Title: Norton's Theorem Slide 1: Introduction - Norton's Theorem: A powerful tool in circuit analysis. - It simplifies complex circuits to a current source and a parallel resistor. Slide 2: Thevenin vs. Norton - Comparison between Thevenin and Norton Theorems. - Thevenin uses voltage sources, Norton uses current sources. Slide 3: Norton's Theorem Components - Key components: Norton current source (I_N) and Norton equivalent resistance (R_N). Slide 4: Finding I_N - Steps to find I_N: Short circuit the load, calculate the current. - I_N = V_S / R_S, where V_S is the short-circuit voltage and R_S is the source resistance. Slide 5: Finding R_N - Steps to find R_N: Disable all independent sources, find equivalent resistance. Slide 6: Practical Application - Norton's Theorem simplifies complex circuits for analysis. - Used in various electronic applications and circuit design. Slide 7: Example Problem - Solve a simple circuit using Norton's Theorem step by step. Slide 8: Conclusion - Norton's Theorem is a valuable tool for circuit analysis. - Simplifies circuit analysis by replacing complex networks with a simple current source and resistor. Slide 9: Q&A - Open the floor for questions from the audience. Slide 10: Thank You - Conclude the presentation and thank the audience for their attention.

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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