Nodal Analysis for ac Circuits

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Kirchhoff's current law serves as the foundation for nodal analysis of alternating current circuit steady-state conditions. The nodal and supernode for an alternating current circuit are identical to those for a direct current circuit, therefore we will not have any trouble here.
Before examining the application of the method to ac networks, the student should first review the appropriate sections on nodal analysis in Chapter 6 since the content of this section will be limited to the general conclusions of Chapter 6.
The fundamental steps are the following:
  • Determine the number of nodes within the network.
  • Pick a reference node and label each remaining node with a subscripted value of voltage: $V_1$, $V_2$, and so on.
  • Apply Kirchhoff's current law at each node except the reference. Assume that all unknown currents leave the node for each application of Kirchhoff's current law.
  • Solve the resulting equations for the nodal voltages
A few examples will refresh your memory about the content of Chapter 6 and the general approach to a nodal-analysis solution.
Example 1: find the voltage across the inductor for the network of [Fig. 1] using nodal analysis.
Fig. 1: Example 1.
Solution:
Steps 1 and 2 are indicated in [Fig. 1].
Fig. 2: Assigning the nodal voltages and subscripted impedances to the network of Fig. 1.
The network is redrawn in [Fig. 2] with subscripted impedances:
Step 3: application of Kirchhoff's current law to node $V_1$:
$$\sum I_i = \sum I_o$$
$$ 0 = I_1 + I_2 + I_3 $$
$${V_1 -E \over Z_1} + {V_1 \over Z_2 } + {V_1 - V_2 \over Z_3} = 0$$
Rearranging terms:
$$V_1[{1 \over Z_1} + { 1 \over Z_2} + { 1\over Z_3}] - V_2[{1 \over Z_3}] = {E_1 \over Z_1}$$
the application of Kirchhoff's current law to node $V_2$ gives,
$$ 0 = I_3 + I_4 + I $$
$${V_2 -V_1 \over Z_3} + {V_2 \over Z_4 }+ I = 0$$
Rearranging terms:
$$V_2[{1 \over Z_3} + { 1 \over Z_4}] - V_1[{1 \over Z_3}] = -I$$
Grouping equations:
\begin{align*} V_1[{1 \over Z_1} + { 1 \over Z_2} + { 1\over Z_3}] & - V_2[{1 \over Z_3}] &= {E_1 \over Z_1}\\ V_1[{1 \over Z_3}] & -V_2[{1 \over Z_3} + { 1 \over Z_4}] &= I\\ \end{align*}
The equations are ready to put the values and solve for $V_1$ using Cramer's Rule.
Example 2: Write the nodal equations for the network of [Fig. 3] having a dependent current source.
Fig. 3: Example 2.
Solution:
Steps 1 and 2 are indicated in [Fig. 3].
Fig. 4: Assigning the nodal voltages and subscripted impedances to the network of Fig. 1.
The network is redrawn in [Fig. 4] with subscripted impedances:
Step 3: application of Kirchhoff's current law to node $V_1$:
$$\sum I_i = \sum I_o$$
$$ 0 = I_1 + I_2 + I_3 $$
$${V_1 -E \over Z_1} + {V_1 \over Z_2 } + {V_1 - V_2 \over Z_3} = 0$$
Rearranging terms:
$$V_1[{1 \over Z_1} + { 1 \over Z_2} + { 1\over Z_3}] - V_2[{1 \over Z_3}] = {E_1 \over Z_1}$$
the application of Kirchhoff's current law to node $V_2$ gives,
$$ 0 = I_3 + I_4 + I $$
$${V_2 -V_1 \over Z_3} + {V_2 \over Z_4 }+ I = 0$$
Rearranging terms:
$$V_2[{1 \over Z_3} + { 1 \over Z_4}] - V_1[{1 \over Z_3}] = -I$$
Grouping equations:
\begin{align*} V_1[{1 \over Z_1} + { 1 \over Z_2} + { 1\over Z_3}] & - V_2[{1 \over Z_3}] &= {E_1 \over Z_1}\\ V_1[{1 \over Z_3}] & -V_2[{1 \over Z_3} + { 1 \over Z_4}] &= I\\ \end{align*}
The equations are ready to put the values and solve for $V_1$ using Cramer's Rule.
Example 3: Apply nodal analysis to the network of [Fig. 5]. Determine the voltage $V_L$.
Fig. 5: Example 3.
Solution: In this case there is no need for a source conversion. The network is redrawn in [Fig. 6] with the chosen nodal voltage and subscripted impedances.
Fig. 6: Assigning the nodal voltage and subscripted impedances for the network of Fig. 5.
Apply the format approach:
$$ Y_1 = { 1 \over Z_1} = { 1 \over 4 kΩ} = 0.25 mS$$
$$ Y_2 = { 1 \over Z_2} = { 1 \over 1 kΩ} = 1 mS$$
$$ Y_3 = { 1 \over Z_3} = { 1 \over 2 kΩ \angle 90^\circ} = -j 0.5 mS$$
At node $V_1$
$$ (Y_1 + Y_2 + Y_3 ) V_1 = -100 I$$
and
$$ \begin{split} V_1 &= { -100 I \over Y_1 + Y_2 + Y_3 }\\ &= { -100 I \over 0.25 mS + 1 mS -j 0.5 mS}\\ &= { -100 I \over 1.25 mS -j 0.5 mS}\\ &= { -100 I \over (1.25 S -j 0.5 S)\times 10^{-3}}\\ &= { -100 \times 10^{3} I \over 1.3463 \angle -21.80^\circ}\\ &= -74.28 \times 10^{3} I \angle 21.80^\circ\\ &= -74.28 \times 10^{3} ({V_i \over 1 kΩ}) \angle 21.80^\circ\\ V_1 &= V_L = -(74.28 V_i) V \angle 21.80^\circ\\ \end {split} $$

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