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.
Norton’s Theorem may be stated as Any Linear Electric Network or complex circuit with Current and Voltage sources can be replaced by an equivalent circuit containing of a single independent Current Source IN and a Parallel Resistance RN.
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 (IN)
3. Open Current Sources, Short Voltage Sources and Open Load Resistor.
4. Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (RN)
5. Now, Redraw the circuit with measured short circuit Current (IN) in, Step(2) as current Source and measured open circuit resistance (RN) 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. IL = IN / (RN / (RN+ RL))
Example: Find RN, IN, the current flowing through and Load Voltage across the load resistor in fig by using Norton’s Theorem.
Step 1. Short the 1.5Ω load resistor as shown in (Fig).
Step 2. Calculate / measure the Short Circuit Current. This is the Norton Current (IN).
We have shorted the AB terminals to determine the Norton current, IN. 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).
RT = 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)]
→ IT= 2Ω + 2Ω = 4Ω.
RT = 4Ω
IT = V / RT
IT = 12V / 4Ω= 3A..
Now we have to find ISC = IN
Apply CDR… (Current Divider Rule)…
ISC = IN = 3A x [(6Ω / (3Ω + 6Ω)] = 2A.
ISC= IN = 2A.
Step 3. Open Current Sources, Short Voltage Sources and Open Load Resistor.
Step 4. Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (RN) 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)
RN = 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]
RN = 3Ω + 1.5Ω
RN = 4.5Ω
Step 5. Connect the RN in Parallel with Current Source IN and re-connect the load resistor. This is shown in fig 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
IL = IN x [RN / (RN+ RL)]
= 2A x (4.5Ω /4.5Ω +1.5kΩ) → = 1.5A
IL = 1. 5A And Load Voltage across Load Resistor…
VL = IL x RL= 1.5A x 1.5Ω
= 2.25V
Norton’s theorem is also applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The Norton equivalent circuit is illustrated in Fig. below, where a linear circuit s replaced by a current source in parallel with an impedance. Keep in mind that the two equivalent circuits are related as
VTh= ZNIN, ZTh= ZN
just as in source transformation. VTh is the open-circuit voltage while IN is the short-circuit current.
