State Norton’s theorem and find the current using Norton’s theorem through a load.
2 Answers
Certainly! Let's break it down step by step.
### Norton’s Theorem
Norton’s Theorem is a way to simplify a complex linear electrical network with multiple sources and resistors into a simpler equivalent circuit. This theorem is particularly useful for analyzing circuits with a load resistor.
**Statement of Norton’s Theorem:**
Norton’s Theorem states that any linear electrical network with multiple voltage sources, current sources, and resistors can be simplified to a simple equivalent circuit consisting of:
1. **A single current source (I_N) in parallel with**
2. **A single resistor (R_N)**
**Where:**
- **I_N** is the Norton equivalent current, which is the current flowing through the load resistor when it is short-circuited (i.e., the current through a wire placed in place of the load resistor).
- **R_N** is the Norton equivalent resistance, which is the resistance seen by the load resistor when all independent voltage sources are replaced by short circuits and all independent current sources are replaced by open circuits.
### Steps to Find Norton Equivalent Circuit
1. **Find Norton Equivalent Current (I_N):**
- **Short-Circuit the Load:** Temporarily remove the load resistor from the circuit and replace it with a wire (a short circuit).
- **Calculate the Current:** Determine the current through this short circuit. This current is the Norton equivalent current \( I_N \).
2. **Find Norton Equivalent Resistance (R_N):**
- **Turn Off All Independent Sources:** Replace all independent voltage sources with short circuits and all independent current sources with open circuits.
- **Calculate the Resistance:** Determine the equivalent resistance across the terminals where the load resistor was connected. This resistance is the Norton equivalent resistance \( R_N \).
3. **Construct the Norton Equivalent Circuit:**
- Place the Norton equivalent current source \( I_N \) in parallel with the Norton equivalent resistance \( R_N \).
### Example to Find Current Through a Load Using Norton’s Theorem
Let’s apply Norton’s Theorem to a sample circuit.
**Example Circuit:**
Consider a circuit with the following components:
- A 12V voltage source \( V \)
- A 4Ω resistor \( R_1 \) in series with the voltage source
- A 6Ω resistor \( R_2 \) connected in parallel with the series combination of \( V \) and \( R_1 \)
- A load resistor \( R_L \) connected across \( R_2 \)
**1. Find Norton Equivalent Current (I_N):**
- **Short-Circuit \( R_L \):** Replace \( R_L \) with a wire.
- **Determine the Current Through the Short-Circuit:**
- The voltage across \( R_2 \) is the same as the voltage across the short circuit.
- The current through \( R_2 \) can be found using the voltage divider rule:
\[
V_{R_2} = V_{source} \cdot \frac{R_2}{R_1 + R_2}
\]
\[
V_{R_2} = 12V \cdot \frac{6Ω}{4Ω + 6Ω} = 12V \cdot \frac{6}{10} = 7.2V
\]
- The current through \( R_2 \) is:
\[
I_N = \frac{V_{R_2}}{R_2} = \frac{7.2V}{6Ω} = 1.2A
\]
**2. Find Norton Equivalent Resistance (R_N):**
- **Turn Off All Independent Sources:** Replace the 12V source with a short circuit.
- **Calculate the Resistance Seen Across the Terminals Where \( R_L \) Was Connected:**
- The resistance seen is the parallel combination of \( R_1 \) and \( R_2 \):
\[
R_N = R_1 \parallel R_2 = \frac{R_1 \cdot R_2}{R_1 + R_2}
\]
\[
R_N = \frac{4Ω \cdot 6Ω}{4Ω + 6Ω} = \frac{24}{10} = 2.4Ω
\]
**3. Norton Equivalent Circuit:**
- The Norton equivalent circuit is a 1.2A current source in parallel with a 2.4Ω resistor.
**To Find Current Through Load \( R_L \):**
- Reconnect the load resistor \( R_L \) to the Norton equivalent circuit.
- The current through \( R_L \) can be calculated using the parallel resistor formula:
\[
I_{R_L} = \frac{I_N}{1 + \frac{R_N}{R_L}}
\]
If \( R_L \) is, say, 3Ω:
\[
I_{R_L} = \frac{1.2A}{1 + \frac{2.4Ω}{3Ω}} = \frac{1.2A}{1 + 0.8} = \frac{1.2A}{1.8} = 0.67A
\]
So, the current through the load resistor \( R_L \) is approximately 0.67A.
This detailed approach shows how Norton’s theorem helps simplify complex circuits and find the current through a load resistor by transforming the original circuit into a simpler equivalent.
### Norton’s Theorem
Norton’s Theorem is a way to simplify a complex linear electrical network with multiple sources and resistors into a simpler equivalent circuit. This theorem is particularly useful for analyzing circuits with a load resistor.
**Statement of Norton’s Theorem:**
Norton’s Theorem states that any linear electrical network with multiple voltage sources, current sources, and resistors can be simplified to a simple equivalent circuit consisting of:
1. **A single current source (I_N) in parallel with**
2. **A single resistor (R_N)**
**Where:**
- **I_N** is the Norton equivalent current, which is the current flowing through the load resistor when it is short-circuited (i.e., the current through a wire placed in place of the load resistor).
- **R_N** is the Norton equivalent resistance, which is the resistance seen by the load resistor when all independent voltage sources are replaced by short circuits and all independent current sources are replaced by open circuits.
### Steps to Find Norton Equivalent Circuit
1. **Find Norton Equivalent Current (I_N):**
- **Short-Circuit the Load:** Temporarily remove the load resistor from the circuit and replace it with a wire (a short circuit).
- **Calculate the Current:** Determine the current through this short circuit. This current is the Norton equivalent current \( I_N \).
2. **Find Norton Equivalent Resistance (R_N):**
- **Turn Off All Independent Sources:** Replace all independent voltage sources with short circuits and all independent current sources with open circuits.
- **Calculate the Resistance:** Determine the equivalent resistance across the terminals where the load resistor was connected. This resistance is the Norton equivalent resistance \( R_N \).
3. **Construct the Norton Equivalent Circuit:**
- Place the Norton equivalent current source \( I_N \) in parallel with the Norton equivalent resistance \( R_N \).
### Example to Find Current Through a Load Using Norton’s Theorem
Let’s apply Norton’s Theorem to a sample circuit.
**Example Circuit:**
Consider a circuit with the following components:
- A 12V voltage source \( V \)
- A 4Ω resistor \( R_1 \) in series with the voltage source
- A 6Ω resistor \( R_2 \) connected in parallel with the series combination of \( V \) and \( R_1 \)
- A load resistor \( R_L \) connected across \( R_2 \)
**1. Find Norton Equivalent Current (I_N):**
- **Short-Circuit \( R_L \):** Replace \( R_L \) with a wire.
- **Determine the Current Through the Short-Circuit:**
- The voltage across \( R_2 \) is the same as the voltage across the short circuit.
- The current through \( R_2 \) can be found using the voltage divider rule:
\[
V_{R_2} = V_{source} \cdot \frac{R_2}{R_1 + R_2}
\]
\[
V_{R_2} = 12V \cdot \frac{6Ω}{4Ω + 6Ω} = 12V \cdot \frac{6}{10} = 7.2V
\]
- The current through \( R_2 \) is:
\[
I_N = \frac{V_{R_2}}{R_2} = \frac{7.2V}{6Ω} = 1.2A
\]
**2. Find Norton Equivalent Resistance (R_N):**
- **Turn Off All Independent Sources:** Replace the 12V source with a short circuit.
- **Calculate the Resistance Seen Across the Terminals Where \( R_L \) Was Connected:**
- The resistance seen is the parallel combination of \( R_1 \) and \( R_2 \):
\[
R_N = R_1 \parallel R_2 = \frac{R_1 \cdot R_2}{R_1 + R_2}
\]
\[
R_N = \frac{4Ω \cdot 6Ω}{4Ω + 6Ω} = \frac{24}{10} = 2.4Ω
\]
**3. Norton Equivalent Circuit:**
- The Norton equivalent circuit is a 1.2A current source in parallel with a 2.4Ω resistor.
**To Find Current Through Load \( R_L \):**
- Reconnect the load resistor \( R_L \) to the Norton equivalent circuit.
- The current through \( R_L \) can be calculated using the parallel resistor formula:
\[
I_{R_L} = \frac{I_N}{1 + \frac{R_N}{R_L}}
\]
If \( R_L \) is, say, 3Ω:
\[
I_{R_L} = \frac{1.2A}{1 + \frac{2.4Ω}{3Ω}} = \frac{1.2A}{1 + 0.8} = \frac{1.2A}{1.8} = 0.67A
\]
So, the current through the load resistor \( R_L \) is approximately 0.67A.
This detailed approach shows how Norton’s theorem helps simplify complex circuits and find the current through a load resistor by transforming the original circuit into a simpler equivalent.
### Norton's Theorem
Norton's Theorem is a fundamental principle used in electrical engineering to simplify complex linear circuits. The theorem states:
**"Any two-terminal linear electrical network containing independent sources (voltage or current sources) and resistors can be replaced with an equivalent circuit consisting of a single current source (called the Norton current) in parallel with a single resistor (called the Norton resistance)."**
This equivalent circuit behaves the same as the original circuit when connected to the same load.
### Steps to Apply Norton’s Theorem
To find the current through a load using Norton’s Theorem, follow these steps:
1. **Identify the Portion of the Circuit to Analyze:**
- Identify the two terminals of the circuit where the load is connected.
2. **Remove the Load Resistor:**
- Temporarily remove the load resistor from the circuit.
3. **Calculate Norton Resistance (R<sub>N</sub>):**
- Deactivate all independent sources in the circuit:
- Replace all independent voltage sources with short circuits (a wire).
- Replace all independent current sources with open circuits (a break in the wire).
- Calculate the equivalent resistance between the two terminals where the load was connected. This resistance is the Norton resistance, \( R_N \).
4. **Calculate Norton Current (I<sub>N</sub>):**
- Reconnect the original sources.
- Calculate the short-circuit current between the two terminals where the load was connected. This current is the Norton current, \( I_N \).
5. **Construct the Norton Equivalent Circuit:**
- Draw a current source \( I_N \) in parallel with the resistance \( R_N \).
6. **Reconnect the Load Resistor:**
- Attach the load resistor back to the Norton equivalent circuit.
7. **Calculate the Load Current (I<sub>L</sub>):**
- Use the parallel circuit analysis to calculate the current through the load resistor.
### Example: Finding the Current Through a Load Using Norton’s Theorem
Let's solve a problem using Norton's theorem.
**Given Circuit:**
Assume you have a circuit with the following components:
- A voltage source \( V_s = 12 \, V \)
- A series resistor \( R_1 = 4 \, \Omega \)
- A parallel combination of resistors \( R_2 = 6 \, \Omega \) and \( R_3 = 3 \, \Omega \)
- A load resistor \( R_L = 2 \, \Omega \)
**Objective:** Find the current through \( R_L \) using Norton’s Theorem.
#### Step 1: Identify the Portion of the Circuit
- The portion of the circuit of interest is where \( R_L \) is connected.
#### Step 2: Remove the Load Resistor
- Remove \( R_L \) from the circuit.
#### Step 3: Calculate Norton Resistance (R<sub>N</sub>)
1. **Deactivate the Voltage Source:** Replace the voltage source \( V_s \) with a short circuit.
2. **Calculate the Equivalent Resistance:**
- The resistors \( R_2 \) and \( R_3 \) are in parallel.
- Calculate their equivalent resistance:
\[
R_{23} = \left(\frac{1}{R_2} + \frac{1}{R_3}\right)^{-1} = \left(\frac{1}{6} + \frac{1}{3}\right)^{-1} = \left(\frac{1}{6} + \frac{2}{6}\right)^{-1} = \left(\frac{3}{6}\right)^{-1} = 2 \, \Omega
\]
- Now, \( R_1 \) is in series with \( R_{23} \):
\[
R_N = R_1 + R_{23} = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega
\]
#### Step 4: Calculate Norton Current (I<sub>N</sub>)
1. **Reconnect the Voltage Source**:
- Now, calculate the short-circuit current between the terminals where \( R_L \) was connected.
- The total resistance seen by the source \( V_s \) is:
\[
R_{total} = R_1 + \left(\frac{R_2 \times R_3}{R_2 + R_3}\right) = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega
\]
- The current through \( R_1 \) is:
\[
I = \frac{V_s}{R_{total}} = \frac{12 \, V}{6 \, \Omega} = 2 \, A
\]
- The current through the parallel combination \( I_{parallel} \) is the same as \( I \) because it's a series circuit before the parallel combination:
\[
I_{parallel} = 2 \, A
\]
- The Norton current is the same as the current through \( R_3 \) because of the parallel network:
\[
I_N = I_{parallel} \times \frac{R_2}{R_2 + R_3} = 2 \, A \times \frac{3 \, \Omega}{6 \, \Omega} = 1 \, A
\]
#### Step 5: Construct the Norton Equivalent Circuit
- The Norton equivalent circuit is a current source \( I_N = 1 \, A \) in parallel with \( R_N = 6 \, \Omega \).
#### Step 6: Reconnect the Load Resistor
- Reconnect \( R_L = 2 \, \Omega \) across the Norton equivalent circuit.
#### Step 7: Calculate the Load Current (I<sub>L</sub>)
- Use parallel circuit analysis to find the current through \( R_L \):
\[
I_L = I_N \times \frac{R_N}{R_N + R_L} = 1 \, A \times \frac{6 \, \Omega}{6 \, \Omega + 2 \, \Omega} = 1 \, A \times \frac{6}{8} = 0.75 \, A
\]
### Final Answer
The current through the load resistor \( R_L = 2 \, \Omega \) using Norton’s Theorem is **0.75 A**.
Norton's Theorem is a fundamental principle used in electrical engineering to simplify complex linear circuits. The theorem states:
**"Any two-terminal linear electrical network containing independent sources (voltage or current sources) and resistors can be replaced with an equivalent circuit consisting of a single current source (called the Norton current) in parallel with a single resistor (called the Norton resistance)."**
This equivalent circuit behaves the same as the original circuit when connected to the same load.
### Steps to Apply Norton’s Theorem
To find the current through a load using Norton’s Theorem, follow these steps:
1. **Identify the Portion of the Circuit to Analyze:**
- Identify the two terminals of the circuit where the load is connected.
2. **Remove the Load Resistor:**
- Temporarily remove the load resistor from the circuit.
3. **Calculate Norton Resistance (R<sub>N</sub>):**
- Deactivate all independent sources in the circuit:
- Replace all independent voltage sources with short circuits (a wire).
- Replace all independent current sources with open circuits (a break in the wire).
- Calculate the equivalent resistance between the two terminals where the load was connected. This resistance is the Norton resistance, \( R_N \).
4. **Calculate Norton Current (I<sub>N</sub>):**
- Reconnect the original sources.
- Calculate the short-circuit current between the two terminals where the load was connected. This current is the Norton current, \( I_N \).
5. **Construct the Norton Equivalent Circuit:**
- Draw a current source \( I_N \) in parallel with the resistance \( R_N \).
6. **Reconnect the Load Resistor:**
- Attach the load resistor back to the Norton equivalent circuit.
7. **Calculate the Load Current (I<sub>L</sub>):**
- Use the parallel circuit analysis to calculate the current through the load resistor.
### Example: Finding the Current Through a Load Using Norton’s Theorem
Let's solve a problem using Norton's theorem.
**Given Circuit:**
Assume you have a circuit with the following components:
- A voltage source \( V_s = 12 \, V \)
- A series resistor \( R_1 = 4 \, \Omega \)
- A parallel combination of resistors \( R_2 = 6 \, \Omega \) and \( R_3 = 3 \, \Omega \)
- A load resistor \( R_L = 2 \, \Omega \)
**Objective:** Find the current through \( R_L \) using Norton’s Theorem.
#### Step 1: Identify the Portion of the Circuit
- The portion of the circuit of interest is where \( R_L \) is connected.
#### Step 2: Remove the Load Resistor
- Remove \( R_L \) from the circuit.
#### Step 3: Calculate Norton Resistance (R<sub>N</sub>)
1. **Deactivate the Voltage Source:** Replace the voltage source \( V_s \) with a short circuit.
2. **Calculate the Equivalent Resistance:**
- The resistors \( R_2 \) and \( R_3 \) are in parallel.
- Calculate their equivalent resistance:
\[
R_{23} = \left(\frac{1}{R_2} + \frac{1}{R_3}\right)^{-1} = \left(\frac{1}{6} + \frac{1}{3}\right)^{-1} = \left(\frac{1}{6} + \frac{2}{6}\right)^{-1} = \left(\frac{3}{6}\right)^{-1} = 2 \, \Omega
\]
- Now, \( R_1 \) is in series with \( R_{23} \):
\[
R_N = R_1 + R_{23} = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega
\]
#### Step 4: Calculate Norton Current (I<sub>N</sub>)
1. **Reconnect the Voltage Source**:
- Now, calculate the short-circuit current between the terminals where \( R_L \) was connected.
- The total resistance seen by the source \( V_s \) is:
\[
R_{total} = R_1 + \left(\frac{R_2 \times R_3}{R_2 + R_3}\right) = 4 \, \Omega + 2 \, \Omega = 6 \, \Omega
\]
- The current through \( R_1 \) is:
\[
I = \frac{V_s}{R_{total}} = \frac{12 \, V}{6 \, \Omega} = 2 \, A
\]
- The current through the parallel combination \( I_{parallel} \) is the same as \( I \) because it's a series circuit before the parallel combination:
\[
I_{parallel} = 2 \, A
\]
- The Norton current is the same as the current through \( R_3 \) because of the parallel network:
\[
I_N = I_{parallel} \times \frac{R_2}{R_2 + R_3} = 2 \, A \times \frac{3 \, \Omega}{6 \, \Omega} = 1 \, A
\]
#### Step 5: Construct the Norton Equivalent Circuit
- The Norton equivalent circuit is a current source \( I_N = 1 \, A \) in parallel with \( R_N = 6 \, \Omega \).
#### Step 6: Reconnect the Load Resistor
- Reconnect \( R_L = 2 \, \Omega \) across the Norton equivalent circuit.
#### Step 7: Calculate the Load Current (I<sub>L</sub>)
- Use parallel circuit analysis to find the current through \( R_L \):
\[
I_L = I_N \times \frac{R_N}{R_N + R_L} = 1 \, A \times \frac{6 \, \Omega}{6 \, \Omega + 2 \, \Omega} = 1 \, A \times \frac{6}{8} = 0.75 \, A
\]
### Final Answer
The current through the load resistor \( R_L = 2 \, \Omega \) using Norton’s Theorem is **0.75 A**.