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

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

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:

- Transformation of the system from time functions into
phasors and complex numbers. Notice that
*the resultant system is independent of time*. - 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.
- 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).

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

Below is an example of a network and its transform.

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

V_{I} - V = R_{1}(**I** -
I_{3})
+ (1/jωC)(I - I_{2})

V = (1/jωC)(I_{2}
- I) + jωL(I_{2}
- I_{3}) + I_{2}R_{2}

0 = I_{3}R_{3} + jωL(I_{3}
- I_{2}) + R_{1}(I_{3} - 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 + (V_{A} - V_{C})/R_{3}
+ (V_{A} - V_{B})/R_{1} = 0

Node B: (V_{B} - V_{A})/R_{1}
+ (V_{B} - V)/(1/jωC)
+ (V_{B} - V_{C})/(jωL) = 0

Node C: (V_{C} - V_{A})/R_{3}
+ (V_{C} - V_{B})/(jωL)
+ V_{C}/R_{2} = 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:

X_{L} = ωL

The units of X_{L} are ohms: radians/second × henries is ohms.

The reactance for a capacitor:

X_{C} = -1/ωC

Note the minus sign. X is real, not imaginary. You have to add the j
to make it imaginary. The units of X_{C} 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:

Z_{L} = jX_{L} = jωL

Z_{C} = jX_{C} = -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. V**s** = 72/55° V, I**s** = 5/-29° A, R = 17 Ω,
Z** _{L}** = j11 Ω, Z

3. V**s** = 7/33° V, I**s** = 4/-179° A, R1 = 175 Ω,
R2 = 80 Ω, R3 = 217 Ω, Z** _{L}** = j101 Ω, Z

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. v_{s}(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).