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April 19, 2023

June 16, 2021

Where IN is the short circuit current (called **Norton's current**) at the terminals to which the load resistance is connected and RN is the resistance (called **Norton's resistance**) measured between the terminals with all energy sources are replaced by their internal resistances.

Here the simple circuit which is transformed from the complex circuit is known as **Norton's equivalent circuit**. However, there are some methods to find this Norton's equivalent circuit which we will know below. Before that let us know a little bit about the constant current source first.

What is a Constant Current source?

Let us consider a voltage source of emf E and internal resistance Ri supplying power to a load resistance RL.

The load current,

The voltage at the terminals of the voltage source

If RL is varied, I remain constant although the terminal voltage Vt changes. Thus a voltage source behaves as a constant current source when its internal resistance is much larger than the load resistance. For an ideal current source, the internal resistance is infinite.

To find Norton's equivalent circuit or simplifying any linear circuits we consider a circuit with a complicated linear network driven by an energy source (VS). Where the network also contains three resistors (R1, R2, and R3) and they are connected with a load (RL).

After Norton's conversion, this complicated linear network exactly looks like this...

Where the circuit contains just a single current source and parallel resistance connected to a load.

• Identify the load resistance (RL).

• Replace the load with a short circuit and calculate the current (called Norton's current IN) through the short circuit.

• Replace the energy source with their internal resistances and calculate the resistance (called Norton's resistance RN) across the open ends.

Here the load resistance is 4Ω.

Now replace the load with a short circuit and calculate the current (called Norton's current IN) through the short circuit.

By applying Kirchhoff's Voltage Law (KVL) in loop 1, we get

By applying Kirchhoff's Voltage Law (KVL) in loop 2, we get

Using equation no - (2) in equation no - (1), we get

Now replace the energy source with their internal resistances and calculate the resistance (called Norton's resistance RN) across the open ends.

Here,

We connect the load resistance RL across the current source IN (Norton's current) shunted by the resistance RN.

Now the voltage across the RN and RL in parallel is

Therefore, the current through the load resistance RL is given by

Here the load resistance is 5Ω.

Now replace the load with a short circuit and calculate the current (called Norton's current IN) through the short circuit.

By applying Kirchhoff's Voltage Law (KVL) in loop 1, we get

By applying **Kirchhoff's Voltage Law (KVL)** in loop 2, we get

Using equation no - (2) in equation no - (1), we get

Now replace the energy source with their internal resistances and calculate the resistance (called Norton's resistance RN) across the open ends.

Here,

**Norton's equivalent circuit**

We connect the load resistance RL across the current source IN (Norton's current) shunted by the resistance RN.

Here RN and RL in parallel, so the equivalent resistance

Now the voltage across the RN and RL in parallel is

Therefore, the current through the load resistance RL is given by

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