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# Concept & verification of network theorems

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### Concept & verification of network theorems

1. 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. INTERNATIONAL JOURNAL OF ELECTRICAL ENGINEERING & TECHNOLOGY (IJEET) ISSN 0976 – 6545(Print) ISSN 0976 – 6553(Online) Volume 4, Issue 4, July-August (2013), pp. 101-107 © IAEME: www.iaeme.com/ijeet.asp Journal Impact Factor (2013): 5.5028 (Calculated by GISI) www.jifactor.com IJEET © I A E M E
2. 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 102 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. 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 103 (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. 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 104 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. 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 105 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. 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. 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 107 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.