Transcript Document

Admin:
• Assignment 6 is posted. Due Monday.
• Collect your exam at the end of class
• Erratum: US mains is 60Hz (I said 50Hz last lecture!)
An example:
1+2j
An example:
Im
1+2j
tan-1(2/1)
θ=
= 63.4°
A=sqrt(12+22) = 2.24
Phasor form = 2.24 63.4°
= 2.24 e j63.4°
= 2.24(cos[63.4°] + sin[63.4°])
Real part = 1 = A cosθ
Imaginary part = 2 = A sinθ
*
Re
When adding or subtracting complex numbers, the rectangular form is simplest:
c1 = a1 + jb1
c2 = a2 + jb2
c1 + c2 = (a1 + a2) + j(b1 + b2)
Example: (1 + 2 j) + (-1 + 4j) = 0 + 6j
When multiplying or dividing, the polar or phasor forms are simplest:
θ=tan-1(b/a)
Example: (1 + 2 j)2 = 2.24 63.4° × 2.24
=5
63.4°
126.8° =5cos(126.8°) + j5sin(126.8°)= -3 + 4j
i-V relationships in AC circuits:
Resistors
Source vs(t)=Asinωt
vR(t)= vs(t)=Asinωt
iR (t ) 
vR (t ) A
 sin t
R
R
vR(t) and iR(t) are in phase
Complex representation: vS(t)=Asinωt=Acos(ωt-90)=real part of [VS(jω)]
where VS(jω) is the complex number A[cos(ωt-90)+jsin(ωt-90 )]=Aej (ωt-90)
Phasor representation: VS(jω) =A(ωt-90)
IS(jω)=(A/R) (ωt-90)
Impedance=complex number of Resistance Z=VS(jω)/IS(jω)=R
Generalized Ohm's Law: VS(jω)=ZIS(jω)
Play animation
Note – no phase
dependence for a resistor
Capacitors in AC circuits
No charge flows through a capacitor
A Capacitor in a DC circuit acts like a break (an open circuit)
But in AC circuits charge build-up and discharge mimics a
current.
V
-V
Capacitors in AC circuits
Capacitive Load
vC = Asin w t
qC = CvC
dqC
= wCA cos(w t)
dt
VC ( jw ) = AÐ(w t - 90) "capacitive reactance”
XC=1/ωC
I C ( jw ) = wCAÐ(w t - 0)
iC =
ZC =
VC ( jw ) 1
=
Ð - 90
w
C
I C ( jw )
cos(90)  j sin(90)   j
=
- j - j. j
1
=
=
wC jwC jwC "capacitive impedance”
• Voltage and current not in phase:
• Current leads voltage by 90 degrees (Physical - current must conduct charge to capacitor
plates in order to raise the voltage)
• Impedance of Capacitor decreases with increasing frequency
Play animation
30-7 AC Circuits with AC Source
Example 30-10: Capacitor reactance.
What is the rms current in the circuit shown if C = 1.0 μF
and Vrms = 120 V?
Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 105 Hz.
ZC =
1
jwC
XC=1/ωC
Note: we can simply use reactance (XC) if dealing with phase
independent quantities like rms. But the answer is not valid at any
given instant in time: for that we need the impedance (ZC)
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