The document presents a seminar on Norton’s Theorem, which states that any linear bilateral circuit with multiple energy sources can be simplified to a constant current source in parallel with a resistor. It outlines steps to implement the theorem and provides numerical examples for calculating Norton’s current and equivalent resistance. The final result indicates the load current obtained using Norton’s equivalent circuit.

1. Course Name: Network Theory (PCCET303T)
Course In-charge: MOHAMMAD WASEEM AKRAM
Course Seminar on
“Norton’s Theorem”
Presented by
✽ 28 CHAITALI UKE ✽ 29 CHAITALI INGALE

2. Edward Lawry Norton
Norton’s Theorem
Statement:
Any linear bilateral complicated circuit with multiple
energy source and resistance can be replaced by a constant
current source in parallel with a resistor connected across
the load.
IN ➳ The short-circuit current through the terminals
RN ➳ Input/Equivalent resistance at the terminals when the independent source are
turned off
➺
Norton's equivalent
circuit

3. Steps to Solve Norton’s Theorem
01 02
04
05
03
Identify the load resistance RL
and short the load terminal of the
circuit.
By utilizing any network
simplification technique, determine the
current flowing through the shorted
branch. This current will be the Norton’s
current, IN.
Now remove RL from the give
circuit and replace all the active
sources with their equivalent
internal resistance.
Further, evaluate the equivalent
resistance across the open ends of the
circuit. This resistance will be the
Norton’s equivalent resistance RN.
Now, draw Norton's equivalent
circuit, comprising of IN in parallel
combination with RN across the load
resistance RL. Then find the current
through the load resistance RL using:

4. Consider the circuit showm below:
Numerical Implementation of Norton’s Theorem
➠
IN
↻
i1 i2 i3
For mesh 1,
50 = 5 i1 + 5 (i1 - i2 )
50 = 10 i1 - 5 i2
For mesh 2,
0 = 10 i2 + 5 (i2 – i1 ) + 5 (i2 – i3 )
0 = 5i1 - 20 i2 - 5 i3
For mesh 3,
0 = 5 (i3 – i2 )
i2 = i3 = N = 2A

5. For calculating RN:
Numerical Implementation of Norton’s Theorem
➠ RN
RN = {( 5 II 5 ) + 10} II 5
RN = {(
5∗5
5+5
) + 10} II 5
RN = {( 2.5) + 10} II 5
RN = ( 12.5) II 5
RN = (
12.5∗5
12.5+5
)
RN = 3.5 Ω

6. Numerical Implementation of Norton’s Theorem
Norton's Equivalent Circuit:
IL
= NRN
RN+RL
=
2 ∗ 3.5
2 + 3.5
= 1.27A

7. THANK YOU
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