the course web site is: ...trueman/elec273files/elec273_10...ย ยท trueman ... numbers. final exam:...
Post on 22-May-2018
218 Views
Preview:
TRANSCRIPT
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?
โข AC circuit analysis uses complex numbers to find the forced response to a sinusoidal generator.
โข 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 ๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐Sinusoidal current ๐๐(๐ก๐ก) = ๐ต๐ต cos ๐๐๐ก๐ก + ๐๐
Real numbers for V and I Complex numbers called phasors๐๐ = ๐ด๐ด๐๐๐๐๐๐ and ๐ผ๐ผ = ๐ต๐ต๐๐๐๐๐๐
Real number resistances R Complex number impedances๐๐๐ ๐ = ๐ ๐ , ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐, and ๐๐๐ถ๐ถ = 1
๐๐๐๐๐ถ๐ถ
Real number node matrix or mesh matrix Complex number node matrix or mesh matrix
Given: โข the frequency ๐๐ Hz, hence ๐๐ = 2๐๐๐๐ rad/secโข The amplitude of the source, 10 voltsFind:โข The amplitude ๐ด๐ด and the phase ๐๐ of the output
voltage ๐ฃ๐ฃ ๐ก๐ก at โsteady stateโ after the transients have died out.
From Alexander and Sadiku.
10 cos๐๐๐ก๐ก+๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos(๐๐๐ก๐ก + ๐๐)โ
๐ ๐ 1๐ ๐ 2๐ถ๐ถ
๐๐
PhasorGiven an AC voltage ๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ , define the phasor representing voltage ๐ฃ๐ฃ(๐ก๐ก) as the complex number๐๐ = ๐ด๐ด๐๐๐๐๐๐
where ๐๐ = โ1and ๐๐ = 2๐๐๐๐ is the frequency in radians per second. The magnitude of the phasor ๐ด๐ด๐๐๐๐๐๐ is the amplitude of the cosine ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ .The angle of the phasor, ๐๐, is the phase angle of the cosine ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ .
It is sometimes convenient to visualize the phasor by drawing it on the complex plane.
The phasor ๐๐ = ๐ด๐ด๐๐๐๐๐๐ is an arrow of length A making an angle of ๐๐to the real axis.
It can be quite useful to draw phasors on the complex plane!
We can use angle notation:๐๐ = ๐ด๐ดโ ๐๐Or exponential notation๐๐ = ๐ด๐ด๐๐๐๐๐๐
Imaginary
Real
๐๐๐ด๐ด
๐๐
๐๐ = ๐ด๐ด๐๐๐๐๐๐
Rotating PhasorThe phasor representing AC voltage ๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ is the complex number ๐๐ = ๐ด๐ด๐๐๐๐๐๐ .
It is sometimes convenient to visualize the relationship between two phasors by defining a rotating phasor as
๐ฃ๐ฃ๐๐ ๐ก๐ก = ๐๐๐๐๐๐๐๐๐๐
Hence๐ฃ๐ฃ๐๐ ๐ก๐ก = ๐ด๐ด๐๐๐๐๐๐๐๐๐๐๐๐๐๐ = ๐ด๐ด๐๐๐๐(๐๐๐๐+๐๐)
The AC voltage we started with is the real part of the rotating phasor.๐ฃ๐ฃ ๐ก๐ก = Re ๐ฃ๐ฃ๐๐ ๐ก๐ก = Re ๐ด๐ด๐๐๐๐ ๐๐๐๐+๐๐ = Re ๐ด๐ดcos ๐๐๐ก๐ก + ๐๐ + j๐ด๐ดsin ๐๐๐ก๐ก + ๐๐ = ๐ด๐ดcos(๐๐๐ก๐ก + ๐๐)
Draw the rotating phasor on the complex plane at t=0.
At ๐ก๐ก = 0, ๐ฃ๐ฃ๐๐ 0 = ๐ด๐ด๐๐๐๐๐๐๐๐๐๐๐๐๐ฅ๐ฅ๐ฅ = ๐ด๐ด๐๐๐๐๐๐
where the โcโ subscript on ๐ฃ๐ฃ๐๐ ๐ก๐ก denotes a complex version of voltage ๐ฃ๐ฃ ๐ก๐ก .
Imaginary
Real
๐ด๐ด
๐๐
๐ด๐ด๐๐๐๐๐๐At ๐ก๐ก = 0๐ฃ๐ฃ๐๐ 0 = ๐ด๐ด๐๐๐๐๐๐
Draw the rotating phasor on the complex plane as time advances:
๐ด๐ด๐๐๐๐(๐๐๐๐+๐๐)
Draw the rotating phasor at time ๐ก๐ก > 0
๐ฃ๐ฃ๐๐ ๐ก๐ก = ๐ด๐ด๐๐๐๐(๐๐๐๐+๐๐)
This amounts to adding angle ๐๐๐ก๐ก to the angle of phasor ๐ด๐ด๐๐๐๐๐๐.
As time increases, ๐๐๐ก๐ก increases. We add a full 2๐๐radians of angle in one AC period ๐๐ = 1
๐๐.
The phasor rotates by one full rotation for each T seconds of time.
Imaginary
Real
๐ด๐ด
๐๐
For ๐ก๐ก > 0๐ฃ๐ฃ๐๐ ๐ก๐ก = ๐ด๐ด๐๐๐๐(๐๐๐๐+๐๐)
๐๐๐ก๐ก
๐๐๐ก๐ก + ๐๐
Phasorโข The phasor representing sinusoidal voltage ๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ is the complex number ๐๐ = ๐ด๐ด๐๐๐๐๐๐
โข The amplitude A of the cosine is the magnitude of the phasor: ๐ด๐ด = ๐๐ = ๐ด๐ด๐๐๐๐๐๐
โข The phase angle ๐๐ of the cosine is the angle of the complex number โphasorโ.
โข We can โrecoverโ the cosine from the phasor by multiplying the phasor by ๐๐๐๐๐๐๐๐ and then using Eulerโs Identity: ๐๐๐๐๐๐ = cos๐๐ + ๐๐ sin๐๐
๐ฃ๐ฃ ๐ก๐ก = Re ๐๐๐๐๐๐๐๐๐๐ = Re ๐ด๐ด๐๐๐๐๐๐๐๐๐๐๐๐๐๐ = Re ๐ด๐ด๐๐๐๐(๐๐๐๐+๐๐)
๐ฃ๐ฃ ๐ก๐ก = Re ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ + ๐๐ sin ๐๐๐ก๐ก + ๐๐๐ฃ๐ฃ ๐ก๐ก = Re ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ + ๐๐ ๐ด๐ดsin ๐๐๐ก๐ก + ๐๐๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐
โข To recover the cosine from the phasor:โข The amplitude A of the cosine is the magnitude of the phasor V โข The phase angle ๐๐ of the cosine is the angle of the complex number
Examples of Phasors1.Find the phasor for ๐ฃ๐ฃ1 ๐ก๐ก = 10 cos ๐๐๐ก๐ก + ๐๐
4Answer: ๐๐1 = 10๐๐๐๐
๐๐4
2.Find the cosine represented by phasor ๐๐2 = 5๐๐โ๐๐๐๐6
Answer: ๐ฃ๐ฃ2 ๐ก๐ก = 5 cos ๐๐๐ก๐ก โ ๐๐6
3.Find the sum: ๐ฃ๐ฃ3 ๐ก๐ก = ๐ฃ๐ฃ1 ๐ก๐ก + ๐ฃ๐ฃ2 ๐ก๐กUsing trig identities is the hard way: ๐ฃ๐ฃ3 ๐ก๐ก = 10 cos ๐๐๐ก๐ก +
๐๐4
+ 5 cos ๐๐๐ก๐ก โ๐๐6
Try it!The easy way: ๐๐3 = ๐๐1 + ๐๐2 = 10๐๐๐๐
๐๐4 + 5๐๐โ๐๐
๐๐6 (see the next slide)
Evaluate the sum ๐๐3 = ๐๐1 + ๐๐2๐๐3 = 10๐๐๐๐
๐๐4 + 5๐๐โ๐๐
๐๐6
Showing ALL the steps:๐๐3 = 10 cos
๐๐4
+ ๐๐๐๐ sin๐๐4
+ 5 cosโ๐๐6
+ ๐๐๐ sinโ๐๐6
๐๐3 = 7.071 + ๐๐๐.071 + 4.330 โ ๐๐2.5๐๐3 = 11.401 + ๐๐4.571Magnitude 11.4012 + 4.5712 = 12.283Angle tan๐๐ = 4.571
11.4๐ฅ1= 0.4049 so ๐๐ = 21.85 deg or 0.3814 rad
The phasor is๐๐3 = 12.283๐๐๐๐๐ฅ.3814
so the time function is๐ฃ๐ฃ3 ๐ก๐ก = 12.283 cos ๐๐๐ก๐ก + 0.3814
Check the result by sketching the phasors on the complex plane:
๐๐1 = 10๐๐๐๐๐๐4 = 10 angle 45 degrees = 7.071+j7.071
๐๐2 = 5๐๐โ๐๐๐๐6= 5 angle -30 degrees=4.330-j2.500
๐๐3 = ๐๐1 + ๐๐2
๐๐3 = 12.283๐๐๐๐21.85ยฐ= 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
๐๐1 = 10๐๐๐๐๐๐4
๐๐2 = 5๐๐โ๐๐๐๐6
45ยฐ
30ยฐ
10
5
๐๐1 = 10๐๐๐๐๐๐4
๐๐2 = 5๐๐โ๐๐๐๐6
๐๐3 = ๐๐1 + ๐๐2
Impedance: the ratio of voltage phasor to current phasor.
โข Consider a component in a circuit driven by a sinusoidal generator cos๐๐๐ก๐ก
โข The voltage ๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ is represented by phasor ๐๐ = ๐ด๐ด๐๐๐๐๐๐
โข The current ๐๐ ๐ก๐ก = ๐ต๐ต cos ๐๐๐ก๐ก + ๐๐ is represented by phasor ๐ผ๐ผ = ๐ต๐ต๐๐๐๐๐๐
โข The impedance is defined as
๐๐ = ๐๐๐ผ๐ผ
ohms
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก)+๐๐โ
๐ผ๐ผ
The Impedance is a Complex Number
โข ๐๐ = ๐ด๐ด๐๐๐๐๐๐
โข ๐ผ๐ผ = ๐ต๐ต๐๐๐๐๐๐
โข The impedance is a complex number:
๐๐ =๐๐๐ผ๐ผ =
๐ด๐ด๐๐๐๐๐๐
๐ต๐ต๐๐๐๐๐๐=๐ด๐ด๐ต๐ต ๐๐
๐๐(๐๐โ๐๐) =๐ด๐ด๐ต๐ต cos ๐๐ โ ๐๐ + ๐๐
๐ด๐ด๐ต๐ต sin ๐๐ โ ๐๐ = ๐ ๐ + ๐๐๐๐
๐๐ = ๐ ๐ + ๐๐๐๐
R=the resistance is the real part of the impedanceX=the reactance is the imaginary part of the impedance.
+๐๐โ
๐ผ๐ผ
Admittance: the reciprocal of the impedanceThe admittance is the ratio of the current phasor to the voltage phasor:๐๐ = ๐ผ๐ผ
๐๐Siemens
Note that๐๐ = 1
๐๐
The admittance is a complex number:๐๐ = ๐บ๐บ + ๐๐๐ต๐ต
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?
๐ฃ๐ฃ ๐ก๐ก = ๐ ๐ ๐๐ ๐ก๐ก๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ volts then the current is
๐๐ ๐ก๐ก =๐ฃ๐ฃ(๐ก๐ก)๐ ๐ =
๐ด๐ด๐ ๐ cos ๐๐๐ก๐ก + ๐๐
with amplitude ๐ด๐ด๐ ๐
and phase ๐๐The voltage phasor is ๐๐ = ๐ด๐ด๐๐๐๐๐๐
The current phasor is ๐ผ๐ผ = ๐ด๐ด๐ ๐ ๐๐๐๐๐๐; magnitude ๐ด๐ด
๐ ๐ and angle ๐๐
The impedance is ๐๐ = ๐๐๐ผ๐ผ
= ๐ด๐ด๐๐๐๐๐๐๐ด๐ด๐ ๐ ๐๐
๐๐๐๐= ๐ ๐ ohms.
The admittance is ๐๐ = 1๐๐
= 1๐ ๐
= ๐บ๐บ Siemens
Time domain Frequency Domain also called the โphasor domainโ
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก) +๐๐โ
๐ผ๐ผ
๐๐ = ๐ ๐ ๐ ๐
What is the impedance of an inductor?๐ฃ๐ฃ ๐ก๐ก = ๐๐
๐๐๐๐๐ก๐ก๐๐ ๐ก๐ก
If ๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ volts then we can find the current using๐๐๐๐๐๐๐๐
= 1๐ฟ๐ฟ๐ฃ๐ฃ ๐ก๐ก so ๐๐๐๐ = 1
๐ฟ๐ฟ๐ฃ๐ฃ ๐ก๐ก ๐๐๐ก๐ก and by integrating both sides
๐๐ =1๐๐๏ฟฝ๐ฃ๐ฃ ๐ก๐ก ๐๐๐ก๐ก =
1๐๐๏ฟฝ๐ด๐ด cos(๐๐๐ก๐ก + ๐๐)๐๐๐ก๐ก =
๐ด๐ด๐๐๐๐
sin(๐๐๐ก๐ก + ๐๐)
To write the phasor we need to change sine to cosine.Use the trig identity:sin๐ผ๐ผ = co๐ ๐ ๐ผ๐ผ โ
๐๐2
With ฮฑ = ๐๐๐ก๐ก + ๐๐, we have
๐๐ =๐ด๐ด๐๐๐๐ sin(๐๐๐ก๐ก + ๐๐) =
๐ด๐ด๐๐๐๐ cos(๐๐๐ก๐ก + ๐๐ โ
๐๐2)
Which has amplitude ๐ด๐ด๐๐๐ฟ๐ฟ
and phase ๐๐ โ ๐๐2
, so the current phasor is
๐ผ๐ผ =๐ด๐ด๐๐๐๐ ๐๐
๐๐ ๐๐โ๐๐2
The impedance is
๐๐ =๐๐๐ผ๐ผ =
๐ด๐ด๐๐๐๐๐๐
๐ด๐ด๐๐๐๐ ๐๐
๐๐(๐๐โ๐๐2)= ๐๐๐๐๐๐๐๐
๐๐2
Time domain
Frequency domain
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก)
+๐๐โ
๐ผ๐ผ
๐๐ = ๐๐๐๐๐๐
๐๐
Impedance of an inductor, continued;๐๐ = ๐๐๐๐๐๐๐๐
๐๐2
We can see from the Argand diagram that ๐๐๐๐๐๐2 = ๐๐.
Another way to prove this is with Eulerโs Identity:๐๐๐๐
๐๐2 = cos
๐๐2
+ ๐๐ sin๐๐2
= 0 + ๐๐๐ = ๐๐
Hence the impedance of an inductor is
๐๐ = ๐๐๐๐๐๐ ohms
The admittance of an inductor is
๐๐ = 1๐๐
= 1๐๐๐๐๐ฟ๐ฟ
SiemensArgand diagram
Re
Im
1 ๐๐2
๐๐ = 1๐๐๐๐๐๐2
Time domain
Frequency domain
+๐๐โ
๐ผ๐ผ
๐๐ = ๐๐๐๐๐๐
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก)
๐๐
What is the impedance of a capacitor?๐๐(๐ก๐ก) = ๐ถ๐ถ
๐๐๐๐๐ก๐ก๐ฃ๐ฃ ๐ก๐ก
If ๐ฃ๐ฃ ๐ก๐ก = ๐ด๐ด cos ๐๐๐ก๐ก + ๐๐ volts then we can calculate the current as
๐๐ = ๐ถ๐ถ๐๐๐๐๐ก๐ก๐ด๐ด cos(๐๐๐ก๐ก + ๐๐) = โฯ๐ถ๐ถ๐ด๐ด sin(๐๐๐ก๐ก + ๐๐)
To write the phasor we need to change sine to cosine using the trig identity:sin๐ผ๐ผ = โco๐ ๐ ๐ผ๐ผ + ๐๐
2With ฮฑ = ๐๐๐ก๐ก + ๐๐, we have๐๐ = โฯ๐ถ๐ถ๐ด๐ด sin(๐๐๐ก๐ก + ๐๐) = ฯ๐ถ๐ถAcos(๐๐๐ก๐ก + ๐๐ +
๐๐2)
which has amplitude ฯ๐ถ๐ถ๐ด๐ด and phase ๐๐ + ๐๐2
, so the current phasor is
๐ผ๐ผ = ฯ๐ถ๐ถ๐ด๐ด๐๐๐๐ ๐๐+๐๐2
The impedance is
๐๐ =๐๐๐ผ๐ผ =
๐ด๐ด๐๐๐๐๐๐
ฯ๐ถ๐ถ๐ด๐ด๐๐๐๐ ๐๐+๐๐2=
1
ฯ๐ถ๐ถ๐๐๐๐๐๐2
Since ๐๐๐๐๐๐2 = ๐๐ we can write the impedance of a capacitor as
๐๐ = 1๐๐๐๐๐ถ๐ถ
ohms.
The admittance of a capacitor is ๐๐ = 1๐๐
= ๐๐๐๐๐ถ๐ถ Siemens
Frequency domain
+๐๐โ
๐ผ๐ผ
๐๐ =1
๐๐๐๐๐ถ๐ถ
Time domain
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก)
๐ถ๐ถ
Component Time Domain Frequency Domain
Resistor
Inductor
Capacitor
Impedance
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก)
๐ ๐
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก)
๐ถ๐ถ
+๐ฃ๐ฃ(๐ก๐ก)โ
๐๐(๐ก๐ก)
๐๐
๐ฃ๐ฃ = ๐ ๐ ๐๐
๐ฃ๐ฃ = ๐๐๐๐๐๐๐๐๐ก๐ก
๐๐ = ๐ถ๐ถ๐๐๐ฃ๐ฃ๐๐๐ก๐ก
๐๐ = ๐ ๐ ๐ผ๐ผ
๐๐ = ๐๐๐๐๐๐๐ผ๐ผ
๐๐ =1๐๐๐๐๐ถ๐ถ
๐ผ๐ผ
Combining ImpedancesImpedances in series:
๐๐ = ๐๐1 + ๐๐2 = ๐๐1๐ผ๐ผ + ๐๐2๐ผ๐ผ=(๐๐1+๐๐2)๐ผ๐ผSo the equivalent impedance is
๐๐ =๐๐๐ผ๐ผ
=(๐๐1+๐๐2)๐ผ๐ผ
๐ผ๐ผ= ๐๐1 + ๐๐2
Series impedances add.
Impedances in parallel:
๐ผ๐ผ = ๐ผ๐ผ1 + ๐ผ๐ผ2 =๐๐๐๐1
+๐๐๐๐2
=1๐๐1
+1๐๐2
๐๐
The equivalent impedance is
๐๐ =๐๐๐ผ๐ผ =
๐๐1๐๐1
+ 1๐๐2
๐๐=
๐๐1๐๐2๐๐1 + ๐๐2
Parallel impedances combine using product over sum.
+
๐๐
โ
+๐๐2โ
๐ผ๐ผ
+๐๐1โ
๐ผ๐ผ2๐ผ๐ผ1
+
๐๐
โ
๐ผ๐ผ
๐๐2๐๐1
๐๐2
๐๐1
Combining Admittances: ๐ผ๐ผ = ๐๐๐๐ so ๐๐ = ๐ผ๐ผ๐๐
Admittances in series:
๐๐ = ๐๐1 + ๐๐2 =๐ผ๐ผ๐๐1
+๐ผ๐ผ๐๐2
=1๐๐1
+1๐๐2
๐ผ๐ผ
So the equivalent admittance is
๐๐ =๐ผ๐ผ๐๐
=I
๐ผ๐ผ๐๐1
+ ๐ผ๐ผ๐๐2
๐ผ๐ผ=
๐๐1๐๐2๐๐1 + ๐๐2
Series admittances combine like parallel resistors: product over sum.
Admittances in parallel:๐ผ๐ผ = ๐ผ๐ผ1 + ๐ผ๐ผ2 = ๐๐1๐๐ + ๐๐2๐๐ = (๐๐1 + ๐๐2)๐๐The equivalent admittance is
๐๐ =๐ผ๐ผ๐๐
=(๐๐1 + ๐๐2)๐๐
๐๐= ๐๐1 + ๐๐2
Parallel admittances add.
+
๐๐
โ
+๐๐2โ
๐ผ๐ผ
+๐๐1โ
๐ผ๐ผ2๐ผ๐ผ1
+
๐๐
โ
๐ผ๐ผ
๐๐2๐๐1
๐๐2
๐๐1
Series-Parallel Combinations Find the input impedance ๐๐๐๐๐๐ at an operating frequency of 60 Hz.The radian frequency is ๐๐ = 2๐๐๐๐ = 2๐๐๐๐๐๐ = ๐2๐๐๐ = 376.99 rad.sec
Change the components to impedances:
๐๐๐ฟ๐ฟ1 = ๐๐๐๐๐๐1 = ๐๐๐๐๐2๐๐๐๐๐2.๐๐๐๐10โ3 = ๐๐๐
๐๐๐ ๐ 1 = ๐ ๐ 1 = 2
๐๐๐ ๐ 2 = ๐ ๐ 2 = 2
๐๐๐ ๐ 3 = ๐ ๐ 3 = 3
๐๐๐ฟ๐ฟ2 = ๐๐๐๐๐๐2 = ๐๐๐๐๐2๐๐๐๐๐๐.3๐๐๐10โ3 = ๐๐2
๐๐๐ถ๐ถ =1๐๐๐๐๐ถ๐ถ =
1๐๐๐๐120๐๐๐๐1326๐๐10โ6 =
2๐๐ =
2๐๐โ๐๐โ๐๐ = โ๐๐2
Method:1.Change the components into impedances.2.Combine impedances in series and in parallel.
๐๐๐๐๐๐
๐ ๐ 1 = 2 ฮฉ
๐ ๐ 2 = 2 ฮฉ ๐ ๐ 3 = 3 ฮฉ๐๐1 = 2.65 mH
๐๐2 = 5.30 mH
๐ถ๐ถ = 1,326 ยตF
๐ ๐ 1 = 2 ฮฉ
๐ ๐ 2 = 2 ฮฉ
๐ ๐ 3 = 3 ฮฉ๐๐1 = 2.65 mH
๐๐2 = 5.30 mH
๐ถ๐ถ = 1,326 ยตF
Find the input impedance, continued:
๐๐1is j2 ohms in parallel with 3 ohms:
๐๐1 =๐๐2๐๐3๐๐2 + 3 =
๐๐๐3 + ๐๐2
3 โ 2๐๐3 โ ๐๐2 =
๐๐๐๐ + 123๐๐3 + 2๐๐2 =
12 + ๐๐๐๐13
๐๐2is -j2 ohms in parallel with 2 ohms:
๐๐2 =โ๐๐2๐๐2โ๐๐2 + 2 =
โ๐๐๐2 โ ๐๐2
2 + 2๐๐2 + ๐๐2 =
โ๐๐๐ + 82๐๐2 + 2๐๐2 =
8 โ ๐๐๐8 = 1 โ ๐๐
๐๐๐๐๐๐2 ฮฉ
2 ฮฉ 3 ฮฉ๐๐2 ฮฉj1 ฮฉ
โ๐๐2 ฮฉ
๐๐1 ๐๐2 ฮฉ 3 ฮฉ
๐๐2
2 ฮฉ
โ๐๐2 ฮฉ
๐๐๐๐๐๐2 ฮฉ
j1 ฮฉ ๐๐1๐๐2
Combine ๐๐1 and ๐๐2 in series:
๐๐3 = ๐๐1 + ๐๐2๐๐๐๐๐๐2 ฮฉ
j1 ฮฉ ๐๐1๐๐2 ๐๐1
๐๐2
๐๐1 =12 + ๐๐๐๐
13๐๐2 = 1 โ ๐๐
๐๐3 = ๐๐1 + ๐๐2 =12 + ๐๐๐๐
13 + (1 โ ๐๐) =12 + ๐๐๐๐
13 +13 โ ๐๐๐3
13๐๐3 =
25 + ๐๐513
Find the input impedance, continued:Define ๐๐4as j1 ohms in parallel with ๐๐3:
๐๐4 =๐๐๐๐ 25 + ๐๐๐
13๐๐ + 25 + ๐๐๐
13
=โ5 + ๐๐2๐25 + ๐๐๐๐
25 โ ๐๐๐๐25 โ ๐๐๐๐
=โ125 + ๐๐๐๐ + ๐๐๐2๐ + 450
2๐๐๐2๐ + ๐๐๐๐๐๐=
325 โ ๐๐๐๐๐949
๐๐๐๐๐๐ is 2 ohms in series with ๐๐4:
๐๐๐๐๐๐ = 2 + ๐๐4 = 2 +325 โ ๐๐๐๐๐
949 =1989 + 325 โ ๐๐๐๐๐
949=
2223 โ ๐๐๐๐๐949
๐๐๐๐๐๐ = 2.342 โ ๐๐๐.7534 ฮฉ
This is our final answer!
๐๐3๐๐๐๐๐๐2 ฮฉ
j1 ฮฉ
๐๐4๐๐๐๐๐๐2 ฮฉ
Spice with a Sinusoidal Generator
To calculate the input impedance with LTSpice:โข Drive the circuit with a 1 volt generator at 60 Hzโข Find the current in ๐ ๐ 1 flowing into the circuit ๐ผ๐ผ๐๐๐๐โข Calculate the input impedance as
๐๐๐๐๐๐ = 1๐ผ๐ผ๐๐๐๐
Construct the circuit.
How do we specify a sinusoidal excitation?
๐๐๐๐๐๐2 ฮฉ
2 ฮฉ 3 ฮฉ๐๐2 ฮฉj1 ฮฉ
โ๐๐2 ฮฉ
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 โACโ source of amplitude 1 volt.
Choose AC Analysis and give the frequency: Click on Stimulate and then Edit Stimulation Card.
Choose AC Analysis:
โข Specify the number of points per octave as 1.โข 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 โdot commandโ .ac oct 1 60 60.Click on the โrunโ 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 ๐ ๐ 1, so๐ผ๐ผ๐๐๐๐ =I(R1)=0.4064 angle -17.8 degrees
๐๐๐๐๐๐ =1๐ผ๐ผ๐๐๐๐
=1
0.4064โ โ 17.8ยฐ = 2.342 + ๐๐๐.7522
which agrees with our calculation.
We calculated๐๐๐๐๐๐ = 2.342 โ ๐๐๐.7534
To calculate the input impedance with LTSpice:โข Drive the circuit with a 1 volt generator at 60 Hzโข Find the current in ๐ ๐ 1 flowing into the circuit ๐ผ๐ผ๐๐๐๐โข Calculate the input impedance as
๐๐๐๐๐๐ = 1๐ผ๐ผ๐๐๐๐
๐๐๐๐๐๐2 ฮฉ
2 ฮฉ 3 ฮฉ๐๐2 ฮฉj1 ฮฉ
โ๐๐2 ฮฉ
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 ๐๐๐๐๐๐ is real. The frequency is ๐๐ = 4 rad/sec.Solution
Convert the circuit elements to impedances at ๐๐ = 4 rad/sec : ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ and ๐๐๐ถ๐ถ = 1๐๐๐๐๐ถ๐ถ
.
๐๐๐๐๐๐ is 1 ohm in parallel with ๐๐2๐๐๐๐๐๐=1||๐๐2 = ๐๐2
1+๐๐2If ๐๐2 is real then ๐๐๐๐๐๐ is real.So choose C to make ๐๐2 real.
Define ๐๐2 by omitting the 1-ohm resistor:
1 ฮฉ1 ฮฉ
C1 H
1 H
๐๐๐๐๐๐
๐๐๐๐๐๐ 1 ฮฉ๐๐๐ ฮฉ
๐๐๐ ฮฉ1 ฮฉ
1๐๐๐๐ถ๐ถ
๐๐2๐๐๐ ฮฉ
1๐๐๐๐ถ๐ถ
1 ฮฉ๐๐๐ ฮฉ
4114
141441
12
jCj
jY
jZjZ
++
+=+=+=
Choose the value of C to make ๐๐2 real.
1)41(44142 ++
++=
jCjjjZ
CjCjjZ
4)161(4142 +โ
++=
CjCjCjCjZ
4)161(41)4)161((4
2 +โ+++โ
=
Define ๐๐1 to be 1๐๐4๐ถ๐ถ
in parallel with 1 + ๐๐๐ .
Since parallel admittances add,
๐๐1 = ๐๐๐๐ถ๐ถ +1
1 + ๐๐๐The input impedance ๐๐2 is j4 in series with ๐๐1
๐๐2๐๐๐ ฮฉ
1๐๐๐๐ถ๐ถ
1 ฮฉ๐๐๐ ฮฉ
๐๐2๐๐๐ ฮฉ
๐๐1
CjCCjCZ
4)161()4644()161(
2 +โ+โ+โ
=
โโโโ
+โโ+โ
=CjCCjC
CjCCjCZ
4)161(4)161(
4)161()648()161(
2
22
2
2 16)161())161(4)161)(648((4)648()161(
CCCCCCjCCCZ
+โโโโโ+โ+โ
=
To make ๐๐2 real, set the imaginary part to zero:
0))161(4)161)(648(( =โโโโ CCCC
0)4648)(161( =โโโ CCCThere is a common factor of (1 โ 16C) so factor it out:
0)688)(161( =โโ CC
0)161( =โ C 0)688( =โ 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, ๐ฃ๐ฃ๐ ๐ . The frequency is 400 Hz.
Find the impedances:๐๐ = 2๐๐๐๐ = 2๐๐๐๐๐๐๐ = 2,513 radians/sec๐๐๐ ๐ = ๐ ๐ = 5 ohms๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ = ๐๐๐๐2๐๐3๐๐๐.๐๐๐๐๐๐10โ3 = ๐๐2 ohms
+๐ฃ๐ฃ๐ ๐ โ
๐ ๐ = 5 ฮฉ
๐๐ = 07958 mH
10 cos๐๐๐ก๐ก+๐๐๐ ๐ โ
5 ฮฉ
๐๐2 ฮฉ
10
Solve the circuit using complex arithmetic:
Node analysis: 10 โ ๐๐๐ ๐ ๐๐2 โ
๐๐๐ ๐ 5 = 0
Multiply by j10:50 โ 5๐๐๐ ๐ โ ๐๐2๐๐๐ ๐ = 05 + ๐๐2 ๐๐๐ ๐ = 50
๐๐๐ ๐ =50
5 + ๐๐25 โ ๐๐25 โ ๐๐2 =
450 โ ๐๐๐๐๐29 = 8.621 โ ๐๐3.448
Magnitude 8.6212 + (โ3.448)2=9.284Angle tanโ1 โ3.448
8.621=-21.8 degrees
Hence๐๐๐ ๐ = 9.284โ โ 21.8ยฐ
Mesh analysis: 10 โ ๐๐2๐ผ๐ผ โ ๐๐ผ๐ผ = 0๐ผ๐ผ = 1๐ฅ
5+๐๐25โ๐๐25โ๐๐2
= 5๐ฅโ๐๐2๐ฅ29
= 1.724 โ ๐๐๐.6896 amps
๐๐๐ ๐ = ๐๐ผ๐ผ = 5 1.724 โ j0.6896 = 8.621 โ j3.448Same as by node analysis!๐๐๐ ๐ = 9.284โ โ 21.8ยฐ volts
Amplitude = ๐๐๐ ๐ =9.284 volts
Phase = the angle of ๐๐๐ ๐ which is -21.8 degrees.
๐ฃ๐ฃ๐ ๐ ๐ก๐ก = 9.284 cos ๐๐๐ก๐ก โ 21.8ยฐ
+๐๐๐ ๐ โ
5 ฮฉ
๐๐2 ฮฉ
10+๐๐๐ ๐ โ
5 ฮฉ
๐๐2 ฮฉ
10
10 ๐๐๐ ๐
๐ผ๐ผ
Example: Solve a Circuit Using Phasors
โข Solve this circuit to find the amplitude and phase of the load voltage ๐๐ at ๐๐ =2 GHz.โข The generator in this circuit is 10 cos๐๐๐ก๐ก volts at frequency ๐๐ =2 GHz so the radian frequency is ๐๐ =
2๐๐๐๐ = 1.2๐๐๐๐๐101๐ฅ 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 ฮฉ
2.1 nH
11.8 pF
50 ฮฉ
10 cos๐๐๐ก๐ก+๐๐โ
73 ฮฉ
2.1 nH
11.8 pF
50 ฮฉ
10 cos๐๐๐ก๐ก+๐๐โ 73 ฮฉ
๐๐2๐.39 ฮฉ
โ๐๐๐.744 ฮฉ
50 ฮฉ
10โ ๐ยฐ+๐๐โ
10โ ๐ยฐ+๐๐โ
50 + ๐๐2๐.39 ฮฉ
0.6178 โ ๐๐๐.687 ฮฉ
Homework: solve this circuit yourself without looking at the lecture notes.
10โ ๐ยฐ+๐๐โ
50 + ๐๐2๐.39 ฮฉ
0.6178 โ ๐๐๐.687 ฮฉ
๐ผ๐ผ ๐ผ๐ผ
Power into a Resistor in AC Circuits
If a circuit is driven by an A.C. voltage, then how much power flows?๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก + ๐๐๐๐ ๐ก๐ก = ๐ผ๐ผ๐๐ cos ๐๐๐ก๐ก + ๐๐
Instantaneous power: ๐๐ ๐ก๐ก = ๐ฃ๐ฃ ๐ก๐ก ๐๐(๐ก๐ก)
Average power: ๐๐๐๐๐๐ = 1๐๐ โซ๐ฅ
๐๐ ๐๐ ๐ก๐ก ๐๐๐ก๐ก
Is there a convenient way to calculate the power directly from the phasors V and I?
๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก + ๐๐
๐๐(๐ก๐ก) = ๐ผ๐ผ๐๐ cos ๐๐๐ก๐ก + ๐๐
๐๐(๐ก๐ก)+๐ฃ๐ฃ(๐ก๐ก)โ
Phasor ๐๐ = ๐๐๐๐โ ๐๐
Phasor ๐ผ๐ผ = ๐ผ๐ผ๐๐โ ๐๐
๐ผ๐ผ+๐๐โ
Instantaneous and Average Power Delivered to a Resistor
The AC voltage ๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก is connected across a resistor R.Find the instantaneous power and the average power delivered to the resistor.
The instantaneous power is defined as๐๐ ๐ก๐ก = ๐ฃ๐ฃ ๐ก๐ก ๐๐ ๐ก๐ก
๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก
๐๐ ๐ก๐ก =๐๐๐๐๐ ๐ cos ๐๐๐ก๐ก
The instantaneous power is:
๐๐ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก๐๐๐๐๐ ๐ cos ๐๐๐ก๐ก =
๐๐๐๐2
๐ ๐ cos2 ๐๐๐ก๐ก
Trig identity: cos2๐๐ = 12
1 + cos(2๐๐)
๐๐(๐ก๐ก) =๐๐๐๐2
2๐ ๐ 1 + cos2๐๐๐ก๐ก
๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก ๐๐(๐ก๐ก) = ๐ผ๐ผ๐๐ cos ๐๐๐ก๐ก๐ ๐
Time t
Instantaneous Power ๐๐(๐ก๐ก)๐๐๐๐2
๐ ๐
๐๐๐๐2
2๐ ๐
๐๐2
๐๐2
Average Power Delivered to a Resistor
The instantaneous power is: ๐๐ ๐ก๐ก = ๐ฃ๐ฃ ๐ก๐ก ๐๐ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก ๐๐๐๐๐ ๐
cos ๐๐๐ก๐ก = ๐๐๐๐2
๐ ๐ cos2 ๐๐๐ก๐ก = 1
2๐๐๐๐2
๐ ๐ 1 + cos2๐๐๐ก๐ก
The average power delivered to the resistor is defined as the average of p(t) over one AC cycle of length T:
๐๐๐๐๐๐ =1๐๐๏ฟฝ๐ฅ
๐๐๐๐ ๐ก๐ก ๐๐๐ก๐ก
so
๐๐๐๐๐๐ =1๐๐๏ฟฝ๐ฅ
๐๐ ๐๐๐๐2
2๐ ๐ 1 + cos2๐๐๐ก๐ก ๐๐๐ก๐ก =1๐๐๏ฟฝ๐ฅ
๐๐ ๐๐๐๐2
2๐ ๐ ๐๐๐ก๐ก +1๐๐๏ฟฝ๐ฅ
๐๐ ๐๐๐๐2
2๐ ๐ cos2๐๐๐ก๐ก ๐๐๐ก๐ก =๐๐๐๐2
2๐ ๐ + 0 =๐๐๐๐2
2๐ ๐
๐๐๐๐๐๐ = ๐๐๐๐2
2๐ ๐ where ๐๐๐๐ is the amplitude of the AC voltage.
This is often written as
๐๐๐๐๐๐ =๐๐๐๐2
2๐ ๐
=๐๐๐๐2
2
๐ ๐ = ๐๐๐ ๐ ๐ ๐ ๐ ๐
2
๐ ๐ where ๐๐๐ ๐ ๐ ๐ ๐ ๐ = ๐๐๐๐
2is the โRMS valueโ of the AC voltage.
๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก
๐๐ ๐ก๐ก =๐๐๐๐๐ ๐
cos ๐๐๐ก๐ก
What is meant by โRMS valueโ?
๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก ๐๐(๐ก๐ก) = ๐ผ๐ผ๐๐ cos ๐๐๐ก๐ก๐ ๐
Amplitude and RMS ValueSuppose an AC voltage is given by๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐กThen the amplitude of the voltage is ๐๐๐๐.
The โroot mean squareโ value or โRMSโ value of a periodic function ๐ฃ๐ฃ(๐ก๐ก) with period ๐๐ is defined as
๐๐๐๐๐๐๐๐ =1๐๐๏ฟฝ๐ฅ
๐๐๐ฃ๐ฃ2 ๐ก๐ก ๐๐๐ก๐ก
This is the average value or โmean valueโ of the square of the voltage function.
For a sinusoid ๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก we can evaluate the RMS value of ๐ฃ๐ฃ(๐ก๐ก) as
๐๐๐๐๐๐๐๐2 = 1
๐๐ โซ๐ฅ๐๐ ๐๐๐๐ cos ๐๐๐ก๐ก 2๐๐๐ก๐ก = ๐๐๐๐2
๐๐ โซ๐ฅ๐๐ 12
1 + cos 2๐๐๐ก๐ก ๐๐๐ก๐ก=๐๐๐๐2
๐๐ โซ๐ฅ๐๐ 12๐๐๐ก๐ก + ๐๐๐๐2
๐๐ โซ๐ฅ๐๐ 12
cos 2๐๐๐ก๐ก ๐๐๐ก๐ก = ๐๐๐๐2
๐๐๐๐2
= ๐๐๐๐2
2
๐๐๐๐๐๐๐๐ =๐๐๐๐
2
Thus the average power delivered to a resistor is
๐๐๐๐๐๐ = ๐๐๐๐2
2๐ ๐ =
๐๐๐๐2
2
๐ ๐ = ๐๐๐ ๐ ๐ ๐ ๐ ๐
2
๐ ๐
Phasors Relative to RMS ValueWe can write phasors โrelative to amplitudeโ or we can write phasors โrelative to RMS valueโ.๐ฃ๐ฃ ๐ก๐ก = ๐๐๐๐ cos ๐๐๐ก๐ก
Phasors relative to amplitude: ๐๐ = ๐๐๐๐๐๐๐๐๐๐The magnitude of the phasor ๐๐ = ๐๐๐๐ is the amplitude of the cosine voltage ๐๐๐๐ cos ๐๐๐ก๐ก
Phasors relative to RMS value:๐๐ = ๐๐๐๐๐๐๐๐๐๐๐๐๐๐
The magnitude of the phasor ๐๐ = ๐๐๐๐๐๐๐๐ is the RMS value of the cosine voltage, so ๐๐๐๐ = 2๐๐๐๐๐๐๐๐ and the cosine is ๐ฃ๐ฃ ๐ก๐ก = 2๐๐๐๐๐๐๐๐ cos ๐๐๐ก๐ก
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 ๐๐๐๐ = 2๐๐๐๐๐๐๐๐ = 1.๐๐๐๐๐๐๐๐ = 155.5 volts.
Most โDVMsโ or โdigital voltmetersโ 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 โPhasor Domainโ1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign mesh currents ๐ผ๐ผ1, ๐ผ๐ผ2, ๐ผ๐ผ3, โฆ3. 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 are๐ฃ๐ฃ1 ๐ก๐ก = 10 cos๐๐๐ก๐ก๐ฃ๐ฃ2 ๐ก๐ก = 16 sin๐๐๐ก๐กWrite mesh equations for this circuit.Solve the equations to find the values of the mesh currents.
+๐ฃ๐ฃ1(๐ก๐ก)โ
+๐ฃ๐ฃ2(๐ก๐ก)โ
2 ฮฉ5 ฮฉ
1 ฮฉ
663.1 ยตF
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 ๐๐ =60 Hz. The radian frequency is ๐๐ = 2๐๐๐๐ = 2๐๐๐๐๐๐ = ๐2๐๐๐ = 376.99 โ 377 radians/second. For the 5.3 mH inductance, ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ = ๐๐๐๐3๐๐๐๐๐.3๐๐10โ3 = ๐๐๐.9981 โ ๐๐2 ohmsThe 10.6 mH inductance has an impedance of ๐๐๐ ohms.The 663.1 microfarad capacitance has an impedance of ๐๐๐ถ๐ถ = 1
๐๐๐๐๐ถ๐ถ= 1
๐๐๐๐๐ถ๐ถโ๐๐โ๐๐
= โ๐๐๐๐๐ถ๐ถ
= โ๐๐377๐ฅ๐ฅ663.1๐ฅ๐ฅ1๐ฅโ6
= โ๐๐๐ ohms.
Write the sources as phasors:๐ฃ๐ฃ1 ๐ก๐ก = 10 cos๐๐๐ก๐ก becomes phasor ๐๐1 = 10โ ๐ยฐ๐ฃ๐ฃ2 ๐ก๐ก = 16 sin๐๐๐ก๐กChange sine to cosine using the trig identity sin๐๐๐ก๐ก = cos(๐๐๐ก๐ก โ 90ยฐ) so ๐ฃ๐ฃ2 ๐ก๐ก = 16 sin๐๐๐ก๐ก = 16cos(๐๐๐ก๐ก โ 90ยฐ) which becomes phasor ๐๐2 = 16โ โ 90ยฐSince 1โ โ 90ยฐ = โ๐๐ we can write ๐๐2 = 16โ โ 90ยฐ = โ๐๐๐๐
+๐ฃ๐ฃ1(๐ก๐ก)โ
+๐ฃ๐ฃ2(๐ก๐ก)โ
2 ฮฉ5 ฮฉ
1 ฮฉ
663.1 ยตF
5.3 mH 10.6 mH ๐๐1 =10
๐๐2 =16โ โ 90ยฐ= โ๐๐๐๐
2 ฮฉ5 ฮฉ
1 ฮฉ
โj4 ฮฉ
๐๐2 ฮฉ ๐๐๐ ฮฉ
๐๐2=โ๐๐๐๐
๐๐1 = 10
2 + j2 ฮฉ 1 + j4 ฮฉ
5 โ j4 ฮฉ
๐ผ๐ผ1 ๐ผ๐ผ2
Mesh path FABEF:+10 โ 2 + ๐๐2 ๐ผ๐ผ1 โ 5 โ ๐๐๐ ๐ผ๐ผ1 โ ๐ผ๐ผ2 = 0
Mesh path EBCDE:+ 5 โ ๐๐๐ ๐ผ๐ผ1 โ ๐ผ๐ผ2 โ 1 + ๐๐๐ ๐ผ๐ผ2 โ โ๐๐๐๐ = 0
Collect terms:โ2 โ ๐๐2 โ 5 + ๐๐๐ ๐ผ๐ผ1 + 5 โ ๐๐๐ ๐ผ๐ผ2 = โ10โ7 + ๐๐2 ๐ผ๐ผ1 + 5 โ ๐๐๐ ๐ผ๐ผ2 = โ10
5 โ ๐๐๐ ๐ผ๐ผ1 + โ5 + ๐๐๐ โ 1 โ ๐๐๐ ๐ผ๐ผ2 = โ๐๐๐65 โ ๐๐๐ ๐ผ๐ผ1 + โ6 ๐ผ๐ผ2 = โ๐๐๐6
Step 2: Assign mesh currents ๐ผ๐ผ1 and ๐ผ๐ผ2:
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:
๐๐2= โ๐๐๐๐๐๐1 = 10
2 + j2 ฮฉ 1 + j4 ฮฉ
5 โ j4 ฮฉ
๐ผ๐ผ1 ๐ผ๐ผ2
10 โ๐๐๐๐
๐ผ๐ผ1 ๐ผ๐ผ2
+ 2 + j2 ๐ผ๐ผ1 โ + 1 + j4 ๐ผ๐ผ2 โ
+5 โ j4 (๐ผ๐ผ1 โ ๐ผ๐ผ2)โ
A B C
DEF
Solve the equations: Solve the equations by hand:โ7 + ๐๐2 ๐ผ๐ผ1 + 5 โ ๐๐๐ ๐ผ๐ผ2 = โ105 โ ๐๐๐ ๐ผ๐ผ1 + โ6 ๐ผ๐ผ2 = โ๐๐๐6
Eliminate ๐ผ๐ผ2:โ7 + ๐๐25 โ ๐๐๐
๐ผ๐ผ1 + ๐ผ๐ผ2 =โ10
5 โ ๐๐๐5 โ ๐๐๐โ6 ๐ผ๐ผ1 + ๐ผ๐ผ2 =
โ๐๐๐๐โ6
Evaluate the coefficients:(โ1.049 โ ๐๐๐.4390)๐ผ๐ผ1 + ๐ผ๐ผ2 = (โ1.220 โ ๐๐๐.9756)(โ0.8333 + ๐๐๐.6667)๐ผ๐ผ1 + ๐ผ๐ผ2 = ๐๐2.667Subtract:(โ0.2157 โ ๐๐1.106)๐ผ๐ผ1 = (โ1.220 โ ๐๐3.643)๐ผ๐ผ1 = 3.380 โ ๐๐๐.4438 = 3.๐๐๐โ โ 7.5ยฐ
Evaluate ๐ผ๐ผ2: ๐ผ๐ผ2 = ๐๐2.667 โ (โ0.8333 + ๐๐๐.6667)๐ผ๐ผ1๐ผ๐ผ2 = 2.521+j0.0435=2.521 โ ๐.0ยฐ
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 ๐ผ๐ผ1, ๐ผ๐ผ2, ๐ผ๐ผ3, โฆ3. Write a KVL equation for each mesh path and an equation for each current source.
The operating frequency is 60 Hz.The sources are๐ฃ๐ฃ1 ๐ก๐ก = 10 cos๐๐๐ก๐ก๐ฃ๐ฃ2 ๐ก๐ก = 2 sin๐๐๐ก๐กWrite mesh equations for this circuit.
Step 1: Convert the sources.๐ฃ๐ฃ๐๐ ๐ก๐ก = 10 cos๐๐๐ก๐ก becomes phasor ๐๐1 = 10โ ๐ยฐ๐๐๐๐ ๐ก๐ก = 2 sin๐๐๐ก๐ก = 2 cos(๐๐๐ก๐ก โ 90ยฐ) becomes ๐๐2 = 2โ โ 90ยฐ = โ๐๐2
10 cos๐๐๐ก๐ก
2 ฮฉ 1 ฮฉ
331.6 ยตF
5.3 mH 10.6 mH
2 sin๐๐๐ก๐ก
Step 1, continued: Convert the components to impedances.
The operating frequency is ๐๐ =60 Hz. The radian frequency is ๐๐ = 2๐๐๐๐ = 2๐๐๐๐๐๐ = ๐2๐๐๐ = 376.99 โ 377radians/second. For the 5.3 mH inductance, ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ = ๐๐๐๐3๐๐๐๐๐.3๐๐10โ3 = ๐๐๐.9981 โ ๐๐2 ohms
The 10.6 mH inductance has an impedance of ๐๐๐ ohms.
The 331.6 microfarad capacitance has an impedance of ๐๐๐ถ๐ถ = 1
๐๐๐๐๐ถ๐ถ= 1
๐๐๐๐๐ถ๐ถโ๐๐โ๐๐
= โ๐๐๐๐๐ถ๐ถ
= โ๐๐377๐ฅ๐ฅ331.6๐ฅ๐ฅ1๐ฅโ6
= โ๐๐๐ ohms.
Step 2: Assign mesh currents ๐ผ๐ผ1 and ๐ผ๐ผ2.
10 cos๐๐๐ก๐ก
2 ฮฉ 1 ฮฉ
331.6 ยตF
5.3 mH 10.6 mH
2 sin๐๐๐ก๐ก 10
2 ฮฉ 1 ฮฉ j4 ฮฉ
-j8 ฮฉโ๐๐2j2 ฮฉ
2+j2 ฮฉ
10 โ๐๐2
1-j4 ฮฉ
๐ผ๐ผ1 ๐ผ๐ผ2
Step 3: Write a KVL equation for each mesh.Write a โconstraint equationโ for each current source.
Constraint equation for the current source: ๐ผ๐ผ2 โ ๐ผ๐ผ1 = โ2๐๐
Supermesh Path ABCDEFA: +10 โ 2 + ๐๐2 ๐ผ๐ผ1 โ 1 โ ๐๐๐ ๐ผ๐ผ2 = 0
2 + ๐๐2 ๐ผ๐ผ1 + 1 โ ๐๐๐ ๐ผ๐ผ2 = 10
Hence the equations are:โ๐ผ๐ผ1 + ๐ผ๐ผ2 = โ2๐๐2 + ๐๐2 ๐ผ๐ผ1 + 1 โ ๐๐๐ ๐ผ๐ผ2 = 10
Homework:1.Solve the equations for ๐ผ๐ผ1 and ๐ผ๐ผ2๐ผ๐ผ1 =5.02 amps, angle 40.0 degrees๐ผ๐ผ2 =4.04 amps, angle 17.7 degrees
2.Verify the solution with LTSpice.
2+j2 ฮฉ
10 โ๐๐2
1-j4 ฮฉ
๐ผ๐ผ1 ๐ผ๐ผ2
A
B C D
EF
Mesh Equations in the โPhasor Domainโ1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign mesh currents ๐ผ๐ผ1, ๐ผ๐ผ2, ๐ผ๐ผ3, โฆ3. Write a KVL equation for each mesh path. Write an equation for each current source.
Write mesh equations at frequency ๐๐ = 100 rad/sec
The sources are๐ฃ๐ฃ๐๐ ๐ก๐ก = 14.14cos(๐๐๐ก๐ก + 45ยฐ)๐๐๐๐ ๐ก๐ก = 2sin(๐๐๐ก๐ก)
Solve the mesh equations.
Verify your solution with LTSpice.
Example
SolutionConvert the components to impedances: ๐๐ = 100 rad/sec.
๐๐ = 20 mH ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ = ๐๐๐๐๐๐๐2๐๐๐10โ3 = ๐๐2๐ถ๐ถ = 0.5 mF ๐๐๐ถ๐ถ = 1
๐๐๐๐๐ถ๐ถ= โ๐๐
1๐ฅ๐ฅ๐ฅ๐ฅ๐ฅ.5๐ฅ๐ฅ1๐ฅโ3= โ๐๐2๐
8 ฮฉ 20 mH 0.5 mF 8 j2 -j20 8-j18
30 mH
8 ฮฉ
j3
33+j3
9 ฮฉ50 mH
9j5
9+j5
Draw the circuit with phasors and impedances:8 ฮฉ 20 mH 0.5 mF 8 j2 -j20 8-j18
30 mH
8 ฮฉ
j3
33+j3
9 ฮฉ50 mH
9j5
9+j5 Convert the sources to phasors: ๐ฃ๐ฃ๐๐ ๐ก๐ก = 14.14cos(๐๐๐ก๐ก + 45ยฐ) becomes phasor ๐๐๐๐ = 14.14โ ๐๐ยฐ๐๐๐๐ ๐ก๐ก = 2 sin ๐๐๐ก๐ก = 2cos(๐๐๐ก๐ก โ 90ยฐ) becomes phasor ๐ผ๐ผ๐๐ = 2โ โ 90ยฐ = โ๐๐2
๐ผ๐ผ๐ ๐ = โ๐๐2
๐๐๐๐ =10 + ๐๐10
๐ผ๐ผ3
9+j5
8-j18 -j9
3+j312
๐ผ๐ผ1 ๐ผ๐ผ2
๐ผ๐ผ๐ ๐ = โ๐๐2
๐๐๐๐ =10 + ๐๐10
๐ผ๐ผ3
9+j5
8-j18 -j9
3+j312
๐ผ๐ผ1 ๐ผ๐ผ2
Write the mesh equations:
Constraint equation: ๐ผ๐ผ2 โ ๐ผ๐ผ1 = โ๐๐2
Mesh path ABCDEA: โ12๐ผ๐ผ1 โ 8 โ ๐๐๐๐ ๐ผ๐ผ1 โ ๐ผ๐ผ3 โ โ๐๐๐ ๐ผ๐ผ2 โ ๐ผ๐ผ3 โ 3 + ๐๐3 ๐ผ๐ผ2 โ 10 + ๐๐๐๐ = 0
Mesh path BFGDCB: โ 9 + ๐๐๐ ๐ผ๐ผ3 + โ๐๐๐ ๐ผ๐ผ2 โ ๐ผ๐ผ3 + 8 โ ๐๐๐๐ ๐ผ๐ผ1 โ ๐ผ๐ผ3 = 0
A
B DC
E
FG
= 10 + ๐๐10
Collect terms:Mesh path ABCDEA: โ12๐ผ๐ผ1 โ 8 โ ๐๐๐๐ ๐ผ๐ผ1 โ ๐ผ๐ผ3 โ โ๐๐๐ ๐ผ๐ผ2 โ ๐ผ๐ผ3 โ 3 + ๐๐3 ๐ผ๐ผ2 โ (10 + ๐๐๐๐) = 0
โ12 โ 8 + ๐๐๐๐ ๐ผ๐ผ1 + ๐๐๐ โ 3 โ ๐๐3 ๐ผ๐ผ2 + 8 โ ๐๐๐๐ โ ๐๐๐ ๐ผ๐ผ3 = 10 + ๐๐๐๐
โ20 + ๐๐๐๐ ๐ผ๐ผ1 + โ3 + ๐๐6 ๐ผ๐ผ2 + 8 โ ๐๐27 ๐ผ๐ผ3 = 10 + ๐๐๐๐
Mesh path BFGDCB: โ 9 + ๐๐๐ ๐ผ๐ผ3 โ โ๐๐๐ ๐ผ๐ผ3 โ ๐ผ๐ผ2 โ 8 โ ๐๐๐๐ ๐ผ๐ผ3 โ ๐ผ๐ผ1 = 0
8 โ ๐๐๐๐ ๐ผ๐ผ1 โ ๐๐๐๐ผ๐ผ2 + โ9 โ ๐๐๐ + ๐๐๐ โ 8 + ๐๐๐๐ ๐ผ๐ผ3 = 0
8 โ ๐๐๐๐ ๐ผ๐ผ1 โ ๐๐๐๐ผ๐ผ2 + โ17 + ๐๐22 ๐ผ๐ผ3 = 0
Current generator: ๐ผ๐ผ2 โ ๐ผ๐ผ1 = โ๐๐2
The mesh equations:
โ20 + ๐๐๐๐ ๐ผ๐ผ1 + โ3 + ๐๐๐ ๐ผ๐ผ2 + 8 โ ๐๐2๐ ๐ผ๐ผ3 = 10 + ๐๐๐๐8 โ ๐๐๐๐ ๐ผ๐ผ1 โ ๐๐๐๐ผ๐ผ2 + โ17 + ๐๐22 ๐ผ๐ผ3 = 0๐ผ๐ผ2 โ ๐ผ๐ผ1 = โ๐๐2
Solve the mesh equations:
๐ผ๐ผ1 = 0.5919โ ๐๐๐.2ยฐ๐ผ๐ผ2 = 1.884โ โ 107.2ยฐ๐ผ๐ผ3 = 0.8574โ โ 172.0ยฐ
Node Equations in the โPhasor Domainโ1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign a datum node and node voltages ๐๐1, ๐๐2, ๐๐3, โฆ3. Write a KCL equation for each node.
Example - really easy!
The operating frequency is 60 Hz.
๐ฃ๐ฃ1 ๐ก๐ก = 10 cos๐๐๐ก๐ก๐ฃ๐ฃ2 ๐ก๐ก = 16 sin๐๐๐ก๐ก
Write node equations for this circuit. Find the voltage across 5 ohms in series with 663.1 microfarads.
+๐ฃ๐ฃ1(๐ก๐ก)โ
+๐ฃ๐ฃ2(๐ก๐ก)โ
2 ฮฉ5 ฮฉ
1 ฮฉ
663.1 ยตF
5.3 mH 10.6 mH
Convert the circuit into the โphasor domainโ:
We converted this circuit to the phasor domain on a previous slide!โข Sources:๐ฃ๐ฃ1 ๐ก๐ก = 10 cos๐๐๐ก๐ก becomes phasor ๐๐1 = 10โ ๐ยฐ๐ฃ๐ฃ2 ๐ก๐ก = 16 sin๐๐๐ก๐ก = 16cos(๐๐๐ก๐ก โ 90ยฐ) becomes phasor ๐๐2 = 16โ โ 90ยฐ = โ๐๐๐๐โข The operating frequency is ๐๐ =60 Hz. The radian frequency is
๐๐ = 2๐๐๐๐ = 2๐๐๐๐๐๐ = ๐2๐๐๐ = 376.99 โ 377 radians/second. โข For the 5.3 mH inductance, ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ = ๐๐๐๐3๐๐๐๐๐.3๐๐10โ3 =
๐๐๐.9981 โ ๐๐2 ohmsโข The 10.6 mH inductance has an impedance of ๐๐๐ ohms.โข The 663.1 microfarad capacitance has an impedance of ๐๐๐ถ๐ถ =
1๐๐๐๐๐ถ๐ถ
= 1๐๐๐๐๐ถ๐ถ
โ๐๐โ๐๐
= โ๐๐๐๐๐ถ๐ถ
= โ๐๐377๐ฅ๐ฅ663.1๐ฅ๐ฅ1๐ฅโ6
= โ๐๐๐ ohms.
+๐ฃ๐ฃ1(๐ก๐ก)โ
+๐ฃ๐ฃ2(๐ก๐ก)โ
2 ฮฉ5 ฮฉ
1 ฮฉ
663.1 ยตF
5.3 mH 10.6 mH๐๐๐๐
2 ฮฉ5 ฮฉ
1 ฮฉ
โj4 ฮฉ
๐๐2 ฮฉ ๐๐๐ ฮฉ10
๐๐๐๐10
2 + j2 ฮฉ 1 + j4 ฮฉ
5 โ j4 ฮฉ
Choose a datum and a node voltage:
Write a KCL equation at the node:
10 โ ๐๐12 + ๐๐2 โ
๐๐15 โ ๐๐๐ โ
๐๐1 โ โ๐๐๐๐1 + ๐๐๐ = 0
Collect terms:โ๐๐1
2 + ๐๐2 โ๐๐1
5 โ ๐๐๐ โ๐๐1
1 + ๐๐๐ =โ10
2 + ๐๐2 +๐๐๐๐
1 + ๐๐๐โ1
2 + ๐๐2 โ1
5 โ ๐๐๐ โ1
1 + ๐๐๐ ๐๐1 =โ10
2 + ๐๐2 +๐๐๐๐
1 + ๐๐๐
One way to solve this quation is to use a common denominator for all the terms:โ 5 โ ๐๐๐ 1 + ๐๐๐ โ 2 + ๐๐2 1 + ๐๐๐ โ (2 + ๐๐2)(5 โ ๐๐๐)
2 + ๐๐2 (5 โ ๐๐๐)(1 + ๐๐๐) ๐๐1
=โ10 1 + ๐๐๐ + ๐๐๐๐(2 + ๐๐2)
(2 + ๐๐2)(1 + ๐๐๐)A better way is to do the division term by term; see next page.
๐๐๐๐10
2 + j2 ฮฉ 1 + j4 ฮฉ
5 โ j4 ฮฉ
๐๐1
Solve:โ1
2 + ๐๐2โ
15 โ ๐๐๐
โ1
1 + ๐๐๐๐๐1 =
โ102 + ๐๐2
+๐๐๐๐
1 + ๐๐๐
Change the coefficients from ratios to simple rectangular form using your calculator:
โ0.25 + ๐๐๐.25 โ 0.1220 โ ๐๐๐.09756 โ 0.05882 + ๐๐๐.2353 ๐๐1 = โ2.5 + j2.5 + 3.765 + j0.9412
Do the addition of real parts and of imaginary parts:โ0.4302 + ๐๐๐.3877 ๐๐1 = 1.265 + ๐๐3.441
Do the division:
๐๐1 =1.265 + ๐๐3.441
โ0.4302 + ๐๐๐.3877 = 2.355 โ ๐๐๐.876 = 6.33๐โ โ 68.1ยฐ
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. Iโd use a short computer program to solve the equation.
Node Equations with a current generator
Example #2
Procedure for writing node equations in the โfrequency domainโ:1. Convert all the components to impedances at the operating frequency. Convert the sources to phasors. 2. Assign a datum node and node voltages ๐๐1, ๐๐2, ๐๐3, โฆ3. Write a KCL equation for each node.
The operating frequency is 60 Hz.The sources are๐ฃ๐ฃ1 ๐ก๐ก = 10 cos๐๐๐ก๐ก๐ฃ๐ฃ2 ๐ก๐ก = 2 sin๐๐๐ก๐กWrite a node equation for the voltage across the current generator.
10 cos๐๐๐ก๐ก
2 ฮฉ 1 ฮฉ
331.6 ยตF
5.3 mH 10.6 mH
2 sin๐๐๐ก๐ก
Step 1: Convert to the phasor domain.
We did this conversion for mesh analysis, see above!
Convert the sources.๐ฃ๐ฃ๐๐ ๐ก๐ก = 10 cos๐๐๐ก๐ก becomes phasor ๐๐1 = 10โ ๐ยฐ๐๐๐๐ ๐ก๐ก = 2 sin๐๐๐ก๐ก = 2 cos(๐๐๐ก๐ก โ 90ยฐ) becomes ๐๐2 = 2โ โ 90ยฐ = โ2๐๐
The operating frequency is ๐๐ =60 Hz. The radian frequency is ๐๐ = 2๐๐๐๐ = 2๐๐๐๐๐๐ = ๐2๐๐๐ = 376.99 โ 377 radians/second.
For the 5.3 mH inductance, ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ = ๐๐๐๐3๐๐๐๐๐.3๐๐10โ3 = ๐๐๐.9981 โ ๐๐2 ohms
The 10.6 mH inductance has an impedance of ๐๐๐ ohms.
The 331.6 microfarad capacitance has an impedance of ๐๐๐ถ๐ถ = 1
๐๐๐๐๐ถ๐ถ= 1
๐๐๐๐๐ถ๐ถโ๐๐โ๐๐
= โ๐๐๐๐๐ถ๐ถ
= โ๐๐377๐ฅ๐ฅ331.6๐ฅ๐ฅ1๐ฅโ6
= โ๐๐๐ ohms.
10 cos๐๐๐ก๐ก
2 ฮฉ 1 ฮฉ
331.6 ยตF
5.3 mH 10.6 mH
2 sin๐๐๐ก๐ก 10
2 ฮฉ 1 ฮฉ j4 ฮฉ
-j8 ฮฉโ๐๐2j2 ฮฉ
2+j2 ฮฉ
10 โ๐๐2
1-j4 ฮฉ๐๐1
Write the node equation at ๐๐1:
10 โ ๐๐12 + ๐๐2 + โ2๐๐ โ
๐๐11 โ ๐๐๐ = 0
โ๐๐12 + ๐๐2 โ
๐๐11 โ ๐๐๐ = 2๐๐ โ
102 + ๐๐2
โ12 + ๐๐2 โ
11 โ ๐๐๐ ๐๐1 = 2๐๐ โ
102 + ๐๐2
Homework: Solve the equation.Verify the solution with Spice.
Step 2: Assign a datum node and node voltages ๐๐1, ๐๐2, ๐๐3, โฆStep 3: Write a KCL equation for each node.
2+j2 ฮฉ
10 โ๐๐2
1-j4 ฮฉ๐๐1
Example: Node equations with a supernode.
The frequency is 1950 MHz. Write node equations for this circuit.Find the amplitude and phase of ๐๐๐๐.
Remark: โข A voltage generator embedded inside a node is called a โsupernodeโ. โข Only one unknown node voltage is needed for a supernode.โข We write a KCL equation for the closed surface that encloses the voltage generator.
2 cos ๐๐๐ก๐ก5 cos ๐๐๐ก๐ก + 45ยฐ
81.63 pH 2 ฮฉ
5 ฮฉ16.32 pF
10 ฮฉ
+๐๐๐๐โ
2 cos ๐๐๐ก๐ก5 cos ๐๐๐ก๐ก + 45ยฐ
81.63 pH 2 ฮฉ
5 ฮฉ 16.32 pF10 ฮฉ
Convert to phasors and impedances:
Sources:๐๐๐๐ ๐ก๐ก = 2 cos๐๐๐ก๐ก is represented by phasor ๐ผ๐ผ๐๐ = 2๐ฃ๐ฃ๐๐ ๐ก๐ก = 5 cos(๐๐๐ก๐ก + 45ยฐ) is represented by phasor ๐๐๐๐ = 5โ ๐๐ยฐ
๐๐ = 1950 MHz so ๐๐ = 2๐๐๐๐ =1.22๐๐๐101๐ฅ rad/sec
๐๐๐๐๐๐ = ๐๐๐๐๐.22๐๐๐101๐ฅ๐๐๐.๐๐3๐๐10โ12 = ๐๐๐ohm1
๐๐๐๐๐ถ๐ถ= โ๐๐
1.225๐ฅ๐ฅ1๐ฅ10๐ฅ๐ฅ16.32๐ฅ๐ฅ1๐ฅโ12= โ๐๐๐ ohms
25โ 45ยฐ
๐๐๐ ฮฉ 2 ฮฉ
5 ฮฉ
โ๐๐๐ ฮฉ10 ฮฉ
+๐๐๐๐โ
Combine impedances:
25โ 45ยฐ
๐๐๐ ฮฉ 2 ฮฉ
5 ฮฉโ๐๐๐ ฮฉ10 ฮฉ
2 10 ฮฉ5โ 45ยฐ
Series:๐๐๐ in series with 2 is equivalent to (2 + ๐๐๐) ohms
Parallel:5 in parallel with โ๐๐๐ is equivalent to๐๐๐ โ๐๐๐
5 โ ๐๐๐=โ๐๐2๐5 โ ๐๐๐
= 2.5 โ ๐๐2.5
2 + ๐๐๐ ฮฉ
2.5 โ ๐๐2.5 ฮฉ+๐๐๐๐โ
Node equations:
2 10 ฮฉ5โ 45ยฐ 2 + ๐๐๐ ฮฉ
2.5 โ ๐๐2.5 ฮฉ
๐๐1 ๐๐2 ๐๐๐ฅ
Constraint equation: ๐๐2 โ ๐๐1 = 5โ ๐๐ยฐ
At the supernode:
2 โ๐๐110 โ
๐๐2 โ ๐๐๐ฅ2 + ๐๐๐ = 0
At the ๐๐๐ฅ node: ๐๐2 โ ๐๐๐ฅ2 + ๐๐๐ โ
๐๐๐ฅ2.5 โ ๐๐2.5 = 0
+๐๐๐๐โ
Use three nodes:
For the supernode: ๐๐1 and ๐๐2.
For the output, use ๐๐๐ฅ.
There are three nodes so three equations to solve, in three unknowns.
Solve to find:๐๐๐๐ = 5.772 โ โ 30.5ยฐ
Amplitude of ๐๐๐ฅ is 5.772 voltsPhase of ๐๐๐ฅ is -30.5 degrees.
Can we solve this circuit โsmarterโ?
2 10 ฮฉ5โ 45ยฐ 2 + ๐๐๐ ฮฉ
2.5 โ ๐๐2.5 ฮฉ
+๐๐๐๐โ
๐๐1Use only one unknown node voltage, ๐๐1.
Build the constraint equation into the supernode directly.
Once the value of ๐๐1 has been found, use the voltage-divider relationship to find ๐๐๐ฅ:
๐๐๐ฅ = 2.5 โ ๐๐2.5๐๐1 + 5โ ๐๐ยฐ
2 + ๐๐๐ + 2.5 โ ๐๐2.5
We can solve the node equation to find๐๐1 = 5.884โ โ 44.2ยฐAnd then the voltage divider to find๐๐๐ฅ = 5.772โ โ 44.2ยฐ
This agrees with the โdumberโ solution!
๐๐1 + 5โ ๐๐ยฐ
We only need one node equation to find node voltage, ๐๐1.At the supernode:
2 โ๐๐110 โ
๐๐1 + 5โ ๐๐ยฐ2 + ๐๐๐ + 2.5 โ ๐๐2.5 = 0
So
โ๐๐110 โ
๐๐14.5 โ ๐๐๐.5 = โ2 +
5โ ๐๐ยฐ4.5 โ ๐๐๐.5
Node Equations in the โPhasor DomainโWe did this example using mesh analysis. Letโs do it by node analysis.
Convert the components to impedances: ๐๐ = 100 rad/sec.
The frequency is ๐๐ = 100 rad/sec.๐ฃ๐ฃ๐๐ ๐ก๐ก = 14.14cos(๐๐๐ก๐ก + 45ยฐ)๐๐๐๐ ๐ก๐ก = 2 sin ๐๐๐ก๐ก = 2cos(๐๐๐ก๐ก โ 90ยฐ)
Write a set of node equations in matrix form.
๐๐ = 20 mH ๐๐๐ฟ๐ฟ = ๐๐๐๐๐๐ = ๐๐๐๐๐๐๐2๐๐๐10โ3 = ๐๐2๐ถ๐ถ = 0.5 mF ๐๐๐ถ๐ถ = 1
๐๐๐๐๐ถ๐ถ= โ๐๐
1๐ฅ๐ฅ๐ฅ๐ฅ๐ฅ.5๐ฅ๐ฅ1๐ฅโ3= โ๐๐2๐
8 ฮฉ 20 mH 0.5 mF 8 j2 -j20 8-j18
30 mH
8 ฮฉ
j3
33+j3
9 ฮฉ50 mH
9j5
9+j5
Draw the circuit with phasors and impedances:Convert the sources to phasors: ๐ฃ๐ฃ๐๐ ๐ก๐ก = 14.14cos(๐๐๐ก๐ก + 45ยฐ) becomes phasor ๐๐๐๐ = 14.14โ ๐๐ยฐ๐๐๐๐ ๐ก๐ก = 2 sin ๐๐๐ก๐ก = 2cos(๐๐๐ก๐ก โ 90ยฐ) becomes phasor ๐ผ๐ผ๐๐ = 2โ โ 90ยฐ = โ๐๐2
Write Node EquationsNode 1:
โ๐๐112
โ๐๐1 โ ๐๐28 โ ๐๐๐๐
โ๐๐1 โ ๐๐39 + ๐๐๐
= 0
Node 2:๐๐1 โ ๐๐28 โ ๐๐๐๐
+ (โ๐๐2) โ๐๐2 โ ๐๐3โ๐๐9
= 0
Node 3:๐๐1 โ ๐๐39 + ๐๐๐ +
๐๐2 โ ๐๐3โ๐๐๐ โ
๐๐3 โ 14.14โ 45ยฐ3 + ๐๐3 = 0
Collect terms in the node equations:Node 1: Multiply by 12(8 โ ๐๐๐๐)(9 + ๐๐๐)
โ๐๐112
โ๐๐1 โ ๐๐28 โ ๐๐๐๐
โ๐๐1 โ ๐๐39 + ๐๐๐
= 0
โ(8 โ ๐๐๐๐)(9 + ๐๐๐)๐๐1 โ 12(9 + ๐๐๐)(๐๐1โ๐๐2) โ 12(8 โ ๐๐๐๐)(๐๐1 โ ๐๐3) = 0โ(72 โ ๐๐๐๐2 + ๐๐๐๐ + 90)๐๐1 โ (108 + ๐๐60)(๐๐1โ๐๐2) โ (96 โ ๐๐216)(๐๐1 โ ๐๐3) = 0โ(162 โ ๐๐๐22)๐๐1 โ (108 + ๐๐๐๐)(๐๐1โ๐๐2) โ (96 โ ๐๐2๐๐)(๐๐1 โ ๐๐3) = 0โ162 + ๐๐๐22 โ 108 โ ๐๐๐๐ โ 96 + ๐๐2๐๐ ๐๐1 + 108 + ๐๐๐๐ ๐๐2 + (96 โ ๐๐2๐๐)๐๐3 = 0โ366 + ๐๐2๐๐ ๐๐1 + 108 + ๐๐๐๐ ๐๐2 + (96 โ ๐๐2๐๐)๐๐3 = 0
Node 2:๐๐1 โ ๐๐28 โ ๐๐๐๐ + (โ๐๐2) โ
๐๐2 โ ๐๐3โ๐๐9 = 0
(โ๐๐๐)(๐๐1 โ ๐๐2) + (8 โ ๐๐๐๐)(โ๐๐๐)(โ๐๐2) โ (8 โ ๐๐๐๐)(๐๐2 โ ๐๐3) = 0(โ๐๐๐)(๐๐1 โ ๐๐2) + (โ๐๐๐2 โ 162)(โ๐๐2) โ (8 โ ๐๐๐๐)(๐๐2 โ ๐๐3) = 0(โ๐๐๐)(๐๐1 โ ๐๐2) + (โ144 + ๐๐32๐) โ (8 โ ๐๐๐๐)(๐๐2 โ ๐๐3) = 0โ๐๐๐ ๐๐1 + (๐๐๐ โ 8 + ๐๐๐๐)๐๐2+(8 โ ๐๐๐๐)๐๐3 = โ โ144 + ๐๐32๐โ๐๐๐ ๐๐1 + (โ8 + ๐๐27)๐๐2+(8 โ ๐๐๐๐)๐๐3 = โ โ144 + ๐๐32๐
Arrange the equations into matrix form:Node 3:๐๐1 โ ๐๐39 + ๐๐๐
+๐๐2 โ ๐๐3โ๐๐๐
โ๐๐3 โ 14.1โ 45ยฐ
3 + ๐๐3= 0
(โ๐๐๐)(3 + ๐๐3)(๐๐1 โ ๐๐3) + (9 + ๐๐๐)(3 + ๐๐3)(๐๐2 โ ๐๐3) โ (9 + ๐๐๐)(โ๐๐๐)(๐๐3 โ 14.1โ 45ยฐ) = 0(โ๐๐27 + 27)(๐๐1 โ ๐๐3) + (27 + ๐๐15 + ๐๐2๐ โ 15)(๐๐2 โ ๐๐3) โ (โ๐๐๐๐ + 45)(๐๐3 โ 14.1โ 45ยฐ) = 0(โ๐๐2๐ + 27)(๐๐1 โ ๐๐3) + (12 + ๐๐42)(๐๐2 โ ๐๐3) โ (โ๐๐๐๐ + 45)๐๐3 = (โ๐๐๐๐ + 45)(โ14.๐โ ๐๐ยฐ)(โ๐๐2๐ + 27)๐๐1 + (12 + ๐๐๐2)๐๐2 + (๐๐2๐ โ 27 โ 12 โ ๐๐๐2 + ๐๐๐๐ โ 45)๐๐3 = (โ๐๐๐๐ + 45)(โ14.๐โ ๐๐ยฐ)โ๐๐2๐ + 27 ๐๐1 + 12 + ๐๐๐2 ๐๐2 + โ84 + ๐๐66 ๐๐3 = โ1256 + ๐๐3๐๐.9
Hence the node equations are:โ366 + ๐๐2๐๐ ๐๐1 + 108 + ๐๐๐๐ ๐๐2 + (96 โ ๐๐2๐๐)๐๐3 = 0โ๐๐๐ ๐๐1 + (โ8 + ๐๐2๐)๐๐2+(8 โ ๐๐๐๐)๐๐3 = โ โ144 + ๐๐32๐โ๐๐2๐ + 27 ๐๐1 + 12 + ๐๐๐2 ๐๐2 + โ84 + ๐๐๐๐ ๐๐3 = โ1256 + ๐๐3๐๐.9
Arrange the node equations into a matrix equation:โ366 + ๐๐2๐๐ 108 + ๐๐๐๐ (96 โ ๐๐2๐๐)
โ๐๐๐ (โ8 + ๐๐2๐) (8 โ ๐๐๐๐)โ๐๐2๐ + 27 12 + ๐๐๐2 โ84 + ๐๐๐๐
๐๐1๐๐2๐๐3
=0
144 โ ๐๐32๐โ1256 + ๐๐3๐๐.9
๐๐1 = 7.091โ โ 19.8ยฐ๐๐2 = 1.448โ 168.9ยฐ๐๐3 = 14.00โ ๐2.0ยฐ
top related