# Nodal Analysis for ac Circuits   Whatsapp  Before examining the application of the method to ac networks, the student should first review the appropriate sections on nodal analysis in Chapter 7 since the content of this section will be limited to the general conclusions of Chapter 7.
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 7 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. 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 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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