Question

Use nodal analysis to determine the voltage across 5 resistance and the current in the 12 V source.

Answer 1

Transform the 12-volt and 4-ohm resistor into current-source and parallel resistor. 

Mark the nodes O. A. B and C on the diagram. Salf-and mutual conductance terms are to be wirtten down next.

 At A, G_aa = 1/4 + 1/2 + 1/4 = 1 mho 

At B, G_bb = 1/2 + 1/5 + 1/100 = 0.71 mho 

At C, G_cc = 1/4 + 1/5 + 1/20 = 0/50 mho 

Between A and B, G_ab=0.5 mho, 

Between B and C, G_be=0.2 mho, 

Between A and C, G_ac= 0.25mho.

Current Source matrix: At node A, 3 amp incoming and 9 amp outgoing currents give a net outgoing current of 6 amp. At node C, incoming current =9 amp. At node B, no current source is connected. Hence, the current -source matrix is: [-6, 0, 9]

The potentials of three nodes to be found are ; VA, VB, VC

V_A = Δ_a/Δ =+0/6075/0.095625 =6.353 volts 

V_B = Δ_b/Δ = 1.125/0.095625 =11.765 volts 

V_c = Δ_c/Δ = 2.475/0.95625 = 25.882 volts 

Hence, voltage across 5-ohm resistor = V_c - V_B =14.18 volts. Obviously. B is positive w.r. to A. From these node potentials, current through 100-ohm resistor is 0.118 amp; (i) current through 20 ohm resistor is 1.294 amp. 

(ii) Current through 5-ohm resistor = 14.18/5 = 2.836 amp. 

(iii) Current through 4-ohm resistor between C and A = 19.53/4 =4.883 amp 

Check: Apply KCL at node C 

Incoming current = 9 amp, from the source. 

Outgoing currents as calculated in (i), (ii) and (iii) above = 1.294 + 2.836 + 4.883 ≅ 9 amp 

(iv) Current through 2-ohm resistor =(V_B - V_A)/2 = 2.706 amp, from B to A. 

(v) Current in A-O branch = 6.353/4 = 1.588 amp

In Fig. the transformed equivalent circuit is shown. The 3-amp current source (0 to A) and the current of 1.588 amp in A-O branch have to be interpreted with reference to the actual circuit, shown in Fig., where in a node D exists at a potential of 12 volts w.r. to the reference node. The 4-ohm resistor between D and A carries an upward current of {(12 - 6.353)/4 =} 1.412 amp, which is nothing but 3 amp into the node and 1.588 amp away from the node, as in Fig. , at node A. The current in the 12-V source is thus 1.412 amp. 

Check. KCL at node A should give a check that incoming currents should add-up to 9 amp. 

1.412 + 2.706 + 4.883 ≅ 9 amp