Worked examples: AC circuits by nodal analysis

WORKED EXAMPLES

AC CIRCUITS BY NODAL ANALYSIS

Example 13 Determine the real power output of the source in the circuit shown in figure, by nodal analysis and verify the results by using loop analysis.

Solution: Nodal Method

There are two nodes 1 and 2. Let the node 2 be the reference one. Then applying KCL at node 1 we get the following equation:

Loop Current Method :

Let the loop currents be as shown

I₁=10A....(i)

Applying KVL to loop 2, we get

It is the same answer as in the nodal method. Thus the solution is verified.

Example 14 In the network shown in figure, find the node voltages V₁ and V₂. Find also the current supplied by the source.

Solution: The above circuit is re-drawn as below after converting the voltage source into its equivalent current source.

Example 15 Using nodal analysis, find the current through the 4 ohm resistor in the circuit shown in figure.

Solution: The voltage source is converted into its equivalent current source and the network is drawn below:

V₁ = Δ₁/Δ = 9.06 ∠ 65.5° / 0.513 ∠ 22.3°

= 17.66 ∠Δ 87.8° volts

=  (0.677 + j17.6) volts

V₂ = Δ₂/Δ = 17.77 ∠ 39.3° /  0.513 ∠ 22.3°

= 34.64 ∠ 61.6° volts

= (16.47+ j30.47) volts

 Voltage across 4Ω = V₂ - V₁

= (16.47 + j30.47) - (0.677 + j17.6)

= (15.793 + j12.87)

= 20.37 ∠ 39.17° volts

 Current through 4 Ω = 20.37 ∠39.17o / 4

= 5.0925 ∠ 39.17°A

This current flows from node 2 to node 1.

Example 16 Given the nodes 1 and 2 in network of figure find the ratio of node voltage V₁/V₂

Solution: The network is re-drawn as below after converting the voltage source into current source:

Putting in the matrix form, we get

V₁ / V₂ =  0.2 V_g (0.35 – j0.2) /  0.05v_g

=(0.35 - j0.2) / 0.25

= (0.35 - j0.2) / 0.25

= 0.403 ∠ -30° / 0.25

= 1.612 ∠ -30°