The course web site is: http://users.encs.concordia.ca/~trueman/web_page_273.htm

ELEC273 Lecture Notes Set 10 Phasors and Impedance

Homework on complex numbers.

Final Exam: Friday December 15, 2017 from 9:00 to 12:00 (confirmed)

What is AC Circuit Analysis?

• AC circuit analysis uses complex numbers to find the forced response to a sinusoidal generator.

• AC circuit analysis is very similar to DC circuit analysis that we studied earlier in the course. DC Circuit Analysis AC Circuit Analysis

Constant voltage V and current I Sinusoidal voltage 𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃Sinusoidal current 𝑖𝑖(𝑡𝑡) = 𝐵𝐵 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙

Real numbers for V and I Complex numbers called phasors𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃 and 𝐼𝐼 = 𝐵𝐵𝑒𝑒𝑗𝑗𝜙𝜙

Real number resistances R Complex number impedances𝑍𝑍𝑅𝑅 = 𝑅𝑅, 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔, and 𝑍𝑍𝐶𝐶 = 1

𝑗𝑗𝜔𝜔𝐶𝐶

Real number node matrix or mesh matrix Complex number node matrix or mesh matrix

Given: • the frequency 𝑓𝑓 Hz, hence 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 rad/sec• The amplitude of the source, 10 voltsFind:• The amplitude 𝐴𝐴 and the phase 𝜃𝜃 of the output

voltage 𝑣𝑣 𝑡𝑡 at “steady state” after the transients have died out.

From Alexander and Sadiku.

10 cos𝜔𝜔𝑡𝑡+𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos(𝜔𝜔𝑡𝑡 + 𝜃𝜃)−

𝑅𝑅1𝑅𝑅2𝐶𝐶

𝜔𝜔

PhasorGiven an AC voltage 𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 , define the phasor representing voltage 𝑣𝑣(𝑡𝑡) as the complex number𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

where 𝑗𝑗 = −1and 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 is the frequency in radians per second. The magnitude of the phasor 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃 is the amplitude of the cosine 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 .The angle of the phasor, 𝜃𝜃, is the phase angle of the cosine 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 .

It is sometimes convenient to visualize the phasor by drawing it on the complex plane.

The phasor 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃 is an arrow of length A making an angle of 𝜃𝜃to the real axis.

It can be quite useful to draw phasors on the complex plane!

We can use angle notation:𝑉𝑉 = 𝐴𝐴∠𝜃𝜃Or exponential notation𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

Imaginary

Real

𝑉𝑉𝐴𝐴

𝜃𝜃

𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

Rotating PhasorThe phasor representing AC voltage 𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 is the complex number 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃 .

It is sometimes convenient to visualize the relationship between two phasors by defining a rotating phasor as

𝑣𝑣𝑐𝑐 𝑡𝑡 = 𝑉𝑉𝑒𝑒𝑗𝑗𝜔𝜔𝜔𝜔

Hence𝑣𝑣𝑐𝑐 𝑡𝑡 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃𝑒𝑒𝑗𝑗𝜔𝜔𝜔𝜔 = 𝐴𝐴𝑒𝑒𝑗𝑗(𝜔𝜔𝜔𝜔+𝜃𝜃)

The AC voltage we started with is the real part of the rotating phasor.𝑣𝑣 𝑡𝑡 = Re 𝑣𝑣𝑐𝑐 𝑡𝑡 = Re 𝐴𝐴𝑒𝑒𝑗𝑗 𝜔𝜔𝜔𝜔+𝜃𝜃 = Re 𝐴𝐴cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 + j𝐴𝐴sin 𝜔𝜔𝑡𝑡 + 𝜃𝜃 = 𝐴𝐴cos(𝜔𝜔𝑡𝑡 + 𝜃𝜃)

Draw the rotating phasor on the complex plane at t=0.

At 𝑡𝑡 = 0, 𝑣𝑣𝑐𝑐 0 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃𝑒𝑒𝑗𝑗𝜔𝜔𝑥𝑥𝑥 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

where the “c” subscript on 𝑣𝑣𝑐𝑐 𝑡𝑡 denotes a complex version of voltage 𝑣𝑣 𝑡𝑡 .

Imaginary

Real

𝐴𝐴

𝜃𝜃

𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃At 𝑡𝑡 = 0𝑣𝑣𝑐𝑐 0 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

Draw the rotating phasor on the complex plane as time advances:

𝐴𝐴𝑒𝑒𝑗𝑗(𝜔𝜔𝜔𝜔+𝜃𝜃)

Draw the rotating phasor at time 𝑡𝑡 > 0

𝑣𝑣𝑐𝑐 𝑡𝑡 = 𝐴𝐴𝑒𝑒𝑗𝑗(𝜔𝜔𝜔𝜔+𝜃𝜃)

This amounts to adding angle 𝜔𝜔𝑡𝑡 to the angle of phasor 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃.

As time increases, 𝜔𝜔𝑡𝑡 increases. We add a full 2𝜋𝜋radians of angle in one AC period 𝑇𝑇 = 1

𝑓𝑓.

The phasor rotates by one full rotation for each T seconds of time.

Imaginary

Real

𝐴𝐴

𝜃𝜃

For 𝑡𝑡 > 0𝑣𝑣𝑐𝑐 𝑡𝑡 = 𝐴𝐴𝑒𝑒𝑗𝑗(𝜔𝜔𝜔𝜔+𝜃𝜃)

𝜔𝜔𝑡𝑡

𝜔𝜔𝑡𝑡 + 𝜃𝜃

Phasor• The phasor representing sinusoidal voltage 𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 is the complex number 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

• The amplitude A of the cosine is the magnitude of the phasor: 𝐴𝐴 = 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

• The phase angle 𝜃𝜃 of the cosine is the angle of the complex number “phasor”.

• We can “recover” the cosine from the phasor by multiplying the phasor by 𝑒𝑒𝑗𝑗𝜔𝜔𝜔𝜔 and then using Euler’s Identity: 𝑒𝑒𝑗𝑗𝜙𝜙 = cos𝜙𝜙 + 𝑗𝑗 sin𝜙𝜙

𝑣𝑣 𝑡𝑡 = Re 𝑉𝑉𝑒𝑒𝑗𝑗𝜔𝜔𝜔𝜔 = Re 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃𝑒𝑒𝑗𝑗𝜔𝜔𝜔𝜔 = Re 𝐴𝐴𝑒𝑒𝑗𝑗(𝜔𝜔𝜔𝜔+𝜃𝜃)

𝑣𝑣 𝑡𝑡 = Re 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 + 𝑗𝑗 sin 𝜔𝜔𝑡𝑡 + 𝜃𝜃𝑣𝑣 𝑡𝑡 = Re 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 + 𝑗𝑗 𝐴𝐴sin 𝜔𝜔𝑡𝑡 + 𝜃𝜃𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃

• To recover the cosine from the phasor:• The amplitude A of the cosine is the magnitude of the phasor V • The phase angle 𝜃𝜃 of the cosine is the angle of the complex number

Examples of Phasors1.Find the phasor for 𝑣𝑣1 𝑡𝑡 = 10 cos 𝜔𝜔𝑡𝑡 + 𝜋𝜋

4Answer: 𝑉𝑉1 = 10𝑒𝑒𝑗𝑗

𝜋𝜋4

2.Find the cosine represented by phasor 𝑉𝑉2 = 5𝑒𝑒−𝑗𝑗𝜋𝜋6

Answer: 𝑣𝑣2 𝑡𝑡 = 5 cos 𝜔𝜔𝑡𝑡 − 𝜋𝜋6

3.Find the sum: 𝑣𝑣3 𝑡𝑡 = 𝑣𝑣1 𝑡𝑡 + 𝑣𝑣2 𝑡𝑡Using trig identities is the hard way: 𝑣𝑣3 𝑡𝑡 = 10 cos 𝜔𝜔𝑡𝑡 +

𝜋𝜋4

+ 5 cos 𝜔𝜔𝑡𝑡 −𝜋𝜋6

Try it!The easy way: 𝑉𝑉3 = 𝑉𝑉1 + 𝑉𝑉2 = 10𝑒𝑒𝑗𝑗

𝜋𝜋4 + 5𝑒𝑒−𝑗𝑗

𝜋𝜋6 (see the next slide)

Evaluate the sum 𝑉𝑉3 = 𝑉𝑉1 + 𝑉𝑉2𝑉𝑉3 = 10𝑒𝑒𝑗𝑗

𝜋𝜋4 + 5𝑒𝑒−𝑗𝑗

𝜋𝜋6

Showing ALL the steps:𝑉𝑉3 = 10 cos

𝜋𝜋4

+ 𝑗𝑗𝑗𝑗 sin𝜋𝜋4

+ 5 cos−𝜋𝜋6

+ 𝑗𝑗𝑗 sin−𝜋𝜋6

𝑉𝑉3 = 7.071 + 𝑗𝑗𝑗.071 + 4.330 − 𝑗𝑗2.5𝑉𝑉3 = 11.401 + 𝑗𝑗4.571Magnitude 11.4012 + 4.5712 = 12.283Angle tan𝜃𝜃 = 4.571

11.4𝑥1= 0.4049 so 𝜃𝜃 = 21.85 deg or 0.3814 rad

The phasor is𝑉𝑉3 = 12.283𝑒𝑒𝑗𝑗𝑥.3814

so the time function is𝑣𝑣3 𝑡𝑡 = 12.283 cos 𝜔𝜔𝑡𝑡 + 0.3814

Check the result by sketching the phasors on the complex plane:

𝑉𝑉1 = 10𝑒𝑒𝑗𝑗𝜋𝜋4 = 10 angle 45 degrees = 7.071+j7.071

𝑉𝑉2 = 5𝑒𝑒−𝑗𝑗𝜋𝜋6= 5 angle -30 degrees=4.330-j2.500

𝑉𝑉3 = 𝑉𝑉1 + 𝑉𝑉2

𝑉𝑉3 = 12.283𝑒𝑒𝑗𝑗21.85°= 12.283 angle 21.85 degrees

Drawing the phasors tip to tail in the complex plane shows that 12.283 angle 21.85 degrees is a reasonable answer.

ImIm

Re

Re

𝑉𝑉1 = 10𝑒𝑒𝑗𝑗𝜋𝜋4

𝑉𝑉2 = 5𝑒𝑒−𝑗𝑗𝜋𝜋6

45°

30°

10

5

𝑉𝑉1 = 10𝑒𝑒𝑗𝑗𝜋𝜋4

𝑉𝑉2 = 5𝑒𝑒−𝑗𝑗𝜋𝜋6

𝑉𝑉3 = 𝑉𝑉1 + 𝑉𝑉2

Impedance: the ratio of voltage phasor to current phasor.

• Consider a component in a circuit driven by a sinusoidal generator cos𝜔𝜔𝑡𝑡

• The voltage 𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 is represented by phasor 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

• The current 𝑖𝑖 𝑡𝑡 = 𝐵𝐵 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙 is represented by phasor 𝐼𝐼 = 𝐵𝐵𝑒𝑒𝑗𝑗𝜙𝜙

• The impedance is defined as

𝑍𝑍 = 𝑉𝑉𝐼𝐼

ohms

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡)+𝑉𝑉−

𝐼𝐼

The Impedance is a Complex Number

• 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

• 𝐼𝐼 = 𝐵𝐵𝑒𝑒𝑗𝑗𝜙𝜙

• The impedance is a complex number:

𝑍𝑍 =𝑉𝑉𝐼𝐼 =

𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

𝐵𝐵𝑒𝑒𝑗𝑗𝜙𝜙=𝐴𝐴𝐵𝐵 𝑒𝑒

𝑗𝑗(𝜃𝜃−𝜙𝜙) =𝐴𝐴𝐵𝐵 cos 𝜃𝜃 − 𝜙𝜙 + 𝑗𝑗

𝐴𝐴𝐵𝐵 sin 𝜃𝜃 − 𝜙𝜙 = 𝑅𝑅 + 𝑗𝑗𝑗𝑗

𝑍𝑍 = 𝑅𝑅 + 𝑗𝑗𝑗𝑗

R=the resistance is the real part of the impedanceX=the reactance is the imaginary part of the impedance.

+𝑉𝑉−

𝐼𝐼

Admittance: the reciprocal of the impedanceThe admittance is the ratio of the current phasor to the voltage phasor:𝑌𝑌 = 𝐼𝐼

𝑉𝑉Siemens

Note that𝑌𝑌 = 1

𝑍𝑍

The admittance is a complex number:𝑌𝑌 = 𝐺𝐺 + 𝑗𝑗𝐵𝐵

G=the conductance is the real part of the admittanceB=the susceptance is the imaginary part of the admittance.

What is the impedance of a resistor?

𝑣𝑣 𝑡𝑡 = 𝑅𝑅𝑖𝑖 𝑡𝑡𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 volts then the current is

𝑖𝑖 𝑡𝑡 =𝑣𝑣(𝑡𝑡)𝑅𝑅 =

𝐴𝐴𝑅𝑅 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃

with amplitude 𝐴𝐴𝑅𝑅

and phase 𝜃𝜃The voltage phasor is 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

The current phasor is 𝐼𝐼 = 𝐴𝐴𝑅𝑅𝑒𝑒𝑗𝑗𝜃𝜃; magnitude 𝐴𝐴

𝑅𝑅and angle 𝜃𝜃

The impedance is 𝑍𝑍 = 𝑉𝑉𝐼𝐼

= 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃𝐴𝐴𝑅𝑅𝑒𝑒

𝑗𝑗𝜃𝜃= 𝑅𝑅 ohms.

The admittance is 𝑌𝑌 = 1𝑍𝑍

= 1𝑅𝑅

= 𝐺𝐺 Siemens

Time domain Frequency Domain also called the “phasor domain”

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡) +𝑉𝑉−

𝐼𝐼

𝑍𝑍 = 𝑅𝑅𝑅𝑅

What is the impedance of an inductor?𝑣𝑣 𝑡𝑡 = 𝜔𝜔

𝑑𝑑𝑑𝑑𝑡𝑡𝑖𝑖 𝑡𝑡

If 𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 volts then we can find the current using𝑑𝑑𝑖𝑖𝑑𝑑𝜔𝜔

= 1𝐿𝐿𝑣𝑣 𝑡𝑡 so 𝑑𝑑𝑖𝑖 = 1

𝐿𝐿𝑣𝑣 𝑡𝑡 𝑑𝑑𝑡𝑡 and by integrating both sides

𝑖𝑖 =1𝜔𝜔�𝑣𝑣 𝑡𝑡 𝑑𝑑𝑡𝑡 =

1𝜔𝜔�𝐴𝐴 cos(𝜔𝜔𝑡𝑡 + 𝜃𝜃)𝑑𝑑𝑡𝑡 =

𝐴𝐴𝜔𝜔𝜔𝜔

sin(𝜔𝜔𝑡𝑡 + 𝜃𝜃)

To write the phasor we need to change sine to cosine.Use the trig identity:sin𝛼𝛼 = co𝑠𝑠 𝛼𝛼 −

𝜋𝜋2

With α = 𝜔𝜔𝑡𝑡 + 𝜃𝜃, we have

𝑖𝑖 =𝐴𝐴𝜔𝜔𝜔𝜔 sin(𝜔𝜔𝑡𝑡 + 𝜃𝜃) =

𝐴𝐴𝜔𝜔𝜔𝜔 cos(𝜔𝜔𝑡𝑡 + 𝜃𝜃 −

𝜋𝜋2)

Which has amplitude 𝐴𝐴𝜔𝜔𝐿𝐿

and phase 𝜃𝜃 − 𝜋𝜋2

, so the current phasor is

𝐼𝐼 =𝐴𝐴𝜔𝜔𝜔𝜔 𝑒𝑒

𝑗𝑗 𝜃𝜃−𝜋𝜋2

The impedance is

𝑍𝑍 =𝑉𝑉𝐼𝐼 =

𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

𝐴𝐴𝜔𝜔𝜔𝜔 𝑒𝑒

𝑗𝑗(𝜃𝜃−𝜋𝜋2)= 𝜔𝜔𝜔𝜔𝑒𝑒𝑗𝑗

𝜋𝜋2

Time domain

Frequency domain

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡)

+𝑉𝑉−

𝐼𝐼

𝑍𝑍 = 𝑗𝑗𝜔𝜔𝜔𝜔

𝜔𝜔

Impedance of an inductor, continued;𝑍𝑍 = 𝜔𝜔𝜔𝜔𝑒𝑒𝑗𝑗

𝜋𝜋2

We can see from the Argand diagram that 𝑒𝑒𝑗𝑗𝜋𝜋2 = 𝑗𝑗.

Another way to prove this is with Euler’s Identity:𝑒𝑒𝑗𝑗

𝜋𝜋2 = cos

𝜋𝜋2

+ 𝑗𝑗 sin𝜋𝜋2

= 0 + 𝑗𝑗𝑗 = 𝑗𝑗

Hence the impedance of an inductor is

𝑍𝑍 = 𝑗𝑗𝜔𝜔𝜔𝜔 ohms

The admittance of an inductor is

𝑌𝑌 = 1𝑍𝑍

= 1𝑗𝑗𝜔𝜔𝐿𝐿

SiemensArgand diagram

Re

Im

1 𝜋𝜋2

𝑗𝑗 = 1𝑒𝑒𝑗𝑗𝜋𝜋2

Time domain

Frequency domain

+𝑉𝑉−

𝐼𝐼

𝑍𝑍 = 𝑗𝑗𝜔𝜔𝜔𝜔

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡)

𝜔𝜔

What is the impedance of a capacitor?𝑖𝑖(𝑡𝑡) = 𝐶𝐶

𝑑𝑑𝑑𝑑𝑡𝑡𝑣𝑣 𝑡𝑡

If 𝑣𝑣 𝑡𝑡 = 𝐴𝐴 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃 volts then we can calculate the current as

𝑖𝑖 = 𝐶𝐶𝑑𝑑𝑑𝑑𝑡𝑡𝐴𝐴 cos(𝜔𝜔𝑡𝑡 + 𝜃𝜃) = −ω𝐶𝐶𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜃𝜃)

To write the phasor we need to change sine to cosine using the trig identity:sin𝛼𝛼 = −co𝑠𝑠 𝛼𝛼 + 𝜋𝜋

2With α = 𝜔𝜔𝑡𝑡 + 𝜃𝜃, we have𝑖𝑖 = −ω𝐶𝐶𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜃𝜃) = ω𝐶𝐶Acos(𝜔𝜔𝑡𝑡 + 𝜃𝜃 +

𝜋𝜋2)

which has amplitude ω𝐶𝐶𝐴𝐴 and phase 𝜃𝜃 + 𝜋𝜋2

, so the current phasor is

𝐼𝐼 = ω𝐶𝐶𝐴𝐴𝑒𝑒𝑗𝑗 𝜃𝜃+𝜋𝜋2

The impedance is

𝑍𝑍 =𝑉𝑉𝐼𝐼 =

𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃

ω𝐶𝐶𝐴𝐴𝑒𝑒𝑗𝑗 𝜃𝜃+𝜋𝜋2=

1

ω𝐶𝐶𝑒𝑒𝑗𝑗𝜋𝜋2

Since 𝑒𝑒𝑗𝑗𝜋𝜋2 = 𝑗𝑗 we can write the impedance of a capacitor as

𝑍𝑍 = 1𝑗𝑗𝜔𝜔𝐶𝐶

ohms.

The admittance of a capacitor is 𝑌𝑌 = 1𝑍𝑍

= 𝑗𝑗𝜔𝜔𝐶𝐶 Siemens

Frequency domain

+𝑉𝑉−

𝐼𝐼

𝑍𝑍 =1

𝑗𝑗𝜔𝜔𝐶𝐶

Time domain

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡)

𝐶𝐶

Component Time Domain Frequency Domain

Resistor

Inductor

Capacitor

Impedance

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡)

𝑅𝑅

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡)

𝐶𝐶

+𝑣𝑣(𝑡𝑡)−

𝑖𝑖(𝑡𝑡)

𝜔𝜔

𝑣𝑣 = 𝑅𝑅𝑖𝑖

𝑣𝑣 = 𝜔𝜔𝑑𝑑𝑖𝑖𝑑𝑑𝑡𝑡

𝑖𝑖 = 𝐶𝐶𝑑𝑑𝑣𝑣𝑑𝑑𝑡𝑡

𝑉𝑉 = 𝑅𝑅𝐼𝐼

𝑉𝑉 = 𝑗𝑗𝜔𝜔𝜔𝜔𝐼𝐼

𝑉𝑉 =1𝑗𝑗𝜔𝜔𝐶𝐶

𝐼𝐼

Combining ImpedancesImpedances in series:

𝑉𝑉 = 𝑉𝑉1 + 𝑉𝑉2 = 𝑍𝑍1𝐼𝐼 + 𝑍𝑍2𝐼𝐼=(𝑍𝑍1+𝑍𝑍2)𝐼𝐼So the equivalent impedance is

𝑍𝑍 =𝑉𝑉𝐼𝐼

=(𝑍𝑍1+𝑍𝑍2)𝐼𝐼

𝐼𝐼= 𝑍𝑍1 + 𝑍𝑍2

Series impedances add.

Impedances in parallel:

𝐼𝐼 = 𝐼𝐼1 + 𝐼𝐼2 =𝑉𝑉𝑍𝑍1

+𝑉𝑉𝑍𝑍2

=1𝑍𝑍1

+1𝑍𝑍2

𝑉𝑉

The equivalent impedance is

𝑍𝑍 =𝑉𝑉𝐼𝐼 =

𝑉𝑉1𝑍𝑍1

+ 1𝑍𝑍2

𝑉𝑉=

𝑍𝑍1𝑍𝑍2𝑍𝑍1 + 𝑍𝑍2

Parallel impedances combine using product over sum.

+

𝑉𝑉

−

+𝑉𝑉2−

𝐼𝐼

+𝑉𝑉1−

𝐼𝐼2𝐼𝐼1

+

𝑉𝑉

−

𝐼𝐼

𝑍𝑍2𝑍𝑍1

𝑍𝑍2

𝑍𝑍1

Combining Admittances: 𝐼𝐼 = 𝑌𝑌𝑉𝑉 so 𝑉𝑉 = 𝐼𝐼𝑌𝑌

Admittances in series:

𝑉𝑉 = 𝑉𝑉1 + 𝑉𝑉2 =𝐼𝐼𝑌𝑌1

+𝐼𝐼𝑌𝑌2

=1𝑌𝑌1

+1𝑌𝑌2

𝐼𝐼

So the equivalent admittance is

𝑌𝑌 =𝐼𝐼𝑉𝑉

=I

𝐼𝐼𝑌𝑌1

+ 𝐼𝐼𝑌𝑌2

𝐼𝐼=

𝑌𝑌1𝑌𝑌2𝑌𝑌1 + 𝑌𝑌2

Series admittances combine like parallel resistors: product over sum.

Admittances in parallel:𝐼𝐼 = 𝐼𝐼1 + 𝐼𝐼2 = 𝑌𝑌1𝑉𝑉 + 𝑌𝑌2𝑉𝑉 = (𝑌𝑌1 + 𝑌𝑌2)𝑉𝑉The equivalent admittance is

𝑌𝑌 =𝐼𝐼𝑉𝑉

=(𝑌𝑌1 + 𝑌𝑌2)𝑉𝑉

𝑉𝑉= 𝑌𝑌1 + 𝑌𝑌2

Parallel admittances add.

+

𝑉𝑉

−

+𝑉𝑉2−

𝐼𝐼

+𝑉𝑉1−

𝐼𝐼2𝐼𝐼1

+

𝑉𝑉

−

𝐼𝐼

𝑌𝑌2𝑌𝑌1

𝑌𝑌2

𝑌𝑌1

Series-Parallel Combinations Find the input impedance 𝑍𝑍𝑖𝑖𝑖𝑖 at an operating frequency of 60 Hz.The radian frequency is 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 2𝜋𝜋𝜋𝜋𝜋𝑗 = 𝑗2𝑗𝜋𝜋 = 376.99 rad.sec

Change the components to impedances:

𝑍𝑍𝐿𝐿1 = 𝑗𝑗𝜔𝜔𝜔𝜔1 = 𝑗𝑗𝜋𝜋𝑗2𝑗𝜋𝜋𝜋𝜋2.𝜋𝑗𝜋𝜋10−3 = 𝑗𝑗𝑗

𝑍𝑍𝑅𝑅1 = 𝑅𝑅1 = 2

𝑍𝑍𝑅𝑅2 = 𝑅𝑅2 = 2

𝑍𝑍𝑅𝑅3 = 𝑅𝑅3 = 3

𝑍𝑍𝐿𝐿2 = 𝑗𝑗𝜔𝜔𝜔𝜔2 = 𝑗𝑗𝜋𝜋𝑗2𝑗𝜋𝜋𝜋𝜋𝑗.3𝑗𝜋𝜋10−3 = 𝑗𝑗2

𝑍𝑍𝐶𝐶 =1𝑗𝑗𝜔𝜔𝐶𝐶 =

1𝑗𝑗𝜋𝜋120𝜋𝜋𝜋𝜋1326𝜋𝜋10−6 =

2𝑗𝑗 =

2𝑗𝑗−𝑗𝑗−𝑗𝑗 = −𝑗𝑗2

Method:1.Change the components into impedances.2.Combine impedances in series and in parallel.

𝑍𝑍𝑖𝑖𝑖𝑖

𝑅𝑅1 = 2 Ω

𝑅𝑅2 = 2 Ω 𝑅𝑅3 = 3 Ω𝜔𝜔1 = 2.65 mH

𝜔𝜔2 = 5.30 mH

𝐶𝐶 = 1,326 µF

𝑅𝑅1 = 2 Ω

𝑅𝑅2 = 2 Ω

𝑅𝑅3 = 3 Ω𝜔𝜔1 = 2.65 mH

𝜔𝜔2 = 5.30 mH

𝐶𝐶 = 1,326 µF

Find the input impedance, continued:

𝑍𝑍1is j2 ohms in parallel with 3 ohms:

𝑍𝑍1 =𝑗𝑗2𝜋𝜋3𝑗𝑗2 + 3 =

𝑗𝑗𝜋3 + 𝑗𝑗2

3 − 2𝑗𝑗3 − 𝑗𝑗2 =

𝑗𝑗𝑗𝑗 + 123𝜋𝜋3 + 2𝜋𝜋2 =

12 + 𝑗𝑗𝑗𝑗13

𝑍𝑍2is -j2 ohms in parallel with 2 ohms:

𝑍𝑍2 =−𝑗𝑗2𝜋𝜋2−𝑗𝑗2 + 2 =

−𝑗𝑗𝑗2 − 𝑗𝑗2

2 + 2𝑗𝑗2 + 𝑗𝑗2 =

−𝑗𝑗𝑗 + 82𝜋𝜋2 + 2𝜋𝜋2 =

8 − 𝑗𝑗𝑗8 = 1 − 𝑗𝑗

𝑍𝑍𝑖𝑖𝑖𝑖2 Ω

2 Ω 3 Ω𝑗𝑗2 Ωj1 Ω

−𝑗𝑗2 Ω

𝑍𝑍1 𝑗𝑗2 Ω 3 Ω

𝑍𝑍2

2 Ω

−𝑗𝑗2 Ω

𝑍𝑍𝑖𝑖𝑖𝑖2 Ω

j1 Ω 𝑍𝑍1𝑍𝑍2

Combine 𝑍𝑍1 and 𝑍𝑍2 in series:

𝑍𝑍3 = 𝑍𝑍1 + 𝑍𝑍2𝑍𝑍𝑖𝑖𝑖𝑖2 Ω

j1 Ω 𝑍𝑍1𝑍𝑍2 𝑍𝑍1

𝑍𝑍2

𝑍𝑍1 =12 + 𝑗𝑗𝑗𝑗

13𝑍𝑍2 = 1 − 𝑗𝑗

𝑍𝑍3 = 𝑍𝑍1 + 𝑍𝑍2 =12 + 𝑗𝑗𝑗𝑗

13 + (1 − 𝑗𝑗) =12 + 𝑗𝑗𝑗𝑗

13 +13 − 𝑗𝑗𝑗3

13𝑍𝑍3 =

25 + 𝑗𝑗513

Find the input impedance, continued:Define 𝑍𝑍4as j1 ohms in parallel with 𝑍𝑍3:

𝑍𝑍4 =𝑗𝑗𝜋𝜋 25 + 𝑗𝑗𝑗

13𝑗𝑗 + 25 + 𝑗𝑗𝑗

13

=−5 + 𝑗𝑗2𝑗25 + 𝑗𝑗𝑗𝑗

25 − 𝑗𝑗𝑗𝑗25 − 𝑗𝑗𝑗𝑗

=−125 + 𝑗𝑗𝑗𝑗 + 𝑗𝑗𝜋2𝑗 + 450

2𝑗𝜋𝜋2𝑗 + 𝑗𝑗𝜋𝜋𝑗𝑗=

325 − 𝑗𝑗𝑗𝑗𝑗949

𝑍𝑍𝑖𝑖𝑖𝑖 is 2 ohms in series with 𝑍𝑍4:

𝑍𝑍𝑖𝑖𝑖𝑖 = 2 + 𝑍𝑍4 = 2 +325 − 𝑗𝑗𝑗𝑗𝑗

949 =1989 + 325 − 𝑗𝑗𝑗𝑗𝑗

949=

2223 − 𝑗𝑗𝑗𝑗𝑗949

𝑍𝑍𝑖𝑖𝑖𝑖 = 2.342 − 𝑗𝑗𝑗.7534 Ω

This is our final answer!

𝑍𝑍3𝑍𝑍𝑖𝑖𝑖𝑖2 Ω

j1 Ω

𝑍𝑍4𝑍𝑍𝑖𝑖𝑖𝑖2 Ω

Spice with a Sinusoidal Generator

To calculate the input impedance with LTSpice:• Drive the circuit with a 1 volt generator at 60 Hz• Find the current in 𝑅𝑅1 flowing into the circuit 𝐼𝐼𝑖𝑖𝑖𝑖• Calculate the input impedance as

𝑍𝑍𝑖𝑖𝑖𝑖 = 1𝐼𝐼𝑖𝑖𝑖𝑖

Construct the circuit.

How do we specify a sinusoidal excitation?

𝑍𝑍𝑖𝑖𝑖𝑖2 Ω

2 Ω 3 Ω𝑗𝑗2 Ωj1 Ω

−𝑗𝑗2 Ω

To specify that the generator is an AC source, right-click on the generator circle to pop up a menu:

Specify AC 1, meaning that the generator is a sinusoidal or “AC” source of amplitude 1 volt.

Choose AC Analysis and give the frequency: Click on Stimulate and then Edit Stimulation Card.

Choose AC Analysis:

• Specify the number of points per octave as 1.• Specify both the starting and the stopping frequency as 60 Hz (we only want one frequency).

We are ready to run the simulation:

The system adds the “dot command” .ac oct 1 60 60.Click on the “run” button to solve the circuit.

Spice calculates the amplitude and phase of the voltages and currents:

The current in R1 is reported as I(R1)=0.4064 angle -17.8 degrees

Find the input impedance with LTSpice:

The input current is the current in 𝑅𝑅1, so𝐼𝐼𝑖𝑖𝑖𝑖 =I(R1)=0.4064 angle -17.8 degrees

𝑍𝑍𝑖𝑖𝑖𝑖 =1𝐼𝐼𝑖𝑖𝑖𝑖

=1

0.4064∠ − 17.8° = 2.342 + 𝑗𝑗𝑗.7522

which agrees with our calculation.

We calculated𝑍𝑍𝑖𝑖𝑖𝑖 = 2.342 − 𝑗𝑗𝑗.7534

To calculate the input impedance with LTSpice:• Drive the circuit with a 1 volt generator at 60 Hz• Find the current in 𝑅𝑅1 flowing into the circuit 𝐼𝐼𝑖𝑖𝑖𝑖• Calculate the input impedance as

𝑍𝑍𝑖𝑖𝑖𝑖 = 1𝐼𝐼𝑖𝑖𝑖𝑖

𝑍𝑍𝑖𝑖𝑖𝑖2 Ω

2 Ω 3 Ω𝑗𝑗2 Ωj1 Ω

−𝑗𝑗2 Ω

The course web site is: http://users.encs.concordia.ca/~trueman/web_page_273.htm

ELEC273 Lecture Notes Set 10 Phasors and Impedance

Homework on complex numbers.

Final Exam: Friday December 15, 2017 from 9:00 to 12:00 (confirmed)

Input Impedance Example (from a final exam)

Find the value of capacitance C such that the input impedance 𝑍𝑍𝑖𝑖𝑖𝑖 is real. The frequency is 𝜔𝜔 = 4 rad/sec.Solution

Convert the circuit elements to impedances at 𝜔𝜔 = 4 rad/sec : 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 and 𝑍𝑍𝐶𝐶 = 1𝑗𝑗𝜔𝜔𝐶𝐶

.

𝑍𝑍𝑖𝑖𝑖𝑖 is 1 ohm in parallel with 𝑍𝑍2𝑍𝑍𝑖𝑖𝑖𝑖=1||𝑍𝑍2 = 𝑍𝑍2

1+𝑍𝑍2If 𝑍𝑍2 is real then 𝑍𝑍𝑖𝑖𝑖𝑖 is real.So choose C to make 𝑍𝑍2 real.

Define 𝑍𝑍2 by omitting the 1-ohm resistor:

1 Ω1 Ω

C1 H

1 H

𝑍𝑍𝑖𝑖𝑖𝑖

𝑍𝑍𝑖𝑖𝑖𝑖 1 Ω𝑗𝑗𝑗 Ω

𝑗𝑗𝑗 Ω1 Ω

1𝑗𝑗𝑗𝐶𝐶

𝑍𝑍2𝑗𝑗𝑗 Ω

1𝑗𝑗𝑗𝐶𝐶

1 Ω𝑗𝑗𝑗 Ω

4114

141441

12

jCj

jY

jZjZ

++

+=+=+=

Choose the value of C to make 𝑍𝑍2 real.

1)41(44142 ++

++=

jCjjjZ

CjCjjZ

4)161(4142 +−

++=

CjCjCjCjZ

4)161(41)4)161((4

2 +−+++−

=

Define 𝑍𝑍1 to be 1𝑗𝑗4𝐶𝐶

in parallel with 1 + 𝑗𝑗𝑗 .

Since parallel admittances add,

𝑌𝑌1 = 𝑗𝑗𝑗𝐶𝐶 +1

1 + 𝑗𝑗𝑗The input impedance 𝑍𝑍2 is j4 in series with 𝑍𝑍1

𝑍𝑍2𝑗𝑗𝑗 Ω

1𝑗𝑗𝑗𝐶𝐶

1 Ω𝑗𝑗𝑗 Ω

𝑍𝑍2𝑗𝑗𝑗 Ω

𝑍𝑍1

CjCCjCZ

4)161()4644()161(

2 +−+−+−

=

−−−−

+−−+−

=CjCCjC

CjCCjCZ

4)161(4)161(

4)161()648()161(

2

22

2

2 16)161())161(4)161)(648((4)648()161(

CCCCCCjCCCZ

+−−−−−+−+−

=

To make 𝑍𝑍2 real, set the imaginary part to zero:

0))161(4)161)(648(( =−−−− CCCC

0)4648)(161( =−−− CCCThere is a common factor of (1 − 16C) so factor it out:

0)688)(161( =−− CC

0)161( =− C 0)688( =− C1176.0

172

688

===C0625.0161==C F

and

F, and

Solution of AC Circuits Using Phasors

Change to phasors and impedance.

To solve an AC circuit:1.Find the impedance of each component in the circuit.2.Draw the circuit with phasors and impedances.3.Use mesh analysis or node analysis to solve the circuit, using complex arithmetic.4.The amplitude of the voltage is the magnitude of the phasor and the phase angle of the voltage is the angle of the phasor.

Find the amplitude and phase of the voltage across the resistor, 𝑣𝑣𝑅𝑅. The frequency is 400 Hz.

Find the impedances:𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 2𝜋𝜋𝜋𝜋𝑗𝑗𝑗 = 2,513 radians/sec𝑍𝑍𝑅𝑅 = 𝑅𝑅 = 5 ohms𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝜋𝜋2𝑗𝑗3𝜋𝜋𝑗.𝑗𝑗𝑗𝑗𝜋𝜋10−3 = 𝑗𝑗2 ohms

+𝑣𝑣𝑅𝑅−

𝑅𝑅 = 5 Ω

𝜔𝜔 = 07958 mH

10 cos𝜔𝜔𝑡𝑡+𝑉𝑉𝑅𝑅−

5 Ω

𝑗𝑗2 Ω

10

Solve the circuit using complex arithmetic:

Node analysis: 10 − 𝑉𝑉𝑅𝑅𝑗𝑗2 −

𝑉𝑉𝑅𝑅5 = 0

Multiply by j10:50 − 5𝑉𝑉𝑅𝑅 − 𝑗𝑗2𝑉𝑉𝑅𝑅 = 05 + 𝑗𝑗2 𝑉𝑉𝑅𝑅 = 50

𝑉𝑉𝑅𝑅 =50

5 + 𝑗𝑗25 − 𝑗𝑗25 − 𝑗𝑗2 =

450 − 𝑗𝑗𝑗𝑗𝑗29 = 8.621 − 𝑗𝑗3.448

Magnitude 8.6212 + (−3.448)2=9.284Angle tan−1 −3.448

8.621=-21.8 degrees

Hence𝑉𝑉𝑅𝑅 = 9.284∠ − 21.8°

Mesh analysis: 10 − 𝑗𝑗2𝐼𝐼 − 𝑗𝐼𝐼 = 0𝐼𝐼 = 1𝑥

5+𝑗𝑗25−𝑗𝑗25−𝑗𝑗2

= 5𝑥−𝑗𝑗2𝑥29

= 1.724 − 𝑗𝑗𝑗.6896 amps

𝑉𝑉𝑅𝑅 = 𝑗𝐼𝐼 = 5 1.724 − j0.6896 = 8.621 − j3.448Same as by node analysis!𝑉𝑉𝑅𝑅 = 9.284∠ − 21.8° volts

Amplitude = 𝑉𝑉𝑅𝑅 =9.284 volts

Phase = the angle of 𝑉𝑉𝑅𝑅 which is -21.8 degrees.

𝑣𝑣𝑅𝑅 𝑡𝑡 = 9.284 cos 𝜔𝜔𝑡𝑡 − 21.8°

+𝑉𝑉𝑅𝑅−

5 Ω

𝑗𝑗2 Ω

10+𝑉𝑉𝑅𝑅−

5 Ω

𝑗𝑗2 Ω

10

10 𝑉𝑉𝑅𝑅

𝐼𝐼

Example: Solve a Circuit Using Phasors

• Solve this circuit to find the amplitude and phase of the load voltage 𝑉𝑉 at 𝑓𝑓 =2 GHz.• The generator in this circuit is 10 cos𝜔𝜔𝑡𝑡 volts at frequency 𝑓𝑓 =2 GHz so the radian frequency is 𝜔𝜔 =

2𝜋𝜋𝑓𝑓 = 1.2𝑗𝜋𝜋𝜋𝜋101𝑥 radians/second.

To solve an AC circuit:1.Find the impedance of each component in the circuit.2.Draw the circuit with phasors and impedances.3.Use mesh analysis or node analysis to solve the circuit, using complex arithmetic.4.The amplitude of the voltage is the magnitude of the phasor and the phase angle of the voltage is the angle of the phasor.

73 Ω

2.1 nH

11.8 pF

50 Ω

10 cos𝜔𝜔𝑡𝑡+𝑉𝑉−

73 Ω

2.1 nH

11.8 pF

50 Ω

10 cos𝜔𝜔𝑡𝑡+𝑉𝑉− 73 Ω

𝑗𝑗2𝜋.39 Ω

−𝑗𝑗𝜋.744 Ω

50 Ω

10∠𝑗°+𝑉𝑉−

10∠𝑗°+𝑉𝑉−

50 + 𝑗𝑗2𝜋.39 Ω

0.6178 − 𝑗𝑗𝜋.687 Ω

Homework: solve this circuit yourself without looking at the lecture notes.

10∠𝑗°+𝑉𝑉−

50 + 𝑗𝑗2𝜋.39 Ω

0.6178 − 𝑗𝑗𝜋.687 Ω

𝐼𝐼 𝐼𝐼

Power into a Resistor in AC Circuits

If a circuit is driven by an A.C. voltage, then how much power flows?𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃𝑖𝑖 𝑡𝑡 = 𝐼𝐼𝑚𝑚 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙

Instantaneous power: 𝑝𝑝 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 𝑖𝑖(𝑡𝑡)

Average power: 𝑃𝑃𝑎𝑎𝑎𝑎 = 1𝑇𝑇 ∫𝑥

𝑇𝑇 𝑝𝑝 𝑡𝑡 𝑑𝑑𝑡𝑡

Is there a convenient way to calculate the power directly from the phasors V and I?

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 + 𝜃𝜃

𝑖𝑖(𝑡𝑡) = 𝐼𝐼𝑚𝑚 cos 𝜔𝜔𝑡𝑡 + 𝜙𝜙

𝑖𝑖(𝑡𝑡)+𝑣𝑣(𝑡𝑡)−

Phasor 𝑉𝑉 = 𝑉𝑉𝑚𝑚∠𝜃𝜃

Phasor 𝐼𝐼 = 𝐼𝐼𝑚𝑚∠𝜙𝜙

𝐼𝐼+𝑉𝑉−

Instantaneous and Average Power Delivered to a Resistor

The AC voltage 𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 is connected across a resistor R.Find the instantaneous power and the average power delivered to the resistor.

The instantaneous power is defined as𝑝𝑝 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 𝑖𝑖 𝑡𝑡

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡

𝑖𝑖 𝑡𝑡 =𝑉𝑉𝑚𝑚𝑅𝑅 cos 𝜔𝜔𝑡𝑡

The instantaneous power is:

𝑝𝑝 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡𝑉𝑉𝑚𝑚𝑅𝑅 cos 𝜔𝜔𝑡𝑡 =

𝑉𝑉𝑚𝑚2

𝑅𝑅 cos2 𝜔𝜔𝑡𝑡

Trig identity: cos2𝜃𝜃 = 12

1 + cos(2𝜃𝜃)

𝑝𝑝(𝑡𝑡) =𝑉𝑉𝑚𝑚2

2𝑅𝑅 1 + cos2𝜔𝜔𝑡𝑡

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 𝑖𝑖(𝑡𝑡) = 𝐼𝐼𝑚𝑚 cos 𝜔𝜔𝑡𝑡𝑅𝑅

Time t

Instantaneous Power 𝑝𝑝(𝑡𝑡)𝑉𝑉𝑚𝑚2

𝑅𝑅

𝑉𝑉𝑚𝑚2

2𝑅𝑅

𝑇𝑇2

𝑇𝑇2

Average Power Delivered to a Resistor

The instantaneous power is: 𝑝𝑝 𝑡𝑡 = 𝑣𝑣 𝑡𝑡 𝑖𝑖 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 𝑉𝑉𝑚𝑚𝑅𝑅

cos 𝜔𝜔𝑡𝑡 = 𝑉𝑉𝑚𝑚2

𝑅𝑅cos2 𝜔𝜔𝑡𝑡 = 1

2𝑉𝑉𝑚𝑚2

𝑅𝑅1 + cos2𝜔𝜔𝑡𝑡

The average power delivered to the resistor is defined as the average of p(t) over one AC cycle of length T:

𝑃𝑃𝑎𝑎𝑎𝑎 =1𝑇𝑇�𝑥

𝑇𝑇𝑝𝑝 𝑡𝑡 𝑑𝑑𝑡𝑡

so

𝑃𝑃𝑎𝑎𝑎𝑎 =1𝑇𝑇�𝑥

𝑇𝑇 𝑉𝑉𝑚𝑚2

2𝑅𝑅 1 + cos2𝜔𝜔𝑡𝑡 𝑑𝑑𝑡𝑡 =1𝑇𝑇�𝑥

𝑇𝑇 𝑉𝑉𝑚𝑚2

2𝑅𝑅 𝑑𝑑𝑡𝑡 +1𝑇𝑇�𝑥

𝑇𝑇 𝑉𝑉𝑚𝑚2

2𝑅𝑅 cos2𝜔𝜔𝑡𝑡 𝑑𝑑𝑡𝑡 =𝑉𝑉𝑚𝑚2

2𝑅𝑅 + 0 =𝑉𝑉𝑚𝑚2

2𝑅𝑅

𝑃𝑃𝑎𝑎𝑎𝑎 = 𝑉𝑉𝑚𝑚2

2𝑅𝑅where 𝑉𝑉𝑚𝑚 is the amplitude of the AC voltage.

This is often written as

𝑃𝑃𝑎𝑎𝑎𝑎 =𝑉𝑉𝑚𝑚2

2𝑅𝑅

=𝑉𝑉𝑚𝑚2

2

𝑅𝑅= 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅

2

𝑅𝑅where 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑉𝑉𝑚𝑚

2is the “RMS value” of the AC voltage.

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡

𝑖𝑖 𝑡𝑡 =𝑉𝑉𝑚𝑚𝑅𝑅

cos 𝜔𝜔𝑡𝑡

What is meant by “RMS value”?

𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 𝑖𝑖(𝑡𝑡) = 𝐼𝐼𝑚𝑚 cos 𝜔𝜔𝑡𝑡𝑅𝑅

Amplitude and RMS ValueSuppose an AC voltage is given by𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡Then the amplitude of the voltage is 𝑉𝑉𝑚𝑚.

The “root mean square” value or “RMS” value of a periodic function 𝑣𝑣(𝑡𝑡) with period 𝑇𝑇 is defined as

𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟 =1𝑇𝑇�𝑥

𝑇𝑇𝑣𝑣2 𝑡𝑡 𝑑𝑑𝑡𝑡

This is the average value or “mean value” of the square of the voltage function.

For a sinusoid 𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 we can evaluate the RMS value of 𝑣𝑣(𝑡𝑡) as

𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟2 = 1

𝑇𝑇 ∫𝑥𝑇𝑇 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡 2𝑑𝑑𝑡𝑡 = 𝑉𝑉𝑚𝑚2

𝑇𝑇 ∫𝑥𝑇𝑇 12

1 + cos 2𝜔𝜔𝑡𝑡 𝑑𝑑𝑡𝑡=𝑉𝑉𝑚𝑚2

𝑇𝑇 ∫𝑥𝑇𝑇 12𝑑𝑑𝑡𝑡 + 𝑉𝑉𝑚𝑚2

𝑇𝑇 ∫𝑥𝑇𝑇 12

cos 2𝜔𝜔𝑡𝑡 𝑑𝑑𝑡𝑡 = 𝑉𝑉𝑚𝑚2

𝑇𝑇𝑇𝑇2

= 𝑉𝑉𝑚𝑚2

2

𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟 =𝑉𝑉𝑚𝑚

2

Thus the average power delivered to a resistor is

𝑃𝑃𝑎𝑎𝑎𝑎 = 𝑉𝑉𝑚𝑚2

2𝑅𝑅=

𝑉𝑉𝑚𝑚2

2

𝑅𝑅= 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅

2

𝑅𝑅

Phasors Relative to RMS ValueWe can write phasors “relative to amplitude” or we can write phasors “relative to RMS value”.𝑣𝑣 𝑡𝑡 = 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡

Phasors relative to amplitude: 𝑉𝑉 = 𝑉𝑉𝑚𝑚𝑒𝑒𝑗𝑗𝜃𝜃The magnitude of the phasor 𝑉𝑉 = 𝑉𝑉𝑚𝑚 is the amplitude of the cosine voltage 𝑉𝑉𝑚𝑚 cos 𝜔𝜔𝑡𝑡

Phasors relative to RMS value:𝑉𝑉 = 𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟𝑒𝑒𝑗𝑗𝜃𝜃

The magnitude of the phasor 𝑉𝑉 = 𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟 is the RMS value of the cosine voltage, so 𝑉𝑉𝑚𝑚 = 2𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟 and the cosine is 𝑣𝑣 𝑡𝑡 = 2𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟 cos 𝜔𝜔𝑡𝑡

Which should you use?

In the lecture notes I use phasors relative to amplitude.

However, sometimes phasors relative to RMS value is preferred.

Everybody knows that the AC line voltage is 110 volts.

Is this amplitude or RMS value?

Answer: in the power industry the custom is to use phasors relative to RMS value so 110 volts is the RMS value.

The amplitude is 𝑉𝑉𝑚𝑚 = 2𝑉𝑉𝑟𝑟𝑚𝑚𝑟𝑟 = 1.𝑗𝑗𝑗𝜋𝜋𝑗𝑗𝑗 = 155.5 volts.

Most “DVMs” or “digital voltmeters” read RMS values on the AC voltage setting.

Power to an Impedance?

We will return to the topic of power in AC circuits later in the course.

The course web site is: http://users.encs.concordia.ca/~trueman/web_page_273.htm

ELEC273 Lecture Notes Set 10 Phasors and Impedance

Final Exam: Friday December 15, 2017 from 9:00 to 12:00 (confirmed)

Mesh Equations in the “Phasor Domain”1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign mesh currents 𝐼𝐼1, 𝐼𝐼2, 𝐼𝐼3, …3. Write a KVL equation for each mesh path and an equation for each current source.

Example #1: really easy!!!!!

The operating frequency is 60 Hz.The sources are𝑣𝑣1 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡𝑣𝑣2 𝑡𝑡 = 16 sin𝜔𝜔𝑡𝑡Write mesh equations for this circuit.Solve the equations to find the values of the mesh currents.

+𝑣𝑣1(𝑡𝑡)−

+𝑣𝑣2(𝑡𝑡)−

2 Ω5 Ω

1 Ω

663.1 µF

5.3 mH 10.6 mH

Step 1: Convert all the components to impedances at the operating frequency. Convert the sources to phasors.

The operating frequency is 𝑓𝑓 =60 Hz. The radian frequency is 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 2𝜋𝜋𝜋𝜋𝜋𝑗 = 𝑗2𝑗𝜋𝜋 = 376.99 ≈ 377 radians/second. For the 5.3 mH inductance, 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝜋𝜋3𝑗𝑗𝜋𝜋𝑗.3𝜋𝜋10−3 = 𝑗𝑗𝑗.9981 ≈ 𝑗𝑗2 ohmsThe 10.6 mH inductance has an impedance of 𝑗𝑗𝑗 ohms.The 663.1 microfarad capacitance has an impedance of 𝑍𝑍𝐶𝐶 = 1

𝑗𝑗𝜔𝜔𝐶𝐶= 1

𝑗𝑗𝜔𝜔𝐶𝐶−𝑗𝑗−𝑗𝑗

= −𝑗𝑗𝜔𝜔𝐶𝐶

= −𝑗𝑗377𝑥𝑥663.1𝑥𝑥1𝑥−6

= −𝑗𝑗𝑗 ohms.

Write the sources as phasors:𝑣𝑣1 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡 becomes phasor 𝑉𝑉1 = 10∠𝑗°𝑣𝑣2 𝑡𝑡 = 16 sin𝜔𝜔𝑡𝑡Change sine to cosine using the trig identity sin𝜔𝜔𝑡𝑡 = cos(𝜔𝜔𝑡𝑡 − 90°) so 𝑣𝑣2 𝑡𝑡 = 16 sin𝜔𝜔𝑡𝑡 = 16cos(𝜔𝜔𝑡𝑡 − 90°) which becomes phasor 𝑉𝑉2 = 16∠ − 90°Since 1∠ − 90° = −𝑗𝑗 we can write 𝑉𝑉2 = 16∠ − 90° = −𝑗𝜋𝑗𝑗

+𝑣𝑣1(𝑡𝑡)−

+𝑣𝑣2(𝑡𝑡)−

2 Ω5 Ω

1 Ω

663.1 µF

5.3 mH 10.6 mH 𝑉𝑉1 =10

𝑉𝑉2 =16∠ − 90°= −𝑗𝑗𝑗𝜋

2 Ω5 Ω

1 Ω

−j4 Ω

𝑗𝑗2 Ω 𝑗𝑗𝑗 Ω

𝑉𝑉2=−𝑗𝑗𝑗𝜋

𝑉𝑉1 = 10

2 + j2 Ω 1 + j4 Ω

5 − j4 Ω

𝐼𝐼1 𝐼𝐼2

Mesh path FABEF:+10 − 2 + 𝑗𝑗2 𝐼𝐼1 − 5 − 𝑗𝑗𝑗 𝐼𝐼1 − 𝐼𝐼2 = 0

Mesh path EBCDE:+ 5 − 𝑗𝑗𝑗 𝐼𝐼1 − 𝐼𝐼2 − 1 + 𝑗𝑗𝑗 𝐼𝐼2 − −𝑗𝑗𝑗𝜋 = 0

Collect terms:−2 − 𝑗𝑗2 − 5 + 𝑗𝑗𝑗 𝐼𝐼1 + 5 − 𝑗𝑗𝑗 𝐼𝐼2 = −10−7 + 𝑗𝑗2 𝐼𝐼1 + 5 − 𝑗𝑗𝑗 𝐼𝐼2 = −10

5 − 𝑗𝑗𝑗 𝐼𝐼1 + −5 + 𝑗𝑗𝑗 − 1 − 𝑗𝑗𝑗 𝐼𝐼2 = −𝑗𝑗𝑗65 − 𝑗𝑗𝑗 𝐼𝐼1 + −6 𝐼𝐼2 = −𝑗𝑗𝑗6

Step 2: Assign mesh currents 𝐼𝐼1 and 𝐼𝐼2:

Step 3:Write a KVL equation for each mesh path and an equation for each current source.

Label the circuit diagram with the voltage across each impedance. Then it is easy to get the signs correct in the mesh equations:

𝑉𝑉2= −𝑗𝑗𝑗𝜋𝑉𝑉1 = 10

2 + j2 Ω 1 + j4 Ω

5 − j4 Ω

𝐼𝐼1 𝐼𝐼2

10 −𝑗𝑗𝑗𝜋

𝐼𝐼1 𝐼𝐼2

+ 2 + j2 𝐼𝐼1 − + 1 + j4 𝐼𝐼2 −

+5 − j4 (𝐼𝐼1 − 𝐼𝐼2)−

A B C

DEF

Solve the equations: Solve the equations by hand:−7 + 𝑗𝑗2 𝐼𝐼1 + 5 − 𝑗𝑗𝑗 𝐼𝐼2 = −105 − 𝑗𝑗𝑗 𝐼𝐼1 + −6 𝐼𝐼2 = −𝑗𝑗𝑗6

Eliminate 𝐼𝐼2:−7 + 𝑗𝑗25 − 𝑗𝑗𝑗

𝐼𝐼1 + 𝐼𝐼2 =−10

5 − 𝑗𝑗𝑗5 − 𝑗𝑗𝑗−6 𝐼𝐼1 + 𝐼𝐼2 =

−𝑗𝑗𝑗𝜋−6

Evaluate the coefficients:(−1.049 − 𝑗𝑗𝑗.4390)𝐼𝐼1 + 𝐼𝐼2 = (−1.220 − 𝑗𝑗𝑗.9756)(−0.8333 + 𝑗𝑗𝑗.6667)𝐼𝐼1 + 𝐼𝐼2 = 𝑗𝑗2.667Subtract:(−0.2157 − 𝑗𝑗1.106)𝐼𝐼1 = (−1.220 − 𝑗𝑗3.643)𝐼𝐼1 = 3.380 − 𝑗𝑗𝑗.4438 = 3.𝑗𝑗𝑗∠ − 7.5°

Evaluate 𝐼𝐼2: 𝐼𝐼2 = 𝑗𝑗2.667 − (−0.8333 + 𝑗𝑗𝑗.6667)𝐼𝐼1𝐼𝐼2 = 2.521+j0.0435=2.521 ∠𝑗.0°

Homework: solve the equations by hand yourself!Use determinants and show that you get the same answer.

Example 2: Mesh Equations with a current source. 1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign mesh currents 𝐼𝐼1, 𝐼𝐼2, 𝐼𝐼3, …3. Write a KVL equation for each mesh path and an equation for each current source.

The operating frequency is 60 Hz.The sources are𝑣𝑣1 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡𝑣𝑣2 𝑡𝑡 = 2 sin𝜔𝜔𝑡𝑡Write mesh equations for this circuit.

Step 1: Convert the sources.𝑣𝑣𝑟𝑟 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡 becomes phasor 𝑉𝑉1 = 10∠𝑗°𝑖𝑖𝑟𝑟 𝑡𝑡 = 2 sin𝜔𝜔𝑡𝑡 = 2 cos(𝜔𝜔𝑡𝑡 − 90°) becomes 𝑉𝑉2 = 2∠ − 90° = −𝑗𝑗2

10 cos𝜔𝜔𝑡𝑡

2 Ω 1 Ω

331.6 µF

5.3 mH 10.6 mH

2 sin𝜔𝜔𝑡𝑡

Step 1, continued: Convert the components to impedances.

The operating frequency is 𝑓𝑓 =60 Hz. The radian frequency is 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 2𝜋𝜋𝜋𝜋𝜋𝑗 = 𝑗2𝑗𝜋𝜋 = 376.99 ≈ 377radians/second. For the 5.3 mH inductance, 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝜋𝜋3𝑗𝑗𝜋𝜋𝑗.3𝜋𝜋10−3 = 𝑗𝑗𝑗.9981 ≈ 𝑗𝑗2 ohms

The 10.6 mH inductance has an impedance of 𝑗𝑗𝑗 ohms.

The 331.6 microfarad capacitance has an impedance of 𝑍𝑍𝐶𝐶 = 1

𝑗𝑗𝜔𝜔𝐶𝐶= 1

𝑗𝑗𝜔𝜔𝐶𝐶−𝑗𝑗−𝑗𝑗

= −𝑗𝑗𝜔𝜔𝐶𝐶

= −𝑗𝑗377𝑥𝑥331.6𝑥𝑥1𝑥−6

= −𝑗𝑗𝑗 ohms.

Step 2: Assign mesh currents 𝐼𝐼1 and 𝐼𝐼2.

10 cos𝜔𝜔𝑡𝑡

2 Ω 1 Ω

331.6 µF

5.3 mH 10.6 mH

2 sin𝜔𝜔𝑡𝑡 10

2 Ω 1 Ω j4 Ω

-j8 Ω−𝑗𝑗2j2 Ω

2+j2 Ω

10 −𝑗𝑗2

1-j4 Ω

𝐼𝐼1 𝐼𝐼2

Step 3: Write a KVL equation for each mesh.Write a “constraint equation” for each current source.

Constraint equation for the current source: 𝐼𝐼2 − 𝐼𝐼1 = −2𝑗𝑗

Supermesh Path ABCDEFA: +10 − 2 + 𝑗𝑗2 𝐼𝐼1 − 1 − 𝑗𝑗𝑗 𝐼𝐼2 = 0

2 + 𝑗𝑗2 𝐼𝐼1 + 1 − 𝑗𝑗𝑗 𝐼𝐼2 = 10

Hence the equations are:−𝐼𝐼1 + 𝐼𝐼2 = −2𝑗𝑗2 + 𝑗𝑗2 𝐼𝐼1 + 1 − 𝑗𝑗𝑗 𝐼𝐼2 = 10

Homework:1.Solve the equations for 𝐼𝐼1 and 𝐼𝐼2𝐼𝐼1 =5.02 amps, angle 40.0 degrees𝐼𝐼2 =4.04 amps, angle 17.7 degrees

2.Verify the solution with LTSpice.

2+j2 Ω

10 −𝑗𝑗2

1-j4 Ω

𝐼𝐼1 𝐼𝐼2

A

B C D

EF

Mesh Equations in the “Phasor Domain”1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign mesh currents 𝐼𝐼1, 𝐼𝐼2, 𝐼𝐼3, …3. Write a KVL equation for each mesh path. Write an equation for each current source.

Write mesh equations at frequency 𝜔𝜔 = 100 rad/sec

The sources are𝑣𝑣𝑟𝑟 𝑡𝑡 = 14.14cos(𝜔𝜔𝑡𝑡 + 45°)𝑖𝑖𝑟𝑟 𝑡𝑡 = 2sin(𝜔𝜔𝑡𝑡)

Solve the mesh equations.

Verify your solution with LTSpice.

Example

SolutionConvert the components to impedances: 𝜔𝜔 = 100 rad/sec.

𝜔𝜔 = 20 mH 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗𝑗𝜋𝜋2𝑗𝜋𝜋10−3 = 𝑗𝑗2𝐶𝐶 = 0.5 mF 𝑍𝑍𝐶𝐶 = 1

𝑗𝑗𝜔𝜔𝐶𝐶= −𝑗𝑗

1𝑥𝑥𝑥𝑥𝑥.5𝑥𝑥1𝑥−3= −𝑗𝑗2𝑗

8 Ω 20 mH 0.5 mF 8 j2 -j20 8-j18

30 mH

8 Ω

j3

33+j3

9 Ω50 mH

9j5

9+j5

Draw the circuit with phasors and impedances:8 Ω 20 mH 0.5 mF 8 j2 -j20 8-j18

30 mH

8 Ω

j3

33+j3

9 Ω50 mH

9j5

9+j5 Convert the sources to phasors: 𝑣𝑣𝑟𝑟 𝑡𝑡 = 14.14cos(𝜔𝜔𝑡𝑡 + 45°) becomes phasor 𝑉𝑉𝑟𝑟 = 14.14∠𝑗𝑗°𝑖𝑖𝑟𝑟 𝑡𝑡 = 2 sin 𝜔𝜔𝑡𝑡 = 2cos(𝜔𝜔𝑡𝑡 − 90°) becomes phasor 𝐼𝐼𝑟𝑟 = 2∠ − 90° = −𝑗𝑗2

𝐼𝐼𝑅𝑅= −𝑗𝑗2

𝑉𝑉𝑟𝑟 =10 + 𝑗𝑗10

𝐼𝐼3

9+j5

8-j18 -j9

3+j312

𝐼𝐼1 𝐼𝐼2

𝐼𝐼𝑅𝑅= −𝑗𝑗2

𝑉𝑉𝑟𝑟 =10 + 𝑗𝑗10

𝐼𝐼3

9+j5

8-j18 -j9

3+j312

𝐼𝐼1 𝐼𝐼2

Write the mesh equations:

Constraint equation: 𝐼𝐼2 − 𝐼𝐼1 = −𝑗𝑗2

Mesh path ABCDEA: −12𝐼𝐼1 − 8 − 𝑗𝑗𝑗𝑗 𝐼𝐼1 − 𝐼𝐼3 − −𝑗𝑗𝑗 𝐼𝐼2 − 𝐼𝐼3 − 3 + 𝑗𝑗3 𝐼𝐼2 − 10 + 𝑗𝑗𝑗𝑗 = 0

Mesh path BFGDCB: − 9 + 𝑗𝑗𝑗 𝐼𝐼3 + −𝑗𝑗𝑗 𝐼𝐼2 − 𝐼𝐼3 + 8 − 𝑗𝑗𝑗𝑗 𝐼𝐼1 − 𝐼𝐼3 = 0

A

B DC

E

FG

= 10 + 𝑗𝑗10

Collect terms:Mesh path ABCDEA: −12𝐼𝐼1 − 8 − 𝑗𝑗𝑗𝑗 𝐼𝐼1 − 𝐼𝐼3 − −𝑗𝑗𝑗 𝐼𝐼2 − 𝐼𝐼3 − 3 + 𝑗𝑗3 𝐼𝐼2 − (10 + 𝑗𝑗𝑗𝑗) = 0

−12 − 8 + 𝑗𝑗𝑗𝑗 𝐼𝐼1 + 𝑗𝑗𝑗 − 3 − 𝑗𝑗3 𝐼𝐼2 + 8 − 𝑗𝑗𝑗𝑗 − 𝑗𝑗𝑗 𝐼𝐼3 = 10 + 𝑗𝑗𝑗𝑗

−20 + 𝑗𝑗𝑗𝑗 𝐼𝐼1 + −3 + 𝑗𝑗6 𝐼𝐼2 + 8 − 𝑗𝑗27 𝐼𝐼3 = 10 + 𝑗𝑗𝑗𝑗

Mesh path BFGDCB: − 9 + 𝑗𝑗𝑗 𝐼𝐼3 − −𝑗𝑗𝑗 𝐼𝐼3 − 𝐼𝐼2 − 8 − 𝑗𝑗𝑗𝑗 𝐼𝐼3 − 𝐼𝐼1 = 0

8 − 𝑗𝑗𝑗𝑗 𝐼𝐼1 − 𝑗𝑗𝑗𝐼𝐼2 + −9 − 𝑗𝑗𝑗 + 𝑗𝑗𝑗 − 8 + 𝑗𝑗𝑗𝑗 𝐼𝐼3 = 0

8 − 𝑗𝑗𝑗𝑗 𝐼𝐼1 − 𝑗𝑗𝑗𝐼𝐼2 + −17 + 𝑗𝑗22 𝐼𝐼3 = 0

Current generator: 𝐼𝐼2 − 𝐼𝐼1 = −𝑗𝑗2

The mesh equations:

−20 + 𝑗𝑗𝑗𝑗 𝐼𝐼1 + −3 + 𝑗𝑗𝜋 𝐼𝐼2 + 8 − 𝑗𝑗2𝑗 𝐼𝐼3 = 10 + 𝑗𝑗𝑗𝑗8 − 𝑗𝑗𝑗𝑗 𝐼𝐼1 − 𝑗𝑗𝑗𝐼𝐼2 + −17 + 𝑗𝑗22 𝐼𝐼3 = 0𝐼𝐼2 − 𝐼𝐼1 = −𝑗𝑗2

Solve the mesh equations:

𝐼𝐼1 = 0.5919∠𝑗𝜋𝑗.2°𝐼𝐼2 = 1.884∠ − 107.2°𝐼𝐼3 = 0.8574∠ − 172.0°

Node Equations in the “Phasor Domain”1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign a datum node and node voltages 𝑉𝑉1, 𝑉𝑉2, 𝑉𝑉3, …3. Write a KCL equation for each node.

Example - really easy!

The operating frequency is 60 Hz.

𝑣𝑣1 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡𝑣𝑣2 𝑡𝑡 = 16 sin𝜔𝜔𝑡𝑡

Write node equations for this circuit. Find the voltage across 5 ohms in series with 663.1 microfarads.

+𝑣𝑣1(𝑡𝑡)−

+𝑣𝑣2(𝑡𝑡)−

2 Ω5 Ω

1 Ω

663.1 µF

5.3 mH 10.6 mH

Convert the circuit into the “phasor domain”:

We converted this circuit to the phasor domain on a previous slide!• Sources:𝑣𝑣1 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡 becomes phasor 𝑉𝑉1 = 10∠𝑗°𝑣𝑣2 𝑡𝑡 = 16 sin𝜔𝜔𝑡𝑡 = 16cos(𝜔𝜔𝑡𝑡 − 90°) becomes phasor 𝑉𝑉2 = 16∠ − 90° = −𝑗𝑗𝑗𝜋• The operating frequency is 𝑓𝑓 =60 Hz. The radian frequency is

𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 2𝜋𝜋𝜋𝜋𝜋𝑗 = 𝑗2𝑗𝜋𝜋 = 376.99 ≈ 377 radians/second. • For the 5.3 mH inductance, 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝜋𝜋3𝑗𝑗𝜋𝜋𝑗.3𝜋𝜋10−3 =

𝑗𝑗𝑗.9981 ≈ 𝑗𝑗2 ohms• The 10.6 mH inductance has an impedance of 𝑗𝑗𝑗 ohms.• The 663.1 microfarad capacitance has an impedance of 𝑍𝑍𝐶𝐶 =

1𝑗𝑗𝜔𝜔𝐶𝐶

= 1𝑗𝑗𝜔𝜔𝐶𝐶

−𝑗𝑗−𝑗𝑗

= −𝑗𝑗𝜔𝜔𝐶𝐶

= −𝑗𝑗377𝑥𝑥663.1𝑥𝑥1𝑥−6

= −𝑗𝑗𝑗 ohms.

+𝑣𝑣1(𝑡𝑡)−

+𝑣𝑣2(𝑡𝑡)−

2 Ω5 Ω

1 Ω

663.1 µF

5.3 mH 10.6 mH𝑗𝑗𝑗𝜋

2 Ω5 Ω

1 Ω

−j4 Ω

𝑗𝑗2 Ω 𝑗𝑗𝑗 Ω10

𝑗𝑗𝑗𝜋10

2 + j2 Ω 1 + j4 Ω

5 − j4 Ω

Choose a datum and a node voltage:

Write a KCL equation at the node:

10 − 𝑉𝑉12 + 𝑗𝑗2 −

𝑉𝑉15 − 𝑗𝑗𝑗 −

𝑉𝑉1 − −𝑗𝑗𝑗𝜋1 + 𝑗𝑗𝑗 = 0

Collect terms:−𝑉𝑉1

2 + 𝑗𝑗2 −𝑉𝑉1

5 − 𝑗𝑗𝑗 −𝑉𝑉1

1 + 𝑗𝑗𝑗 =−10

2 + 𝑗𝑗2 +𝑗𝑗𝑗𝜋

1 + 𝑗𝑗𝑗−1

2 + 𝑗𝑗2 −1

5 − 𝑗𝑗𝑗 −1

1 + 𝑗𝑗𝑗 𝑉𝑉1 =−10

2 + 𝑗𝑗2 +𝑗𝑗𝑗𝜋

1 + 𝑗𝑗𝑗

One way to solve this quation is to use a common denominator for all the terms:− 5 − 𝑗𝑗𝑗 1 + 𝑗𝑗𝑗 − 2 + 𝑗𝑗2 1 + 𝑗𝑗𝑗 − (2 + 𝑗𝑗2)(5 − 𝑗𝑗𝑗)

2 + 𝑗𝑗2 (5 − 𝑗𝑗𝑗)(1 + 𝑗𝑗𝑗) 𝑉𝑉1

=−10 1 + 𝑗𝑗𝑗 + 𝑗𝑗𝑗𝜋(2 + 𝑗𝑗2)

(2 + 𝑗𝑗2)(1 + 𝑗𝑗𝑗)A better way is to do the division term by term; see next page.

𝑗𝑗𝑗𝜋10

2 + j2 Ω 1 + j4 Ω

5 − j4 Ω

𝑉𝑉1

Solve:−1

2 + 𝑗𝑗2−

15 − 𝑗𝑗𝑗

−1

1 + 𝑗𝑗𝑗𝑉𝑉1 =

−102 + 𝑗𝑗2

+𝑗𝑗𝑗𝜋

1 + 𝑗𝑗𝑗

Change the coefficients from ratios to simple rectangular form using your calculator:

−0.25 + 𝑗𝑗𝑗.25 − 0.1220 − 𝑗𝑗𝑗.09756 − 0.05882 + 𝑗𝑗𝑗.2353 𝑉𝑉1 = −2.5 + j2.5 + 3.765 + j0.9412

Do the addition of real parts and of imaginary parts:−0.4302 + 𝑗𝑗𝑗.3877 𝑉𝑉1 = 1.265 + 𝑗𝑗3.441

Do the division:

𝑉𝑉1 =1.265 + 𝑗𝑗3.441

−0.4302 + 𝑗𝑗𝑗.3877 = 2.355 − 𝑗𝑗𝑗.876 = 6.33𝑗∠ − 68.1°

As a working engineer, would I do this calculation by hand?Probably not! I could easily make a mistake, and hand calculation takes a lot of time. I’d use a short computer program to solve the equation.

Node Equations with a current generator

Example #2

Procedure for writing node equations in the “frequency domain”:1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign a datum node and node voltages 𝑉𝑉1, 𝑉𝑉2, 𝑉𝑉3, …3. Write a KCL equation for each node.

The operating frequency is 60 Hz.The sources are𝑣𝑣1 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡𝑣𝑣2 𝑡𝑡 = 2 sin𝜔𝜔𝑡𝑡Write a node equation for the voltage across the current generator.

10 cos𝜔𝜔𝑡𝑡

2 Ω 1 Ω

331.6 µF

5.3 mH 10.6 mH

2 sin𝜔𝜔𝑡𝑡

Step 1: Convert to the phasor domain.

We did this conversion for mesh analysis, see above!

Convert the sources.𝑣𝑣𝑟𝑟 𝑡𝑡 = 10 cos𝜔𝜔𝑡𝑡 becomes phasor 𝑉𝑉1 = 10∠𝑗°𝑖𝑖𝑟𝑟 𝑡𝑡 = 2 sin𝜔𝜔𝑡𝑡 = 2 cos(𝜔𝜔𝑡𝑡 − 90°) becomes 𝑉𝑉2 = 2∠ − 90° = −2𝑗𝑗

The operating frequency is 𝑓𝑓 =60 Hz. The radian frequency is 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 2𝜋𝜋𝜋𝜋𝜋𝑗 = 𝑗2𝑗𝜋𝜋 = 376.99 ≈ 377 radians/second.

For the 5.3 mH inductance, 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝜋𝜋3𝑗𝑗𝜋𝜋𝑗.3𝜋𝜋10−3 = 𝑗𝑗𝑗.9981 ≈ 𝑗𝑗2 ohms

The 10.6 mH inductance has an impedance of 𝑗𝑗𝑗 ohms.

The 331.6 microfarad capacitance has an impedance of 𝑍𝑍𝐶𝐶 = 1

𝑗𝑗𝜔𝜔𝐶𝐶= 1

𝑗𝑗𝜔𝜔𝐶𝐶−𝑗𝑗−𝑗𝑗

= −𝑗𝑗𝜔𝜔𝐶𝐶

= −𝑗𝑗377𝑥𝑥331.6𝑥𝑥1𝑥−6

= −𝑗𝑗𝑗 ohms.

10 cos𝜔𝜔𝑡𝑡

2 Ω 1 Ω

331.6 µF

5.3 mH 10.6 mH

2 sin𝜔𝜔𝑡𝑡 10

2 Ω 1 Ω j4 Ω

-j8 Ω−𝑗𝑗2j2 Ω

2+j2 Ω

10 −𝑗𝑗2

1-j4 Ω𝑉𝑉1

Write the node equation at 𝑉𝑉1:

10 − 𝑉𝑉12 + 𝑗𝑗2 + −2𝑗𝑗 −

𝑉𝑉11 − 𝑗𝑗𝑗 = 0

−𝑉𝑉12 + 𝑗𝑗2 −

𝑉𝑉11 − 𝑗𝑗𝑗 = 2𝑗𝑗 −

102 + 𝑗𝑗2

−12 + 𝑗𝑗2 −

11 − 𝑗𝑗𝑗 𝑉𝑉1 = 2𝑗𝑗 −

102 + 𝑗𝑗2

Homework: Solve the equation.Verify the solution with Spice.

Step 2: Assign a datum node and node voltages 𝑉𝑉1, 𝑉𝑉2, 𝑉𝑉3, …Step 3: Write a KCL equation for each node.

2+j2 Ω

10 −𝑗𝑗2

1-j4 Ω𝑉𝑉1

Example: Node equations with a supernode.

The frequency is 1950 MHz. Write node equations for this circuit.Find the amplitude and phase of 𝑉𝑉𝑜𝑜.

Remark: • A voltage generator embedded inside a node is called a “supernode”. • Only one unknown node voltage is needed for a supernode.• We write a KCL equation for the closed surface that encloses the voltage generator.

2 cos 𝜔𝜔𝑡𝑡5 cos 𝜔𝜔𝑡𝑡 + 45°

81.63 pH 2 Ω

5 Ω16.32 pF

10 Ω

+𝑉𝑉𝑜𝑜−

2 cos 𝜔𝜔𝑡𝑡5 cos 𝜔𝜔𝑡𝑡 + 45°

81.63 pH 2 Ω

5 Ω 16.32 pF10 Ω

Convert to phasors and impedances:

Sources:𝑖𝑖𝑟𝑟 𝑡𝑡 = 2 cos𝜔𝜔𝑡𝑡 is represented by phasor 𝐼𝐼𝑟𝑟 = 2𝑣𝑣𝑟𝑟 𝑡𝑡 = 5 cos(𝜔𝜔𝑡𝑡 + 45°) is represented by phasor 𝑉𝑉𝑟𝑟 = 5∠𝑗𝑗°

𝑓𝑓 = 1950 MHz so 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 =1.22𝑗𝜋𝜋101𝑥 rad/sec

𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝜋𝜋𝑗.22𝑗𝜋𝜋101𝑥𝜋𝜋𝑗.𝑗𝜋3𝜋𝜋10−12 = 𝑗𝑗𝑗ohm1

𝑗𝑗𝜔𝜔𝐶𝐶= −𝑗𝑗

1.225𝑥𝑥1𝑥10𝑥𝑥16.32𝑥𝑥1𝑥−12= −𝑗𝑗𝑗 ohms

25∠45°

𝑗𝑗𝑗 Ω 2 Ω

5 Ω

−𝑗𝑗𝑗 Ω10 Ω

+𝑉𝑉𝑜𝑜−

Combine impedances:

25∠45°

𝑗𝑗𝑗 Ω 2 Ω

5 Ω−𝑗𝑗𝑗 Ω10 Ω

2 10 Ω5∠45°

Series:𝑗𝑗𝑗 in series with 2 is equivalent to (2 + 𝑗𝑗𝑗) ohms

Parallel:5 in parallel with −𝑗𝑗𝑗 is equivalent to𝑗𝜋𝜋 −𝑗𝑗𝑗

5 − 𝑗𝑗𝑗=−𝑗𝑗2𝑗5 − 𝑗𝑗𝑗

= 2.5 − 𝑗𝑗2.5

2 + 𝑗𝑗𝑗 Ω

2.5 − 𝑗𝑗2.5 Ω+𝑉𝑉𝑜𝑜−

Node equations:

2 10 Ω5∠45° 2 + 𝑗𝑗𝑗 Ω

2.5 − 𝑗𝑗2.5 Ω

𝑉𝑉1 𝑉𝑉2 𝑉𝑉𝑥

Constraint equation: 𝑉𝑉2 − 𝑉𝑉1 = 5∠𝑗𝑗°

At the supernode:

2 −𝑉𝑉110 −

𝑉𝑉2 − 𝑉𝑉𝑥2 + 𝑗𝑗𝑗 = 0

At the 𝑉𝑉𝑥 node: 𝑉𝑉2 − 𝑉𝑉𝑥2 + 𝑗𝑗𝑗 −

𝑉𝑉𝑥2.5 − 𝑗𝑗2.5 = 0

+𝑉𝑉𝑜𝑜−

Use three nodes:

For the supernode: 𝑉𝑉1 and 𝑉𝑉2.

For the output, use 𝑉𝑉𝑥.

There are three nodes so three equations to solve, in three unknowns.

Solve to find:𝑉𝑉𝑜𝑜 = 5.772 ∠ − 30.5°

Amplitude of 𝑉𝑉𝑥 is 5.772 voltsPhase of 𝑉𝑉𝑥 is -30.5 degrees.

Can we solve this circuit “smarter”?

2 10 Ω5∠45° 2 + 𝑗𝑗𝑗 Ω

2.5 − 𝑗𝑗2.5 Ω

+𝑉𝑉𝑜𝑜−

𝑉𝑉1Use only one unknown node voltage, 𝑉𝑉1.

Build the constraint equation into the supernode directly.

Once the value of 𝑉𝑉1 has been found, use the voltage-divider relationship to find 𝑉𝑉𝑥:

𝑉𝑉𝑥 = 2.5 − 𝑗𝑗2.5𝑉𝑉1 + 5∠𝑗𝑗°

2 + 𝑗𝑗𝑗 + 2.5 − 𝑗𝑗2.5

We can solve the node equation to find𝑉𝑉1 = 5.884∠ − 44.2°And then the voltage divider to find𝑉𝑉𝑥 = 5.772∠ − 44.2°

This agrees with the “dumber” solution!

𝑉𝑉1 + 5∠𝑗𝑗°

We only need one node equation to find node voltage, 𝑉𝑉1.At the supernode:

2 −𝑉𝑉110 −

𝑉𝑉1 + 5∠𝑗𝑗°2 + 𝑗𝑗𝑗 + 2.5 − 𝑗𝑗2.5 = 0

So

−𝑉𝑉110 −

𝑉𝑉14.5 − 𝑗𝑗𝑗.5 = −2 +

5∠𝑗𝑗°4.5 − 𝑗𝑗𝑗.5

Node Equations in the “Phasor Domain”We did this example using mesh analysis. Let’s do it by node analysis.

Convert the components to impedances: 𝜔𝜔 = 100 rad/sec.

The frequency is 𝜔𝜔 = 100 rad/sec.𝑣𝑣𝑟𝑟 𝑡𝑡 = 14.14cos(𝜔𝜔𝑡𝑡 + 45°)𝑖𝑖𝑟𝑟 𝑡𝑡 = 2 sin 𝜔𝜔𝑡𝑡 = 2cos(𝜔𝜔𝑡𝑡 − 90°)

Write a set of node equations in matrix form.

𝜔𝜔 = 20 mH 𝑍𝑍𝐿𝐿 = 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗𝑗𝜋𝜋2𝑗𝜋𝜋10−3 = 𝑗𝑗2𝐶𝐶 = 0.5 mF 𝑍𝑍𝐶𝐶 = 1

𝑗𝑗𝜔𝜔𝐶𝐶= −𝑗𝑗

1𝑥𝑥𝑥𝑥𝑥.5𝑥𝑥1𝑥−3= −𝑗𝑗2𝑗

8 Ω 20 mH 0.5 mF 8 j2 -j20 8-j18

30 mH

8 Ω

j3

33+j3

9 Ω50 mH

9j5

9+j5

Draw the circuit with phasors and impedances:Convert the sources to phasors: 𝑣𝑣𝑟𝑟 𝑡𝑡 = 14.14cos(𝜔𝜔𝑡𝑡 + 45°) becomes phasor 𝑉𝑉𝑟𝑟 = 14.14∠𝑗𝑗°𝑖𝑖𝑟𝑟 𝑡𝑡 = 2 sin 𝜔𝜔𝑡𝑡 = 2cos(𝜔𝜔𝑡𝑡 − 90°) becomes phasor 𝐼𝐼𝑟𝑟 = 2∠ − 90° = −𝑗𝑗2

Write Node EquationsNode 1:

−𝑉𝑉112

−𝑉𝑉1 − 𝑉𝑉28 − 𝑗𝑗𝑗𝑗

−𝑉𝑉1 − 𝑉𝑉39 + 𝑗𝑗𝑗

= 0

Node 2:𝑉𝑉1 − 𝑉𝑉28 − 𝑗𝑗𝑗𝑗

+ (−𝑗𝑗2) −𝑉𝑉2 − 𝑉𝑉3−𝑗𝑗9

= 0

Node 3:𝑉𝑉1 − 𝑉𝑉39 + 𝑗𝑗𝑗 +

𝑉𝑉2 − 𝑉𝑉3−𝑗𝑗𝑗 −

𝑉𝑉3 − 14.14∠45°3 + 𝑗𝑗3 = 0

Collect terms in the node equations:Node 1: Multiply by 12(8 − 𝑗𝑗𝑗𝑗)(9 + 𝑗𝑗𝑗)

−𝑉𝑉112

−𝑉𝑉1 − 𝑉𝑉28 − 𝑗𝑗𝑗𝑗

−𝑉𝑉1 − 𝑉𝑉39 + 𝑗𝑗𝑗

= 0

−(8 − 𝑗𝑗𝑗𝑗)(9 + 𝑗𝑗𝑗)𝑉𝑉1 − 12(9 + 𝑗𝑗𝑗)(𝑉𝑉1−𝑉𝑉2) − 12(8 − 𝑗𝑗𝑗𝑗)(𝑉𝑉1 − 𝑉𝑉3) = 0−(72 − 𝑗𝑗𝑗𝜋2 + 𝑗𝑗𝑗𝑗 + 90)𝑉𝑉1 − (108 + 𝑗𝑗60)(𝑉𝑉1−𝑉𝑉2) − (96 − 𝑗𝑗216)(𝑉𝑉1 − 𝑉𝑉3) = 0−(162 − 𝑗𝑗𝑗22)𝑉𝑉1 − (108 + 𝑗𝑗𝜋𝑗)(𝑉𝑉1−𝑉𝑉2) − (96 − 𝑗𝑗2𝑗𝜋)(𝑉𝑉1 − 𝑉𝑉3) = 0−162 + 𝑗𝑗𝑗22 − 108 − 𝑗𝑗𝜋𝑗 − 96 + 𝑗𝑗2𝑗𝜋 𝑉𝑉1 + 108 + 𝑗𝑗𝜋𝑗 𝑉𝑉2 + (96 − 𝑗𝑗2𝑗𝜋)𝑉𝑉3 = 0−366 + 𝑗𝑗2𝑗𝑗 𝑉𝑉1 + 108 + 𝑗𝑗𝜋𝑗 𝑉𝑉2 + (96 − 𝑗𝑗2𝑗𝜋)𝑉𝑉3 = 0

Node 2:𝑉𝑉1 − 𝑉𝑉28 − 𝑗𝑗𝑗𝑗 + (−𝑗𝑗2) −

𝑉𝑉2 − 𝑉𝑉3−𝑗𝑗9 = 0

(−𝑗𝑗𝑗)(𝑉𝑉1 − 𝑉𝑉2) + (8 − 𝑗𝑗𝑗𝑗)(−𝑗𝑗𝑗)(−𝑗𝑗2) − (8 − 𝑗𝑗𝑗𝑗)(𝑉𝑉2 − 𝑉𝑉3) = 0(−𝑗𝑗𝑗)(𝑉𝑉1 − 𝑉𝑉2) + (−𝑗𝑗𝑗2 − 162)(−𝑗𝑗2) − (8 − 𝑗𝑗𝑗𝑗)(𝑉𝑉2 − 𝑉𝑉3) = 0(−𝑗𝑗𝑗)(𝑉𝑉1 − 𝑉𝑉2) + (−144 + 𝑗𝑗32𝑗) − (8 − 𝑗𝑗𝑗𝑗)(𝑉𝑉2 − 𝑉𝑉3) = 0−𝑗𝑗𝑗 𝑉𝑉1 + (𝑗𝑗𝑗 − 8 + 𝑗𝑗𝑗𝑗)𝑉𝑉2+(8 − 𝑗𝑗𝑗𝑗)𝑉𝑉3 = − −144 + 𝑗𝑗32𝑗−𝑗𝑗𝑗 𝑉𝑉1 + (−8 + 𝑗𝑗27)𝑉𝑉2+(8 − 𝑗𝑗𝑗𝑗)𝑉𝑉3 = − −144 + 𝑗𝑗32𝑗

Arrange the equations into matrix form:Node 3:𝑉𝑉1 − 𝑉𝑉39 + 𝑗𝑗𝑗

+𝑉𝑉2 − 𝑉𝑉3−𝑗𝑗𝑗

−𝑉𝑉3 − 14.1∠45°

3 + 𝑗𝑗3= 0

(−𝑗𝑗𝑗)(3 + 𝑗𝑗3)(𝑉𝑉1 − 𝑉𝑉3) + (9 + 𝑗𝑗𝑗)(3 + 𝑗𝑗3)(𝑉𝑉2 − 𝑉𝑉3) − (9 + 𝑗𝑗𝑗)(−𝑗𝑗𝑗)(𝑉𝑉3 − 14.1∠45°) = 0(−𝑗𝑗27 + 27)(𝑉𝑉1 − 𝑉𝑉3) + (27 + 𝑗𝑗15 + 𝑗𝑗2𝑗 − 15)(𝑉𝑉2 − 𝑉𝑉3) − (−𝑗𝑗𝑗𝑗 + 45)(𝑉𝑉3 − 14.1∠45°) = 0(−𝑗𝑗2𝑗 + 27)(𝑉𝑉1 − 𝑉𝑉3) + (12 + 𝑗𝑗42)(𝑉𝑉2 − 𝑉𝑉3) − (−𝑗𝑗𝑗𝑗 + 45)𝑉𝑉3 = (−𝑗𝑗𝑗𝑗 + 45)(−14.𝑗∠𝑗𝑗°)(−𝑗𝑗2𝑗 + 27)𝑉𝑉1 + (12 + 𝑗𝑗𝑗2)𝑉𝑉2 + (𝑗𝑗2𝑗 − 27 − 12 − 𝑗𝑗𝑗2 + 𝑗𝑗𝑗𝑗 − 45)𝑉𝑉3 = (−𝑗𝑗𝑗𝑗 + 45)(−14.𝑗∠𝑗𝑗°)−𝑗𝑗2𝑗 + 27 𝑉𝑉1 + 12 + 𝑗𝑗𝑗2 𝑉𝑉2 + −84 + 𝑗𝑗66 𝑉𝑉3 = −1256 + 𝑗𝑗3𝑗𝑗.9

Hence the node equations are:−366 + 𝑗𝑗2𝑗𝑗 𝑉𝑉1 + 108 + 𝑗𝑗𝜋𝑗 𝑉𝑉2 + (96 − 𝑗𝑗2𝑗𝜋)𝑉𝑉3 = 0−𝑗𝑗𝑗 𝑉𝑉1 + (−8 + 𝑗𝑗2𝑗)𝑉𝑉2+(8 − 𝑗𝑗𝑗𝑗)𝑉𝑉3 = − −144 + 𝑗𝑗32𝑗−𝑗𝑗2𝑗 + 27 𝑉𝑉1 + 12 + 𝑗𝑗𝑗2 𝑉𝑉2 + −84 + 𝑗𝑗𝜋𝜋 𝑉𝑉3 = −1256 + 𝑗𝑗3𝑗𝑗.9

Arrange the node equations into a matrix equation:−366 + 𝑗𝑗2𝑗𝑗 108 + 𝑗𝑗𝜋𝑗 (96 − 𝑗𝑗2𝑗𝜋)

−𝑗𝑗𝑗 (−8 + 𝑗𝑗2𝑗) (8 − 𝑗𝑗𝑗𝑗)−𝑗𝑗2𝑗 + 27 12 + 𝑗𝑗𝑗2 −84 + 𝑗𝑗𝜋𝜋

𝑉𝑉1𝑉𝑉2𝑉𝑉3

=0

144 − 𝑗𝑗32𝑗−1256 + 𝑗𝑗3𝑗𝑗.9

𝑉𝑉1 = 7.091∠ − 19.8°𝑉𝑉2 = 1.448∠168.9°𝑉𝑉3 = 14.00∠𝑗2.0°