1 my chapter 21 lecture outline. 2 chapter 21: alternating currents sinusoidal voltages and currents...
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MyChapter 21
LectureOutline
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Chapter 21: Alternating Currents
•Sinusoidal Voltages and Currents
•Capacitors, Resistors, and Inductors in AC Circuits
•Series RLC Circuits
•Resonance
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§21.1 Sinusoidal Currents and Voltage
A power supply can be set to give an EMF of form:
tt ωεε sin)( 0=
This EMF is time dependent, has an amplitude ε0, and varies with angular frequency ω.
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fπω 2=
angular frequency in rads/sec
frequency in cycles/sec or Hz
The current in a resistor is still given by Ohm’s Law:
tItRR
ttI ωωεε
sinsin)(
)( 00 ===
The current has an amplitude of I0 = ε0/R.
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The instantaneous power dissipated in a resistor will be:
( )( ) tIttI
tVtIP R
ωεωεω 20000 sinsinsin
)()(
==
=
The power dissipated depends on t (where in the cycle the current/voltage are).
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What is the average power dissipated by a resistor in one cycle?
The average value sin2ωt over one cycle is 1/2.
.2
100av εIP =The average power is
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What are the averages of V(t) and I(t) over one cycle?
The “problem” here is that the average value of sin ωt over one complete cycle is zero! This is not a useful way to characterize the quantities V(t) and I(t).
To fix this problem we use the root mean square (rms) as the characteristic value over one cycle.
2 and
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rms0
rms
εε ==I
I
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In terms of rms quantities, the power dissipated by a resistor can be written as:
R
222
1
2rms2
rmsrmsrms
0000av
εε
εε
===
==
RII
IIP
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Example (text problem 21.4): A circuit breaker trips when the rms current exceeds 20.0 A. How many 100.0 W light bulbs can run on this circuit without tripping the breaker? (The voltage is 120 V rms.)
Each light bulb draws a current given by:
( )Amps 83.0
V 120 Watts100
rms
rms
rmsrmsav
=
=
=
I
I
IP ε
If 20 amps is the maximum current, and 0.83 amps is the current drawn per light bulb, then you can run 24 light bulbs without tripping the breaker.
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Example (text problem 21.10): A hair dryer has a power rating of 1200 W at 120 V rms. Assume the hair dryer is the only resistance in the circuit.
(a) What is the resistance of the heating element?
( )
Ω=
=
=
12
V 120 Watts1200
2
2rms
av
RR
RP
ε
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(b) What is the rms current drawn by the hair dryer?
( )Amps 10
V 120 Watts1200
rms
rms
rmsrmsav
=
=
=
I
I
IP ε
(c) What is the maximum instantaneous power that the resistance must withstand?
00max2
00 sin εωε IPtIP =⇒= 00av 2
1 εIP =
Pmax = 2Pav = 2400 Watts
Example continued:
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§21.3-4 Capacitors, Resistors and Inductors in AC circuits
)()( tCVtQ C=For a capacitor:
In the circuit: ⎟⎠
⎞⎜⎝
⎛Δ
Δ=
ΔΔ
=t
tVC
t
tQtI C )()()(
Slope of the plot V(t) vs. t
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The current in the circuit and the voltage drop across the capacitor are 1/4 cycle out of phase. Here the current leads the voltage by 1/4 cycle.
Here it is true that VCI. The equality is Vc = IXC where XC is called capacitive reactance. (Think Ohm’s Law!)
CX
ω1
C = Reactance has units of ohms.
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For a resistor in an AC circuit, .)()( RtItV =
The voltage and current will be in phase with each other.
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For an inductor in an AC circuit:
⎟⎠
⎞⎜⎝
⎛Δ
Δ=
t
tILV
)(L
Also, VL = IXL where the inductive reactance is: LX ω=L
Slope of an I(t) vs. t plot
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The current in the circuit and the voltage drop across the inductor are 1/4 cycle out of phase. Here the current lags the voltage by 1/4 cycle.
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Plot of I(t), V(t), and P(t) for a capacitor.
The average power over one cycle is zero. An ideal capacitor dissipates no energy.
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A similar result is found for inductors; no energy is dissipated by an ideal inductor.
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§21.5 Series RLC Circuits
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( )
( ) ⎟⎠
⎞⎜⎝
⎛ −++⎟⎠
⎞⎜⎝
⎛ +=
+=
2sinsin
2sin
sin)(
CRL
0
πωω
πω
φωεε
tVtVtV
tt
Applying Kirchhoff’s loop rule:
0)()()()( CRL =−−− tVtVtVtε
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To find the amplitude (ε0) and phase () of the total voltage we add VL, VR, and VC together by using phasors.
( )( ) ( )
( )IZ
XXRI
IXIXIR
VVV
=
−+=
−+=
−+=
2CL
2
2CL
2
2CL
2R0ε
Z is called impedance.
X
y
VR
VL
VC
ε0
VL VC
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The phase angle between the current in the circuit and the input voltage is:
Z
RV
R
XX
V
VV
==
−=
−=
0
R
CL
R
CL
cos
tan
εφ
φ
> 0 when XL > XC and the voltage leads the current (shown above).
< 0 when XL < XC and the voltage lags the current.
X
y
VR
VL
VC
ε0
VL VC
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Example (text problem 21.86): In an RLC circuit these three elements are connected in series: a resistor of 20.0 Ω, a 35.0 mH inductor, and a 50.0 F capacitor. The AC source has an rms voltage of 100.0 V and an angular frequency of 1.0103 rad/sec. Find…
(a) The reactances of the capacitor and the inductor.
Ω==
Ω==
0.201
0.35
C
L
CX
LX
ω
ω
(b) The impedance.
( ) Ω=−+= 0.252CL
2 XXRZ
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(c) The rms current:
Amps 00.4 25.0
V 0.100
Zrms
rms
rmsrms
=Ω
==
=ε
ε
I
ZI
(d) The current amplitude:
Amps 66.52
2
rms0
0rms
==
=
II
II
Example continued:
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(e) The phase angle:
( ) rads 644.075.0tan
75.020
2035tan
1
CL
==
=Ω
Ω−Ω=
−=
−φ
φR
XX
(f) The rms voltages across each circuit element:
V 0.80
V 140
V 0.80
CrmsC,rms
LrmsL,rms
rmsR,rms
==
==
==
XIV
XIV
RIV
(Or 37°)
Example continued:
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(g) Does the current lead or lag the voltage?
(h) Draw a phasor diagram.
Since XL > XC, is a positive angle. The voltage leads the current.
Example continued:
y
XVR
VL
VC
εrms
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The power dissipated by a resistor is:
φεε cosrmsrmsRrms,rmsav IIP ==
where cos is called the power factor (compare to slide 7; Why is there a difference?).
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§21.6 Resonance in RLC Circuits
A plot of I vs. ω for a series RLC circuit has a peak at ω = ω0.
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The peak occurs at the resonant frequency for the circuit.
( )2CL2 XXRZ
I−+
==εε
The current will be a maximum when Z is a minimum. This occurs when XL = XC (or when Z = R).
LC
CL
XX
1
1
0
00
CL
=
=
=
ω
ωω
This is the resonance frequency for the circuit.
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At resonance:
1cos
0tan CL
==
=−
=
RR
RXX
φ
φ
The phase angle is 0; the voltage and the current are in phase. The current in the circuit is a maximum as is the power dissipated by the resistor.
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Summary
•Difference Between Instantaneous, Average, and rms Values
• Power Dissipation by R, L, and C
•Reactance for R, L, and C
•Impedance and Phase Angle
•Resonance in an RLC Circuit