## Node Voltage Theorem

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# Node Voltage Theorem

Any junction o, two or more branches of a network is called a node. For application of node voltage theorem one of the nodes is taken as reference node or zero-potential node or datum node and the p.d. between each of the other nodes and the reference node is expressed in terms of an unknown voltage (symbolized as V

From circuit diagram for node A

I

15

I

8

I

6

Similarly for node B

I

4

I

15

Now applying Kirchhoff’s first law to nodes A and B we get

I

20-V

8 6

and I4 = I2+ I5

or

4 8 15

Multiplying expressions (vi) and (vii) by 120 and rearranging the terms we get

43 V

53 V

Solving expressions (viii) and (ix) we get

V

1,027 1,027

Substituting values of V

20 -

Current supplied by 20-V battery, I

15

Current through 8 Ω resistor, I

8 8

i.e. Current of 0.0389 A flowing from B to A

Current through 6- Ω resistor, I

6 6X1027

Current through 4- Ω resistor, I

4 4X1027

30-

Current supplied by 30-V resistor,I

15

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_{1}, V_{2 }or V_{A}, Va or V_{x}V_{y}etc.) Like Maxwell circulating current theorem node-voltage theorem reduces the number of equations to be solved to determine the unknown quantities. We shall now look at an example in which two unknown voltages V_{A }and Va will be evaluated and then unknown currents will be determined.**Q**. Find the currents in various branches of the network given below.**Solution :**Let the node C be taken as reference node and the potentials of nodes A and B with reference to node C be V_{A}and Va volts respectively Let the current direction be, as shown on the circuit diagram arbitrarily.From circuit diagram for node A

I

_{1}=__20-V___{A}15

I

_{2 }=__V___{A}-V_{B}8

I

_{2}=__V___{A}6

Similarly for node B

I

_{4}=__V___{B}4

I

_{5}=__30-V___{B}15

Now applying Kirchhoff’s first law to nodes A and B we get

I

_{1}= I_{2 }+ I_{3}20-V

_{A}= V_{A}__+___{ }-V_{B }__V___{A}8 6

and I4 = I2+ I5

or

__V__=_{B}__V__+_{A}-V_{B}__30-V___{B}4 8 15

Multiplying expressions (vi) and (vii) by 120 and rearranging the terms we get

43 V

_{A}-15_{B}=16053 V

_{B}-15 V_{A}=240Solving expressions (viii) and (ix) we get

V

_{A }=__6,040__voltage and V_{B }=__6,360__V1,027 1,027

Substituting values of V

_{A}and Vn in expressions (i), (ii), (iii) (iv) and (v) we get20 -

__6040__Current supplied by 20-V battery, I

_{1 }=__1027__= 0.9412 A15

__6040__-__6320__Current through 8 Ω resistor, I

_{2}=__V__=_{A}- V_{B}__1027 1027__= - 0.0389 A8 8

i.e. Current of 0.0389 A flowing from B to A

Current through 6- Ω resistor, I

_{3}=__V__=_{A}__6040__= 0.98 A6 6X1027

Current through 4- Ω resistor, I

_{4}=__V__=_{B}__6360__=1.5482 A4 4X1027

30-

__6360__Current supplied by 30-V resistor,I

_{5}=__30-V__=_{B}__1027__= 1.5871A15

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