ECE 220 Network Analysis I

Lesson 11. AC Circuits


Objectives

When you finish this lesson, you should be able to:

  1. Use Kirchhoff's Laws for phasors.
  2. Write phasor mesh current and node voltage equations and solve them.
  3. Combine complex impedances in series and parallel.
  4. Find complete solutions of problems in the time domain.

Introduction

The problems in this unit are rather special. They all have sinusoidal forcing functions. They all involve only the steady state solution. This may seem rather restrictive, but a vast number of problems in electricity fall into this category. Almost all problems in AC circuits do. The power of the phasor method lies in the fact that when a sinusoidal forcing function is used in a linear system, the results (voltage and current) are always sinusoidal, and of the same frequency as the forcing function. Thus, if a 120 V, 60 Hz voltage is used to drive a system of resistors, inductors and capacitors, the steady-state (what happens after waiting a long time) current in the system will be a 60 Hz sinusoid. Its amplitude and phase angle are the only unknowns that must be calculated.

Click here here to see how an AC generator works.

Using the phasor method to get a complete solution to a problem requires 3 steps:

  1. Transformation of the system from time functions into phasors and complex numbers. Notice that the resultant system is independent of time.
  2. Solution of the problem using complex numbers. Kirchhoff's Laws, node voltage method, mesh current method, and combining of impedances in series and parallel may all be useful in doing this.
  3. Transformation of the system back into the time domain.

This method may at first seem long and tedious, but it is far easier than solving the original time domain problem using differential equations.

The mesh current and node voltage methods can be used for AC circuits by using the same methods as with DC circuits, with the addition of impedance concepts and phasor voltages and currents.

Try this site for a good explanation of phasors:  http://people.clarkson.edu/~svoboda/eta/phasors/Phasor10.html (created by Dr. James A. Svoboda of Clarkson University).

Network Transformation

  1. Transform sources into phasors using only magnitude and phase angle; e.g. 100cos(200t + 3) V transforms to 100/3 V.
  2. Transform inductors by converting L to jωL. The units of jωL are ohms.
  3. Transform capacitors by converting C to 1/jωC. The units of 1/jωC are ohms.
  4. Resistors do not require transformation.
  5. Transform unknown functions of time (variables) by writing them as general phasors (using boldface or overline); e.g. v2(t) transforms to V2.

Below is an example of a network and its transform.

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You should see a picture here.

Continuing with the same example, I can be calculated as,
I = 85/27/(-j88.9 + 110 + j135) = 85/27/(110 + j46.1) = 85/27/119.3/22.7 = 0.712/4.3 A

An inverse transform can then be done on I to get:
i(t) = 0.712cos(450t + 4.3) A

Mesh Current Example

You should see a picture here.

VI - V = R1(I - I3) + (1/jωC)(I - I2)

V = (1/jωC)(I2 - I) + jωL(I2 - I3) + I2R2

0 = I3R3 + jωL(I3 - I2) + R1(I3 - I)

This is 3 equations and 3 unknowns. The first equation is not required unless it is necessary to find the voltage across the current source. If only the mesh currents are needed, equations 2 and 3 are sufficient.

Node Voltage Example

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Node A: -I + (VA - VC)/R3 + (VA - VB)/R1 = 0

Node B: (VB - VA)/R1 + (VB - V)/(1/jωC) + (VB - VC)/(jωL) = 0

Node C: (VC - VA)/R3 + (VC - VB)/(jωL) + VC/R2 = 0

Impedance

The complex version of resistance is called impedance.  It is a combination of resistance and reactance.  Resistance is the real part of the impedance, and reactance is the magnitude of the imaginary part.  It's expressed this way:

Z = R + jX

R = resistance, the real part of impedance
X = reactance, the imaginary part, without the j.

The reactance for an inductor:

XL = ωL

The units of XL are ohms:  radians/second × henries is ohms.

The reactance for a capacitor:

XC = -1/ωC

Note the minus sign.  X is real, not imaginary.  You have to add the j to make it imaginary.  The units of XC are ohms:  radians/second × farads is inverse ohms.  Capacitors always have negative reactance and negative impedance.

To convert reactance into equivalent impedance, just add the j:

ZL = jXL = jωL
ZC = jXC = -j/ωC

Ohm's Law has a complex version:

V = I Z

It's the same a Ohm's Law for resistors, except that all the numbers are complex.  The analysis for series/parallel combinations of impedances is the same as it is for resistances.


Before going on to the homework, you should complete Tutorial 11 on phasors.


Homework Problems

1. v(t) = 240sin(400t - 52) volts, i(t) = 8cos(400t + 110) amps, R = 100 Ω, L = 220 mH, C = 12 µF.  Transform this network into the frequency domain.

You should see a picture here.

2. Vs = 72/55 V, Is = 5/-29 A, R = 17 Ω, ZL = j11 Ω, ZC = -j6 Ω.  Write the mesh current equations for this network. Write other equations as needed to have enough to solve the circuit. Do not solve.

You should see a picture here.

3. Vs = 7/33 V, Is = 4/-179 A, R1 = 175 Ω, R2 = 80 Ω, R3 = 217 Ω, ZL = j101 Ω, ZC = -j73 Ω.  Write the node voltage equations for this network. Do not solve.

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4. R = 15 Ω, L = 24 mH, C = 9 µF.  Find the impedance between a and b at a frequency of 750 Hz. Give your answer in polar coordinates.

You should see a picture here.

5. vs(t) = 100cos(2200t - 33) volts, R = 22 Ω, L = 24 mH, C = 150 µF.  Find v(t) (steady state).

You should see a picture here.

Bonus (no partial credit).  This problem will test your ability to apply a concept we learned earlier (superposition) to the solution of a phasor problem with two sinusoidal sources.  It is crucial to note that the two sources have different frequencies.  Find v(t) (steady state).

You should see a picture here.