This is another useful theorem to analyze electric circuits like Thevenin’s Theorem, which reduces linear, active circuits and complex networks into a simple equivalent circuit. The main difference between Thevenin’s theorem and Norton’s theorem is that, Thevenin’s theorem provides an equivalent voltage source and an equivalent series resistance, while Norton’s theorem provides an equivalent Current source and an equivalent parallel resistance.

Simple Steps to Analyze Electric Circuit through Norton’s Theorem

1. Short the load resistor
2. Calculate / measure the Short Circuit Current. This is the Norton Current (I_N)
3. Open Current Sources, Short Voltage Sources and Open Load Resistor.
4. Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R_N)
5. Now, Redraw the circuit with measured short circuit Current (I_N) in Step (2) as current Source and measured open circuit resistance (R_N) in step (4) as a parallel resistance and connect the load resistor which we had removed in Step (3). This is the Equivalent Norton Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed. You have done.
6. Now find the Load current flowing through and Load Voltage across Load Resistor by using the Current divider rule. I_L = I_N / (R_N / (R_N+ R_L)) ((For better understanding…check the solved 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.

Step 1.

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

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.

Step 3.

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

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

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.

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 +1.5Ω) → = 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

With Dennis James Matildo