This document discusses and provides examples of applying network theorems, including Thevenin's theorem, Norton's theorem, and the superposition theorem, to calculate currents in branches of circuits. These theorems allow complex networks to be reduced to simpler equivalent circuits. Thevenin's theorem replaces a network with a voltage source in series with a resistance, while Norton's theorem uses a current source in parallel with a resistance. The superposition theorem allows calculating the total current as the sum of the currents from individual voltage sources. Examples are provided to calculate a current using each theorem and nodal analysis, and the results are shown to be the same, demonstrating the utility of network theorems in circuit analysis.

1. International Journal of Electrical Engineering and Technology (IJEET), ISSN 0976 –
6545(Print), ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), © IAEME
101
CONCEPT & VERIFICATION OF NETWORK THEOREMS
Ashok Kumar
Associate Professor
Dr K.N.Modi Institute of Engineering & Technology,
Modinagar Ghaziabad.
ABSTRACT
This paper ensure the utility of network theorms specially Thevenin’s Norton’s and
superposition theorems. These theorems is possibly the most extensively used in network to find the
current in a required branch.
These theorems reduce a complex network as seen from two terminals into a simple circuit so
that when due another circuit (load) is connected at these terminals, Its response can be easily
determined
Key word: KVL.KCL,VDR.CDR, NETWORK THEOREMS
INTRODUCTION
The fundamental concept to some problem by any theorems, for this kirchoff’s current and
voltage Laws, should be known, in this paper I will discuss the problems solved by any one theorems
(Thevenin’s Norton’s and superposition theorems) used and verified into these theorems. This paper
will be more helpfull.
(A) Kirchoff’s current Law: (KCL) According to this law the algebraic sum of the currents meeting
at a junction is equal to zero ie
For example, Consider the fig For KCL, some Assumption should be made positive sign
towards at the node and Negative sing outwords from the node.
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2. International Journal of Electrical Engineering and Technology (IJEET), ISSN 0976 –
6545(Print), ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), © IAEME
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As per kCL, from fig 1
+ i1 + i2 + i3 – i4 – i5 – i6 = 0 --------------1
+ i1 + i2 + i3 = + i4 + i5 + i6 = 0
It is clear from equ 1, according to KCL total sum of incoming current will be equal to the sum of the
outgoing current at a junction.
(B) Kirchoff’s voltage Law (KVL): According to this law the total algebraic sum of induced emf in
a closed coop circuit will be equal to zero or equal to the product of I. R.
The advancement of KVL, gives mesh analysis for example.
Procedure of KVL of Mesh Analysis
1 Assume, the direction of currents is clock wise.
2 Consider, first sign of the battey at the heed of current arrow and positive sogn with resistance.
3 For common, branch with imaLoops, Assume Max current of considering Loop w.r.t to Another
Loop take substracted of current of the direction of loops are same otherwise Addition.
KVL Equation for coop
–V1 + I1R1 + R2 (I1 – I2) = 0
I1 (R1 + R2) – I2.R2 = V1 …1 From equ 1 and 2 can be solved very conveniently for I1, I1
KVL equ for Loop 2,
R2 (I2 – I1) + I2.R. + V2 = 0
–R2I1 + (R2 + R3)I2 = –V2
R2I1 – (R2 + R3)I2 = V2 …2
As per equation 1 and 2 Shows due procedure of KVL equation Formation.
(c) Voltage Divison Rule (VDR): This rule in used where more than one resistances connected in
series to calculate Individual voltage across each resistance for example
V1=VxR1/(R1+R2) and V2=VxR2/(R1+R2)
In this way VDR can be used

3. International Journal of Electrical Engineering and Technology (IJEET), ISSN 0976 –
6545(Print), ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), © IAEME
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(d) Current Division Rule (CDR): This rule will be used when more than resistance are connected in
parallel to calculate Individual branch current like
I1 = IX R2/(R1+R2) and I2 = IX R1/ (R1+R2)
THEVENIN’S THEORM
Statement: In any linear network an equivalent voltage source ie Vth is replaced with in series an
equivalent resistance ie. Rth after killing the source (Voltage source will be short circuited and
current source will be open circuited) living behind their internal resistances.
Finally, the following circuit diagram and formula will be used as per Thevenins’s theorems
-Vth + I (Rth + RL) = 0
I =
୚౪౞
ோ೟೓ାோಽ
(RC =RL)
Where I = Current in the Required branch
Vth = Thevenin’s equivatent voltage
Rth = thevenin’s equaivalent pesitance.
RL = Load Resistance, is the resistance, in which the current is to be calculate.
Steps to solve the problems by using thevenin’s Theoram
1. Open the terminal in which the current is to be found
2. 2 for Vth: Thevenin’s Equivalent voltage can be calculated by using VDR/CDR/KVL or any
other convenient method.
3. For equivalent resistance kill the source and look into the circule from open circuits side,
calculate the total resistance of the circuit by using simplification of series and parelled
system.
4. Reconnect the removal part and make the following circuit and used formula as per
Thevenins’ Theoram
NORTON’S THEOREM
Statement: In any linear bilateral circuilt an equivalent current source is replace ie. IN in parallel
with an equivalent Resistance ie. RN, after killing the source (Voltage source will be short and
current source will be open circuited) living behind their integral resistance.
Finally the following circuit diagram and formula will be used as per Norton’s Theorem.
Where IN = Norton’s current
RN = Norton’s/Thevenin’s equivalent resistance
RN= Load resistance.

4. International Journal of Electrical Engineering and Technology (IJEET), ISSN 0976 –
6545(Print), ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), © IAEME
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ISC = Short circuit current
In Short, we can say Norton’s Therom is reciprocal of Thevevin’s Theorems. As it clear
As per T.T
I =
௏೟೓
ሺோ೟೓ାோಽሻ
I=
ூೞ೎.ோே
ሺோಿାோಽሻ
Steps:
1 Short circiuted the terminals in which the current is be found.
2 Find ISC, For this, by using KVL/ CDR or Any other Convenient method.
3 For RN same as Rth
4 Finally, IN =
ூೞ೎.ோே
ሺோಿାோಽሻ
SUPERPOSITION THEOREM
Statement: In any linear network various source of emf acting simultaneously, the current in each
branch or current in the required branch, will be equal to the algebraic sum of the current in each
branch (or in required branch) due to Individual source of emf.
Steps:
1 Consider one source of emf (may be voltage source or may be current source). After killing the
other source of emf and calculate individual branch curent by using KVL/ CDR or any other
convenient method.
2 Consider, Another source of emf, after killing the previous source of emf and repeat step 1.
3 Take addition if direction of Currents is same, due to Individual source of emf otherwise take
subtraction.
For Example: Calculate current in branch AB by using Thevenin’s theorem and verified with
Norton’s and superposition Theorems.
By Thevenin’s Theorems

5. International Journal of Electrical Engineering and Technology (IJEET), ISSN 0976 –
6545(Print), ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), © IAEME
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1 Open the terminal AB,
By using KVL (Z assumes as ohms)
–150 + 100I + 50 = 0
100I = 100
I = 1 amp
Voc/Vth = V1 – I.R1 OR V2 + I.R2
Vth = 150 – 1 × 55 OR Vth = 50 + 1 × 45
= 95 volt OR Vth = 50 + 45
2 For Rth =
ହହൈସହ
ଵ଴଴
=
ଶଶହ
ଵ଴଴
= 24.75 ohms
I =
ଽହ
ଶଶ.ହାହ଴
Amp
ଽହ
଻ସ.଻ହ
= 1.27 Amp
By Norton’s Theorem
By using KCL at Node A
ISC = I1 + I2
ISC =
ଵହ଴
ହହ
+
ହ଴
ସହ
= 2.72 + 1.11
ISC = 3.83 Amp
RN/Rth =
ହହൈସହ
ଵ଴଴
= 24.75 ohms
IN =
ூೞ೎.ோಿ
ሺோಿାோ೎ሻ
ൌ
଼.ଶଽൈଶସ.଻ହ
ଶସ.଻ହା ହ଴
=1.27Amp
By Superposition Theorem
Case 1: Consider 150V, short circuited 50V
R = 55 +
ହ଴ൈସହ
ହ଴ା ସହ
= 78.68 ohms
I1` =
ଵହ଴
଻଼.଺଼
= 1.906 Amp
IAB` =
ଵ.ଽ଴଺ൈ ସହ
ଽହ
= 0.9028 Amp
I2 = =
ଵ.ଽ଴଺ൈହ଴
ଽହ
ൌ
ଽହ.ଷ
ଽହ
= 1.003 Amp

6. International Journal of Electrical Engineering and Technology (IJEET), ISSN 0976 –
6545(Print), ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), © IAEME
106
Case 2: Consider, 50V, Short circuited 150V
RT2 = Total Resistance
RT2 = 45 +
ହହൈହ଴
ଵ଴ହ
= 45 + 26.19 = 71.19 ohm
I2 =
ହ଴
଻ଵ.ଵଽ
= 0.702 Amp
I”AB =
଴.଻଴ଶൈହହ
ଵ଴ହ
=
ଷ଼.଺ଶ
ଵ଴ହ
= 0.367 Amp
IAB = I ‘AB + I “AB
IAB = 0.90 + 0.367 = 1.267 Amp
By Nodel Analysis
By Applying KCL at Node A
IAB = I1 + I2 … (A)
Where IAB =
௏஺
ହ଴
’ I1 =
ଵହ଴ି௏஺
ହହ
I2 =
ହ଴ି௏஺
ସହ
By Putting the Value of I1’ I2 & IAB in Equation I
௏஺
ହ଴
=
ଵହ଴ି௏ಲ
ହହ
൅
ହ଴ି ௏ಲ
ସହ
275 ×
௏ಲ
ହ଴
= 275 ቀ
ଵହ଴ ି ௏ಲ
ହହ
ቁ ൅ 275 ቀ
ଵହ଴ ି ௏ಲ
ସହ
ቁ
5.5 VA = 5(150 – VA) + 6.11(50 – VA)
5.5 VA = 750 – 5VA + 305.5 – 6.11VA
VA (5.5 + 5 + 6.11) = 1055.5
16.61 VA = 1055.5
VA =
ଵ଴ହହ.ହ
ଵ଺.଺ଵ
= 63.54 Volt
So, IAB =
௏ಲ
ହ଴
=
଺ଷ.ହସ
ହ଴
IAB = 1.27 Amp

7. International Journal of Electrical Engineering and Technology (IJEET), ISSN 0976 –
6545(Print), ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), © IAEME
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CONCLUSION
Network theorem is most useful to determine current in the required branch of the circuit has
more than one source of emf (May be voltage source, or current source or may be both)
By using Thevenin’s, Norton’s, Superposition Theorems and Nodal analysis, Can be solved
very conveniently and having the same answers.
REFERENCES
1. A. Chakarebarti: Circuite Theory Dhanpat Rai Publication Delhi
2. D.P Kothani I. J. Nagrath Basic Electrical engay Tats MC Gram Hill, New Delhi
3. B. L Theraja Electrical technology, S Chand & Co Ltd
4. V.K.Mehta “Basic Electrical Engineering: S Chand & Co Ltd.