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 V₁, V₂ 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

I₁ =  20-V_A
         15
I₂  = V_A-V_B
           8
I₂  = V_A
         6

Similarly for node B

I₄ = V_B
       4
I₅  = 30-V_B
         15

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

I₁  = I₂ + I₃
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  =160
53 V_B-15 V_A =240

Solving expressions (viii) and (ix) we get

V_A = 6,040  voltage and V_B = 6,360  V
        1,027                            1,027

Substituting values of V_A and Vn in expressions (i), (ii), (iii) (iv) and (v) we get
                                                    20 - 6040
Current supplied by 20-V battery, I₁  =   1027  = 0.9412 A
                                                            15
                                                                 6040 - 6320
Current through 8 Ω resistor, I₂ = V_A - V_B = 1027   1027  = - 0.0389 A
                                                      8                8

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

Current through 6- Ω resistor, I₃= V_A = 6040  = 0.98 A
                                                  6     6X1027

Current through 4- Ω  resistor, I₄ = V_B = 6360  =1.5482 A
                                                   4    4X1027
                                                                    30- 6360
Current supplied by 30-V resistor,I₅ = 30-V_B  =      1027 = 1.5871A
                                                         15

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