Norton's Theorem

                Norton's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin's Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Contrasting our original example circuit against the Norton equivalent: it looks something like this:

               Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

               As with Thevenin's Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin's Theorem are the steps used in Norton's Theorem to calculate the Norton source current (INorton) and Norton resistance (RNorton).

              Norton's Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.

Steps to follow for Norton's Theorem:

• (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be.
• (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for parallel circuits.

Example:

Find R_N, I_N, the current flowing through and Load Voltage across the load resistor in fig (1) by using Norton’s Theorem.

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Solution:-

Step 1

Short the 1.5Ω load resistor as shown in (Fig 2)

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Step 2

Calculate / measure the Short Circuit Current. This is the Norton Current (I_N).

We have shorted the AB terminals to determine the Norton current, I_N. The 6Ω and 3Ω are then in parallel and this parallel combination of 6Ω and 3Ω are then in series with 2Ω.

So the Total Resistance of the circuit to the Source is:-

2Ω + (6Ω || 3Ω) ….. (|| = in parallel with)

R_T = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)] → I_T= 2Ω + 2Ω = 4Ω

R_T = 4Ω

I_T = V / R_T

I_T = 12V / 4Ω

I_T = 3A..

Now we have to find I_SC = I_N… Apply CDR… (Current Divider Rule)…

I_SC = I_N = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.

I_SC= I_N = 2A

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Step 3

Open Current Sources, Short Voltage Sources and Open Load Resistor. Fig (4)

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Step 4

Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R_N)

We have Reduced the 12V DC source to zero is equivalent to replace it with a short in step (3), as shown in figure (4)  We can see that 3Ω resistor is in series with a parallel combination of 6Ω resistor and  2Ω resistor. i.e.:

3Ω + (6Ω || 2Ω) ….. (|| = in parallel with)

R_N = 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]

R_N = 3Ω + 1.5Ω

R_N = 4.5Ω

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Step 5

Connect the R_N in Parallel with Current Source I_Nand re-connect the load resistor. This is shown in fig (6) i.e. Norton Equivalent circuit with load resistor.

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Step 6

Now apply the last step i.e. calculate the load current through and Load voltage across load resistor by Ohm’s Law as shown in fig 7.

Load Current through Load Resistor…

I_L = I_N x [R_N / (R_N+ R_L)]

= 2A x (4.5Ω /4.5Ω +1.5kΩ) → = 1.5A

I_L = 1. 5A

And

Load Voltage across Load Resistor…

V_L = I_L x R_L

V_L = 1.5A x 1.5Ω

V_L= 2.25V

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