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# Nodal Analysis for AC Circuits

Nodal analysis for both DC and AC circuits is the same analysis technique. The only difference is you are now dealing with impedance in AC circuits rather than plain resistance in DC circuits. So if you are having problems using Nodal Analysis in DC circuits, then this technique remains a problem in AC circuits. This tutorial aims to be your complete guide in using Nodal Analysis for AC circuits.

### Introduction to Nodal Analysis

The Nodal Analysis technique is derived from Kirchoff’s Current Law (KCL). Recall that KCL tells us that the algebraic sum of currents leaving or entering a junction or node is zero. Algebraic here means we take the direction of the currents into account. A current entering a node is positive while a current leaving a node is negative.

$\sum I=0$

Consider the circuit below:

At the node labeled “V1”, the current I1 is entering the node while the currents I2 and I3 are leaving the node. Thus, by KCL:

$I_{1} - I_{2} - I_{3} = 0$

At the node labeled “V2”, the current I3 is entering the node while the currents I4 and I5 are leaving the node. Thus, again, by KCL:

$I_{3} - I_{4} - I_{5} = 0$

Now what if we want to know the voltages at nodes “V1” and “V2”. Can we do it using KCL alone? Of course we can, but that would take a lot of time!

Here lies the beauty of Nodal Analysis. With this technique, we can directly solve for the node voltages. Let’s see how.

### Representing Current as Function of Voltage

Ohm’s Law tells us that current is the ratio of voltage and resistance:

$I = \frac{V}{R}$

This formula, however, assumes that the reference for the voltage is zero volts or ground. Remember that voltage is the unit of potential difference. Of course, you can have references other than zero volts. The general formula for current from Ohm’s Law should be:

$I = \frac{V_{higher} - V_{lower}}{R}$

Also remember that (conventional) current always go from higher potential to lower potential:

And so going back to our original circuit:

At node 1, where the node currents are:

$I_{1} - I_{2} - I_{3}=0$

I1 can be substituted with

$I_{1} = \frac{V_{s} - V_{1}}{R_{1}}$

Here, it’s safe to assume that VS is the higher potential because it’s the only voltage source!

We can also determine that:

$I_{2} = \frac{V_{1} - 0}{R_{2}} = \frac{V_{1}}{R_{2}}$

Since the other potential is ground or zero volts.

And for I3:

$I_{3} = \frac{V_{1} - V_{2}}{R_{3}}$

Here, we assume that V1 is higher than V2 because V1 is closer to the voltage source.

Our first nodal equation is now:

$\frac{(V_{S} - V_{1})}{R_{1}} - \frac{V_{1}}{R_{2}} - \frac{(V_{1}-V_{2})}{R_{3}} = 0$

We could repeat the same for the second node, and acquire a second equation:

$\frac{(V_{1}-V_{2})}{R_{3}} - \frac{V_{2}}{R_{4}} - \frac{V_{2}}{R_{5}}=0$

We now have two equations with the node voltages as the two unknowns!

### Nodal Analysis for AC

AC circuits now deal with impedance rather than resistance. Recall that impedance is a complex number whose real part is resistance and imaginary part is reactance.

$Z = R + jX$

A resistor’s impedance does not have an imaginary component so its impedance is equal to its resistance.

$Z_{resistor} = R$

But AC circuits now have capacitors and inductors!

A pure capacitor’s impedance has no real component and only contains an imaginary part, known as capacitive reactance:

$X_{C} = \frac{1}{j2\pi fC}$

Where j is the imaginary number and f is the frequency of the source. Similarly, a pure inductor only has an inductive reactance:

$X_{L} = j2\pi fL$

Note that pure capacitors and inductors are only theoretical. Real capacitors and inductors have small resistances which give them impedances with real parts.

Now let’s give values to the resistors in the circuit above. Also, let’s replace some of the resistors with capacitors and inductors:

Using the formulas above, the impedances of the capacitors and inductor will now be:

$X_{C1} = \frac{1}{j2\pi(100)(160\mu F)}\approx -j10$

$X_{C2} = \frac{1}{j2\pi (100)(80\mu F)}\approx -j20$

$X_{L}=j2\pi(100)(32m)\approx j20$

Redrawing the circuit:

Applying the same technique we used with DC circuits, the equations will now become:

At node 1:

$\frac{(V_{S} - V_{1})}{10} - \frac{V_{1}}{-j20} - \frac{(V_{1}-V_{2})}{j20}=0$

At node 2:

$\frac{(V_{1} - V_{2})}{j20} - \frac{V_{2}}{20} - \frac{V_{2}}{-j10}=0$

The next challenge here is using the appropriate technique to solve for the unknowns V1 and V2. Cramer’s Rule is a good tool to used.

Students often have difficulties dealing with complex number equations especially those involving fractions. This is why I often suggest using admittance instead of impedance. An admittance is the reciprocal of impedance:

$Y=\frac{1}{Z}$

Just like impedance, admittance has a real component and an imaginary component.

$Y = G + jB$

The real component of admittance is called conductance (G) while the imaginary component is called susceptance (B).

For pure capacitors and inductors, only susceptance exists.

For the elements of our example AC circuit:

$B_{C1} = \frac{1}{X_{C1}} = \frac{1}{-j20} = j0.05$

$B_{C2} = \frac{1}{X_{C2}} = \frac{1}{-j10} = j0.1$

$B_{L} = \frac{1}{X_{L}} = \frac{1}{j20} = -j0.05$

For resistors, only conductance exists:

$G_{1} = \frac{1}{R_{1}}=\frac{1}{10}=0.1$

$G_{2} = \frac{1}{R_{2}}=\frac{1}{20}=0.05$

So if we are to write the equations above using admittance:

For node 1:

$G_{1}(V_{S}-V_{1})-B_{C1}(V_{1})-B_{L}(V_{1}-V_{2})=0$

$0.1(V_{S}-V_{1})-j0.05(V_{1})-(-j0.05)(V_{1}-V_{2})=0$

For node 2:

$B_{L}(V_{1}-V_{2})-G_{2}V_{2}-B_{C2}V_{2}=0$

$(-j0.05)(V_{1}-V_{2})-0.05V_{2}-j0.1V_{2}=0$

No more fractions!

### A Mnemonic for Nodal Analysis

We can use a general statement to easily recall nodal analysis for AC:

“Sum of admittances connected to the node multiplied by the node voltage, minus the sum of admittances common to this node and the next node multiplied by the next node voltage is equal to the sum of applied currents to this node”

Let’s see if this statement applies to our node equations. Let us look back to our AC circuit:

The first part is:

“Sum of admittances connected to the node multiplied by the node voltage…”

Assuming this is referring to the 1st node then:

$V_{1}(G_{1} + B_{C1} + B_{L})$

$V_{1}(0.1 + j0.05 -j0.05)$

The next statement,

“Minus the sum of admittances common to this node and the next node multiplied by the next node voltage…”

This should be:

$- V_{2}B_{L}$

Finally,

“…is equal to the sum of applied currents to this node”

Applied current here is the Norton equivalent current to the node. There is just one applied current in the first node and that is

$\frac{V_{S}}{R_{1}} = V_{S}G_{1}$

Thus,

$V_{1}(G_{1} + B_{C1} + B_{L}) - V_{2}B_{L} = V_{S}G_{1}$

This is the same as the first nodal equation, just rearranged!

We can also apply the same statement to the second node:

$V_{2}(B_{L} + G_{2} + B_{C2}) - V_{1}(B_{L}) = 0$

Again, this is the same equation we arrived on earlier.

### Conclusion

Hopefully this tutorial achieved its objective which is to make you understand nodal analysis for AC circuits. For questions, reactions and suggestions, please drop a comment below!

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