When you finish this lesson, you should be able to:
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:
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).
Below is an example of a network and its transform.
Continuing with the same example, I can be
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
i(t) = 0.712cos(450t + 4.3°) A
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 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
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.
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.
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.
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.
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.
5. vs(t) = 100cos(2200t - 33°) volts, R = 22 Ω, L = 24 mH, C = 150 µF. Find v(t) (steady state).
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).