Chapter 2 - Inductors, Capacitors and Alternating Current Circuits
CHAPTER 22 ALTERNATING CURRENT - Physics …people.physics.tamu.edu/adair/phys202/CHAPTER 22...
Transcript of CHAPTER 22 ALTERNATING CURRENT - Physics …people.physics.tamu.edu/adair/phys202/CHAPTER 22...
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CHAPTER 22
ALTERNATING CURRENT
Alternating Currents
Phasors
Reactance
Impedance
Power in AC Circuits
R-L-C Circuits
Resonance
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For Resistor
Consider voltage that is sinusoidal.
� = ������
Note that V is the maximum value of voltage and v varies from 0
to a maximum of V.
Also, a current that is sinusoidal.
= �����
Note that is the maximum value of voltage and varies from 0 to
a maximum of .
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Fig 22.1
Can represent using phasors. Phasors are just tools to help
visualize.
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The
phasor, the vector from the origin to the curved line, has the
length equal to the magnitude of the current or voltage or
whatever the phasor represents.
The projection on the horizontal axis represents the
instantaneous value of the quantity.
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AVERAGE VALUES
Values for the ac quantities used to make calculations of averages
are called root-mean-square values.
They can be obtained from the max values, the amplitude, by:
�� = √2
��� = �√2
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Resistor in ac circuit with current
= �����
Instantaneous voltage across resistor is
�� = � = ������
�� = �������
��������ℎ cos ��������
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Inductor in ac Circuit with current = �����
�� = � ΔΔ�
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�� = � ΔΔ�
Current changes rapidly - Voltage large
Current doesn’t change - Voltage Zero
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It can be shown that
(Calculus by differentiating, ! " )
�� = −������ = �����(�� + &)
Where & is the phase angle. For inductor that is 900.
Inductive Reactance,
Defined as (� = ��
By analogy voltage for inductor �� = (�
Now Capacitor in ac Circuit with current = �����
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)* = +*
Therefore
Δ)*Δt = 1. ΔqΔt = 1. = 1. �����
Thus a sinusoidal curve as seen in Fig. 22.8.
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Thus the voltage curve lags the current curve by 900.
And we have (can be derived by calculus)
�0 = 1�. ����
Or
�0 = 1�. cos(�� − 903)
Or �0 = �0cos(�� − 903)
Capacitive reactance
(0 = 1�.
And �0 = (0
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IMPEDANCE
Impedance of a circuit is the sum of the resistance and reactance
in the circuit. The symbol for impedance is Z.
4 = 5�6 + (6
4 = 5�6 + ((� − (0)6
4 = 7�6 + 8�� − 1�.96
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Consider
(Across Resistor)
What is reactance of the capacitor?
(0 = 1�. = 1:2500 <�� �= > (5.0@10ABC) = 80Ω
What is the current in the circuit?
= ��� = 1.2��� :2500 <�� �= > �300Ω
= 4@10AH��� :2500 <�� �= > �
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What is the impedance?
4 = 7�6 +8�� − 1�.96
4 = 7�6 −8 1�.96
4 = 5(300Ω)6 −(80Ω)6 = 290Ω
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Power in ac circuits
Remember this about power for pure resistance
IJKL = 12 � = 1√2 1√2 � = 1√2 1√2� = �� ���
But with capacitors or inductors
MN)O = PQRSTUVW = R√Q S√QTUVW = SXYVRXYVTUVW
cosW is called The Power Factor
Also I�Z[<C����< = �\
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R-L-C Circuits
Let � = 300Ω � = 60^_ . = 0.50`C
� = 50� � = 10,000 <�� �=
What is the reactance of each component, the reactance of the
circuit and the impedance of the circuit?
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(� = �� = :10,000 <�� �= >(60@10AH) = 600Ω
(0 = 1�. = 1:10,000 <�� �= > (0.50@10ABC) = 200Ω
The reactance of the circuit is
( =(� −(0 = 600Ω − 200Ω = 400Ω
The impedance of the circuit is
4 = √�6 +(6 =5(300Ω)6 +(400Ω)6
4 = 5(9@10b) +(1.6@10c) = √2.5@10c
4 = 500Ω
The current is
= d\ = c3dc33e = 0.10f
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The phase angle &(the angle between the voltage and the current
phasor) is
& = arctan(� −(0�
(� −(0 = 600Ω − 200Ω = 400Ω
� = 300Ω
& = arctan400300 = 533
Power factor is
I = cos 533 = 0.60
Or
I = �4 = 300Ω500Ω = 0.60
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Average Power
IjKL =12 � cos& = �� ��� cos&
IjKL =12(0.10f)(50�) cos 533
IjKL = 1.5k
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Resonance
Series
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When Z is minimum
4 = 7�6 +8�� − 1�.96
Z minimum when (� = (0
�3� = lmn0
�3 = 1√�.
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Resonance
.�<<[�� = �4
At resonance 4�^�^�^ thus �^�@^�^
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Parallel
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When (� = (0
Thus again
�3 = 1√�.
Resonance for parallel circuit
4 = 7�6 +8�� − 1�.96
Thus 4 = � its maximum value.
= �4 = �� ��^�^�^��o�[