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Electrical Engineering: Ch 4: Circuit Theorems (19 of 32) Norton's Theorem Example 1

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Electrical Engineering: Ch 4: Circuit Theorems (18 of 32) Norton's Theorem

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Norton's Theorem states that - A linear active network consisting of independent or dependent voltage source and current sources and the various circuit elements can be substituted by an equivalent circuit consisting of a current source in parallel with a resistance.The current source being the short-circuited current across the load terminal and the resistance being the internal resistance of the source network.

Contents:

The Norton's theorems reduce the networks equivalent to the circuit having one current source, parallel resistance and load. Norton's theorem is the converse of Thevenin's Theorem. It consists of the equivalent current source instead of an equivalent voltage source as in Thevenin's theorem. The determination of internal resistance of the source network is identical in both the theorems.

In the final stage that is in the equivalent circuit, the current is placed in parallel to the internal resistance in Norton's Theorem whereas in Thevenin's Theorem the equivalent voltage source is placed in series with the internal resistance.

Explanation of Norton's Theorem

To understand Norton's Theorem in detail, let us consider a circuit diagram given below
In order to find the current through the load resistance I L as shown in the circuit diagram above, the load resistance has to be short-circuited as shown in the diagram below

Now, the value of current I flowing in the circuit is found out by the equation

And the short-circuit current I SC is given by the equation shown below

Now the short circuit is removed, and the independent source is deactivated as shown in the circuit diagram below and the value of the internal resistance is calculated by

As per the Norton's Theorem, the equivalent source circuit would contain a current source in parallel to the internal resistance, the current source being the short-circuited current across the shorted terminals of the load resistor. The Norton's Equivalent circuit is represented as

Finally the load current I L calculated by the equation shown below

Where,

IL is the load current

Isc is the short circuit current

Rint is the internal resistance of the circuit

RL is the load resistance of the circuit

Steps for Solving a Network Utilizing Norton's Theorem

Step 1 - Remove the load resistance of the circuit.

Step 2 - Find the internal resistance R int of the source network by deactivating the constant sources.

Step 3 - Now short the load terminals and find the short circuit current I SC flowing through the shorted load terminals using conventional network analysis methods.

Step 4 - Norton's equivalent circuit is drawn by keeping the internal resistance R int in parallel with the short circuit current I SC.

Step 5 - Reconnect the load resistance R L of the circuit across the load terminals and find the current through it known as load current I L.

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What is a Norton's Theorem? - Circuit Globe

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Norton's Theorem states that - A linear active network consisting of independent or dependent voltage source and current sources and the various circuit elements can be substituted by an equivalent circuit consisting of a current source in parallel with a resistance.The current source being the short-circuited current across the load terminal and the resistance being the internal resistance of the source network.

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Replacing a network by its Norton equivalent can simplify the analysis of a complex circuit. In this example, the Norton current is obtained from the open circuit voltage (the Thevenin voltage) divided by the resistance r. This resistance is the same as the Thevenin resistance, the resistance looking back from AB with V ₁ replaced by a short circuit.

Index

DC Circuits

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Replacing a network by its Norton equivalent can simplify the analysis of a complex circuit. In this example, the Norton current is obtained from the open circuit voltage (the Thevenin voltage) divided by the resistance r. This resistance is the same as the Thevenin resistance, the resistance looking back from AB with V 1 replaced by a short circuit.

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Electrical Engineering: Ch 4: Circuit Theorems (17 of 35) Thevenin's Theorem Ex. 2

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Step by Step Procedure with Examples.

A French Engineer, M.L Thevenin , made one of these quantum leaps in 1893. Thevenin's Theorem is not by itself an analysis tool, but the basis for a very useful method of simplifying active circuits and complex networks because we can solve complex linear circuits and networks especially electronic networks easily and quickly.

Thevenin's Theorem may be stated below:

Any Linear Electric Network or complex circuit with current and voltage sources can be replaced by an equivalent circuit containing of a single independent voltage source V TH and a Series Resistance RTH.

Steps to Analyze Electric Circuit through Thevenin's Theorem

1. Open the load resistor.

2. Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH).

3. Open current sources and short voltage sources.

4. Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (RTH).

5. Now, redraw the circuit with measured open circuit Voltage (VTH) in Step (2) as voltage source and measured open circuit resistance (RTH) in step (4) as a series resistance and connect the load resistor which we had removed in Step (1). This is the equivalent Thevenin circuit of that linear electric network or complex circuit which had to be simplified and analyzed by Thevenin's Theorem. You have done.

6. Now find the Total current flowing through load resistor by using the Ohm's Law:

Solved Example by Thevenin's Theorem:

and the load current flowing through and load voltage across the load resistor in fig (1) by using Thevenin's Theorem.

Click image to enlarge

Open the 5kΩ load resistor (Fig 2).

Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH). Fig (3).

We have already removed the load resistor from figure 1, so the circuit became an open circuit as shown in fig 2. Now we have to calculate the Thevenin's Voltage. Since 3mA current flows in both 12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor as it is open.

So 12V (3mA x 4kΩ) will appear across the 4kΩ resistor. We also know that current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in parallel with 4k resistor. So the same voltage i.e. 12V will appear across the 8kΩ resistor as well as 4kΩ resistor. Therefore 12V will appear across the AB terminals. So,

Open current sources and short voltage sources as shown below. Fig (4)

Calculate / measure the open circuit resistance. This is the Thevenin Resistance (R TH)

We have removed the 48V DC source to zero as equivalent i.e. 48V DC source has been replaced with a short in step 3 (as shown in figure 3). We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:

8kΩ + (4k Ω || 12kΩ) ..... (|| = in parallel with)

Connect the R THin series with Voltage Source V TH and re-connect the load resistor. This is shown in fig (6) i.e. Thevenin circuit with load resistor. This the Thevenin's equivalent circuit

Now apply the last step i.e Ohm's law . Calculate the total load current & load voltage as shown in fig 6.

= 12V / (11kΩ + 5kΩ) → = 12/16kΩ

Now compare this simple circuit with the original circuit of figure 1. Can you see how much easier it will be to measure/calculate the load current for different load resistors by Thevenin's Theorem? Yes and only yes.

You may also read:

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Thevenin's Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

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Thevenin's Theorem may be stated below: Any Linear Electric Network or complex circuit with current and voltage sources can be replaced by an equivalent circuit containing of a single independent voltage source VTH and a Series Resistance RTH. Norton's Theorem Maximum Power Transfer Theorem SUPERNODE Circuit Analysis

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Please read the link below

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The Thevenin voltage e used in Thevenin's Theorem is an ideal voltage source equal to the open circuit voltage at the terminals. In the example below, the resistance R 2 does not affect this voltage and the resistances R 1 and R 3 form a voltage divider, giving