Norton's Theorem
Statement :
Any linear electric network or complex circuit with current and voltage
sources can be replaced by an equivalent circuit containing a single
independent current source IN and a parallel resistance RN.
1
Norton's Theorem
Norton's theorem is an analytical method used to change a complex
circuit into a simple equivalent circuit consisting of a single
resistance in parallel with a current source
2
Thevenin’s theorem & Norton's Theorem
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.
3
Steps to Analyze an Electric Circuit using 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 was 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.
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))
4
Example : 1
1.Find RN, IN, the current flowing through and Load Voltage across the load
resistor in fig. below by using Norton’s Theorem.
5
STEP 1.
Short the load resistor, RL=1.5Ω.
Short
6
STEP 2.
IT = 3A
Short
IN
Calculate the Short Circuit Current. This is the Norton Current (IN).
Determine the Norton current, IN from the shorted terminals AB.
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 (RT )of the circuit to the Source is:-
RT = 2Ω + (6Ω || 3Ω)
RT = 2Ω + *(3Ω x 6Ω) / (3Ω + 6Ω)+ ,
RT = 2Ω + 2Ω = 4Ω.
IT = V ÷ RT
IT = 12V ÷ 4Ω ; IT = 3A
7
IT= 3A
Short
IN
To find ISC = IN (Apply Current Divider Rule)
We Know, IT = 3A
ISC = IN = 3A x *(6Ω ÷ (3Ω + 6Ω)+
ISC = IN = 2A.
IT= 3A
IN =2A
8
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 fig. above . 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
9
STEP 5.
Connect the RN in Parallel with Current Source IN and reconnect the load
resistor. 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 the load resistor by Ohm’s Law as shown in fig. above.
Load Current through Load Resistor
IL = 2A x (4.5Ω ÷ 4.5Ω + 1.5Ω)
IL = 1. 5A
Load Voltage across Load Resistor , VL = IL x RL
VL = 1.5A x 1.5Ω
VL= 2.25V
10
2. Find the current through 10Ω resistor using Norton’s theorem.
IN= 6.85 Ω, RN= 4.66 Ω IL = 2.18 A
11
3.For the given circuit, calculate the current flows through the 5Ω resistor
using Norton’s theorem.
12
