LEARNING OUTCOMES

• The end of this lesson, students should be able to :

• Explain Norton’s Theorem.• Outline the procedure adopted when using

Norton’s Theorem• Determine the short circuit current ISC and

the equivalent Resistance RN (parallel)• Construct circuit base on Norton’s

Theorem.• Solve linear circuits problem using Norton’s

Theorem.

NORTON’S THEOREM

Nortons Theorem states that "Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor. As far as the load resistance, RL is

Nortons Analysis

• The basic procedure for solving Nortons Analysis equations is as follows 4 steps:1. Remove the load resistor RL or

component concerned and short circuit the

terminals. Find IN on the output terminals A and B.2. Short-circuit the two voltage sources and

open circuit terminals A and B, the two resistors are now connected together in parallel. Find RN

3.Having found both the short circuit current, IN and equivalent internal

resistance, RN. Draw the Nortons equivalent.

4. Replace the original load resistor (remove resistor) connected across terminals A and B and calculate current, I

Example 1i.

Refer to the circuit above determine the current Iflows in the 40Ω resistance by using Norton’s Theorem.

I

Following the procedures;

Step 1 & 2. Remove the load resistor RL or component concerned. Short circuit terminals A and B to give us the following circuit.

I1 I2

IN

When the terminals A and B are shorted together the two resistors are connected in parallel across their two voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as:• Determine the short-circuit current,IN flowing in the branch

Therefore IN = 2 Amps

Step 3. Short-circuit the two voltage sources and open circuit terminals A and B, the two resistors are now connected

together in parallel. Determine the value of the internal resistor

RN is found by calculating the total resistance at the terminals A and B giving us the following circuit

RN

Find the Equivalent Resistance (RN)

Therefore RN = 6.67 Ω

Step 4. Having found both the short circuit

current, IN and equivalent internal resistance, RN. Draw the following Nortons equivalent circuit below.

• Nortons equivalent circuit.

Step 4. Replace the original 40Ω load resistor connected across terminals A and B as shown below. Calculate current, I ;

I

Norton Equivalent circuit

By using current divider

I = RN X INRN + RL

= 6.67Ω x 2 A 6.67Ω+40Ω = 0.286 Amps

I = current flowing through 40 is 0.286 Amps

RELATION THEVENIN THEOREM & NORTON THEOREM

1.To convert Thevenin Equivalent circuit to Norton Equivalent circuit

Norton resistance RN = RTHNorton current , IN = VTH /RTH

2. To convert Norton Equivalent circuit to Thevenin Equivalent circuit

Thevenin resistance, RTH = RNThevenin Voltage, VTH = IN X RN

Exercises

1.Refer to the circuit below, calculate I3

by using Norton’s Theorem

10 15

15V20

I1

20V

I3

I2

Answers ;

IN = 3 ARN = 6 ohmI3= 0.692 A

Exercise :Refer to the circuit, convert Thevenin Equivalent to

Norton Equivalent circuit

5.3k

RL =3k

I1

12V