} \]
The voltage \( V_{R_2} \) across \( R_2 \) is:
\[ V_{R_2}^{(V_2)} = I_2 \cdot R_2 = \frac{V_2 \cdot R_2}{R_1 + R_2} \]
**Step 3: Sum the Results**
The total voltage \( V_{R_2} \) across \( R_2 \) is the sum of the voltages found in steps 1 and 2:
\[ V_{R_2} = V_{R_2}^{(V_1)} + V_{R_2}^{(V_2)} = \frac{V_1 \cdot R_2}{R_1 + R_2} + \frac{V_2 \cdot R_2}{R_1 + R_2} \]
\[ V_{R_2} = \frac{R_2 (V_1 + V_2)}{R_1 + R_2} \]
This is the total voltage across \( R_2 \) considering both sources.
### Summary
To find the voltage \( V \) in any network using the Superposition Theorem, analyze the circuit with each source active individually, then sum the results to get the final voltage or current at the desired location. This method is very effective for solving circuits with multiple sources in linear systems.
### Superposition Theorem Overview
The Superposition Theorem states that in a linear circuit with multiple independent sources (voltage or current sources), the voltage or current at any point in the circuit can be determined by considering the effect of each source independently and then summing the effects.
### Steps to Apply the Superposition Theorem
1. **Identify All Independent Sources**:
- Independent sources include voltage sources and current sources.
- For each source, you will need to consider its effect on the circuit separately.
2. **Turn Off All Sources Except One**:
- To turn off a voltage source, replace it with a short circuit.
- To turn off a current source, replace it with an open circuit.
3. **Analyze the Circuit for the Active Source**:
- With only one source active, solve the circuit to find the voltage or current of interest. This might involve using techniques like Ohm’s Law, Kirchhoff's Voltage and Current Laws, or node and mesh analysis.
4. **Repeat for Each Source**:
- Turn on each source one at a time, turn off the others, and find the voltage or current of interest for each case.
5. **Sum the Results**:
- Add the contributions from each source to find the total voltage or current at the point of interest.
### Example Problem
Let's solve a simple example circuit using the Superposition Theorem. Consider a circuit with two independent voltage sources \( V_1 \) and \( V_2 \) connected in series with resistors \( R_1 \) and \( R_2 \). We want to find the voltage \( V \) across \( R_2 \).
#### Given:
- Voltage Source \( V_1 \)
- Voltage Source \( V_2 \)
- Resistor \( R_1 \)
- Resistor \( R_2 \)
**Step 1: Analyze with \( V_1 \) Active and \( V_2 \) Turned Off**
- Replace \( V_2 \) with a short circuit.
- The circuit now has \( V_1 \) in series with \( R_1 \) and \( R_2 \).
The total resistance in the circuit is \( R_1 + R_2 \). The current \( I_1 \) through the circuit is:
\[ I_1 = \frac{V_1}{R_1 + R_2} \]
The voltage \( V_{R_2} \) across \( R_2 \) is:
\[ V_{R_2}^{(V_1)} = I_1 \cdot R_2 = \frac{V_1 \cdot R_2}{R_1 + R_2} \]
**Step 2: Analyze with \( V_2 \) Active and \( V_1 \) Turned Off**
- Replace \( V_1 \) with a short circuit.
- The circuit now has \( V_2 \) in series with \( R_1 \) and \( R_2 \).
The total resistance in the circuit is \( R_1 + R_2 \). The current \( I_2 \) through the circuit is:
\[ I_2 = \frac{V_2}{R_1 + R_2} \]
The voltage \( V_{R_2} \) across \( R_2 \) is:
\[ V_{R_2}^{(V_2)} = I_2 \cdot R_2 = \frac{V_2 \cdot R_2}{R_1 + R_2} \]
**Step 3: Sum the Results**
The total voltage \( V_{R_2} \) across \( R_2 \) is the sum of the voltages found in steps 1 and 2:
\[ V_{R_2} = V_{R_2}^{(V_1)} + V_{R_2}^{(V_2)} = \frac{V_1 \cdot R_2}{R_1 + R_2} + \frac{V_2 \cdot R_2}{R_1 + R_2} \]
\[ V_{R_2} = \frac{R_2 (V_1 + V_2)}{R_1 + R_2} \]
This is the total voltage across \( R_2 \) considering both sources.
### Summary
To find the voltage \( V \) in any network using the Superposition Theorem, analyze the circuit with each source active individually, then sum the results to get the final voltage or current at the desired location. This method is very effective for solving circuits with multiple sources in linear systems.
To find the voltage \( V \) in a given electrical network using the Superposition Theorem, follow these steps. I'll illustrate the process with a general example to make the approach clear. For this example, assume we have a linear circuit with multiple independent sources (current or voltage sources).
### Example Circuit Description
Consider a circuit with:
- A voltage source \( V_1 \) in series with a resistor \( R_1 \).
- A second voltage source \( V_2 \) in series with a resistor \( R_2 \).
- Both voltage sources are connected in parallel across a common resistor \( R_3 \), and we want to find the voltage \( V \) across \( R_3 \).
### Steps to Apply Superposition Theorem
1. **Identify Independent Sources**: In this case, we have two independent sources, \( V_1 \) and \( V_2 \).
2. **Turn Off All Sources Except One**: To use the superposition theorem, we need to analyze the circuit multiple times, each time considering only one independent source while replacing the others with their internal resistances.
- **Turn off \( V_2 \)**: Replace \( V_2 \) with a short circuit (since an ideal voltage source when turned off becomes a short circuit).
- **Analyze the Circuit with \( V_1 \) Only**: Calculate the voltage \( V_{1} \) across \( R_3 \) with \( V_2 \) shorted.
- **Turn off \( V_1 \)**: Replace \( V_1 \) with a short circuit.
- **Analyze the Circuit with \( V_2 \) Only**: Calculate the voltage \( V_{2} \) across \( R_3 \) with \( V_1 \) shorted.
3. **Calculate the Voltage Due to Each Source**:
- **For \( V_1 \)**:
- **Circuit Analysis**: With \( V_2 \) shorted, the voltage \( V_{1} \) across \( R_3 \) can be found using voltage divider or other circuit analysis methods.
- **Example Calculation**: Assuming the resistors are in series, the voltage \( V_{1} \) across \( R_3 \) can be given by:
\[
V_{1} = V_1 \times \frac{R_3}{R_1 + R_2 + R_3}
\]
- **For \( V_2 \)**:
- **Circuit Analysis**: With \( V_1 \) shorted, calculate the voltage \( V_{2} \) across \( R_3 \) in a similar manner.
- **Example Calculation**: The voltage \( V_{2} \) across \( R_3 \) can be:
\[
V_{2} = V_2 \times \frac{R_3}{R_2 + R_3}
\]
4. **Superimpose the Results**: Add the voltages obtained from each source to get the total voltage \( V \) across \( R_3 \).
\[
V = V_{1} + V_{2}
\]
### Example
Let’s assume the following values for simplicity:
- \( V_1 = 10V \), \( R_1 = 1k\Omega \)
- \( V_2 = 5V \), \( R_2 = 2k\Omega \)
- \( R_3 = 3k\Omega \)
**1. With \( V_2 \) Shorted:**
\[
V_{1} = V_1 \times \frac{R_3}{R_1 + R_2 + R_3} = 10V \times \frac{3k\Omega}{1k\Omega + 2k\Omega + 3k\Omega} = 10V \times \frac{3k\Omega}{6k\Omega} = 5V
\]
**2. With \( V_1 \) Shorted:**
\[
V_{2} = V_2 \times \frac{R_3}{R_2 + R_3} = 5V \times \frac{3k\Omega}{2k\Omega + 3k\Omega} = 5V \times \frac{3k\Omega}{5k\Omega} = 3V
\]
**3. Total Voltage Across \( R_3 \):**
\[
V = V_{1} + V_{2} = 5V + 3V = 8V
\]
So, the voltage across \( R_3 \) is \( 8V \).
### Summary
The Superposition Theorem helps simplify complex circuits with multiple sources by breaking the problem into simpler parts, analyzing each part, and then combining the results. Always remember to turn off all sources except one at a time, and replace each turned-off source with its internal resistance (short circuit for voltage sources and open circuit for current sources).
### Example Circuit Description
Consider a circuit with:
- A voltage source \( V_1 \) in series with a resistor \( R_1 \).
- A second voltage source \( V_2 \) in series with a resistor \( R_2 \).
- Both voltage sources are connected in parallel across a common resistor \( R_3 \), and we want to find the voltage \( V \) across \( R_3 \).
### Steps to Apply Superposition Theorem
1. **Identify Independent Sources**: In this case, we have two independent sources, \( V_1 \) and \( V_2 \).
2. **Turn Off All Sources Except One**: To use the superposition theorem, we need to analyze the circuit multiple times, each time considering only one independent source while replacing the others with their internal resistances.
- **Turn off \( V_2 \)**: Replace \( V_2 \) with a short circuit (since an ideal voltage source when turned off becomes a short circuit).
- **Analyze the Circuit with \( V_1 \) Only**: Calculate the voltage \( V_{1} \) across \( R_3 \) with \( V_2 \) shorted.
- **Turn off \( V_1 \)**: Replace \( V_1 \) with a short circuit.
- **Analyze the Circuit with \( V_2 \) Only**: Calculate the voltage \( V_{2} \) across \( R_3 \) with \( V_1 \) shorted.
3. **Calculate the Voltage Due to Each Source**:
- **For \( V_1 \)**:
- **Circuit Analysis**: With \( V_2 \) shorted, the voltage \( V_{1} \) across \( R_3 \) can be found using voltage divider or other circuit analysis methods.
- **Example Calculation**: Assuming the resistors are in series, the voltage \( V_{1} \) across \( R_3 \) can be given by:
\[
V_{1} = V_1 \times \frac{R_3}{R_1 + R_2 + R_3}
\]
- **For \( V_2 \)**:
- **Circuit Analysis**: With \( V_1 \) shorted, calculate the voltage \( V_{2} \) across \( R_3 \) in a similar manner.
- **Example Calculation**: The voltage \( V_{2} \) across \( R_3 \) can be:
\[
V_{2} = V_2 \times \frac{R_3}{R_2 + R_3}
\]
4. **Superimpose the Results**: Add the voltages obtained from each source to get the total voltage \( V \) across \( R_3 \).
\[
V = V_{1} + V_{2}
\]
### Example
Let’s assume the following values for simplicity:
- \( V_1 = 10V \), \( R_1 = 1k\Omega \)
- \( V_2 = 5V \), \( R_2 = 2k\Omega \)
- \( R_3 = 3k\Omega \)
**1. With \( V_2 \) Shorted:**
\[
V_{1} = V_1 \times \frac{R_3}{R_1 + R_2 + R_3} = 10V \times \frac{3k\Omega}{1k\Omega + 2k\Omega + 3k\Omega} = 10V \times \frac{3k\Omega}{6k\Omega} = 5V
\]
**2. With \( V_1 \) Shorted:**
\[
V_{2} = V_2 \times \frac{R_3}{R_2 + R_3} = 5V \times \frac{3k\Omega}{2k\Omega + 3k\Omega} = 5V \times \frac{3k\Omega}{5k\Omega} = 3V
\]
**3. Total Voltage Across \( R_3 \):**
\[
V = V_{1} + V_{2} = 5V + 3V = 8V
\]
So, the voltage across \( R_3 \) is \( 8V \).
### Summary
The Superposition Theorem helps simplify complex circuits with multiple sources by breaking the problem into simpler parts, analyzing each part, and then combining the results. Always remember to turn off all sources except one at a time, and replace each turned-off source with its internal resistance (short circuit for voltage sources and open circuit for current sources).