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STUDIO Unit 15
PHY-2054 College Physics II
Drs. Bindell & Dubey
[AC CIRCUITS]
15 – INTRODUCTION TO AC CIRCUITS
In this unit we will explore how inductors and capacitors function when the voltage applied to
them is sinusoidal in nature. You will learn to understand



The nature of reactance of a circuit element such as a capacitor or an inductor.
The impedance of a combination or RL and C circuit combinations.
The concept of resonance in RLC circuits.
NOTE: The last experiment dealt with the DC on/off response of an RL circuit. The same
experiment could have been done with a capacitor. This was covered in an earlier chapter in
the textbook so be sure to find it and read the material outside of class. We mentioned this
briefly during one of the lecture sessions.
We next begin a discussion when the driving emf is not an on/off square wave but is sinusoidal
in nature. We therefore will find it useful to study the same LR circuit when it is driven by a sine
(or cosine) emf. Refer to the slide notes and the textbook for amplifying information.
ELEMENTS OF AC CIRCUITS
The easiest place to start a look at AC circuits is the resistor. The resistor responds directly to an
applied voltage via Ohm’s law. So if the applied voltage is given by V  V0 cos(t ) then the current
V 1
I

 V0 cos(t )  I 0 cos(t ) .
will be given by
R R
(a) Plot the graph of V as a function of time.
(b) Plot the graph of I as a function of time.
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Normal AC voltages supplied to the home arrive with a frequency of f=60Hz. For this frequency,
what is:
: _____________ (angular frequency in rad/sec)?
T: ______________ (period in seconds)?
Phasors
For analyzing ac circuit we can use a model. In this model, alternating voltage and the
corresponding current can be represented with counterclockwise rotating arrows called phasors.
Following is the phasor diagram for AC current.
We begin by plotting the Current Phasor I
which represents the maximum current
possible.
This
phasor
rotates
counterclockwise on the diagram at an
angular velocity . The angle between the
phasor and the horizontal axis is t and it
increases with time as the phasor rotates. The
projection onto the horizontal axis of the Iphasor is Icos(t) and this projection
represents the instantaneous current.
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Using the same strategy as described for current, draw the phasor diagram representing AC
voltage.
Root-Mean-Square (rms) Value of a Sinusoidally Varying Quantity
To determine the rms current we follow the following steps which are explained in the adjacent
figure:
1. Graph current i
versus time,
2. square the
instantaneous
current i,
3. take the average
(mean) value of
2
i,
4. take the square
root of the
average.
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These steps result in the rms value of current:
In this equation I is the maximum current.
Using the same strategy, write the equation for rms voltage.
Suppose an rms voltage of 120 volts is supplied to a 50 ohm resistor.
(a) Determine the peak voltage.
(b) On the SAME axes, sketch a graph of both the current and the voltage for one cycle.
Label the time axis appropriately and be careful to indicate that the current and voltage
need separate scales and usually would not be plotted together. Choose your units
appropriately.
Time
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Notice that the two curves look very much the same.
What is the maximum current and voltage on the graph?
ANS: ________________________
What is the minimum current and voltage on the graph?
ANS: ________________________
These two curves are said to be in phase with each other. What does this mean?
Following is the phasor diagram for an ac resistor. For a resistor, the current and voltage
phasors are co-linear since the voltage and the current are in phase.
For the inductor, things are a little different. Consider the following circuit:
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iL
We already know that the current and voltage for an inductor are related by VL  L t . But
what do we do with this expression? We can develop the following mathematics assuming that
the current goes like i  I cos(t ).
Note that when the time interval gets very small, the circled term, sin(x)/x, is unity. Using some
trigonometric expressions that you have probably long forgotten, we find if the current goes as
the cos(t), the voltage will go as the sin(t).
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Choose values that produce reasonable values, plot the current and voltage for a pure inductor on
the following graph. Use f = 60Hz and I=1A and choose L to be a reasonable value so we can
plot both on the same paper. This is important!
time
The final expression is the following:
vL  LI sin(t )
This is actually a relatively strange looking expression. But there is an interesting magical way
to deal with relationships of this type: we use a “phasor diagram”.
In “trig” you learned that the following is true:
sin(t 

2
cos(t 
)   cos(t )

)  sin(t )
2
  2 f
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Using these equations, show that vL   LI cos(t 

2
).
Next, we look at the meaning of this in a graphical context. The important relationships are:
vL  LI sin(t )
vL   LI cos(t 

2
).
and
i=I cos(t).
We can use these relationships to generate a graphical interpretation of how the voltage and
current are related in an RL circuit. Again, pay attention to that $500 textbook that you
purchased earlier in the year.
Let’s look at a “phasor” diagram for this simple circuit. For reference, here is the diagram from
the textbook:
The projection onto the horizontal axis of the I-phasor is Icos(t) and this projection represents
the instantaneous current through the inductor. The voltage across the inductor is represented by
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vL   LI cos(t 

)
V  LI is the maximum value for the voltage and it is
2 . Here L
represented in the diagram by a phasor that rotates 90° ahead of the current so that we say that
the voltage leads the current in this device. The instantaneous voltage,
vL , across the inductor is
the projection onto the horizontal axis as well.
Verify that this projection matches the equation above by looking at the geometry of the
diagram:
Looking at the equation VL  LI , we see that it would look like Ohm’s Law if the resistance
was given by L. We call this term the “Inductive Reactance”, X L, and write the “pseudoOhm’s Law as:
VL  IX L
which is a very important relationship. Note that the “resistance” of the inductor increases with
frequency.
(a) Compute the reactance of a 0.450 H inductor at a frequency of 60.0 Hz.
(b) Most circuits involving coils must also contain a resistor. Why is this true?
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You will use the large coil that you have been using in the previous experiment. The circuit that
we will use now is:
Sine
Wave
Why does “i” in this circuit represent the current in the inductor as well?
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When there is resistance in the circuit, the phasor diagram changes a bit. Let’s freeze the
diagram when t=0 to keep the geometry simple. The entire diagram rotates at the angular
frequency , so we don’t lose anything by using this perspective. The current through the
resistor is in phase with the voltage across the resistor so the current phasor I will be in the same
direction as the voltage phasor VR. This allows the following diagram:
Note that as we had before, the inductor voltage leads the current (and VR by 90°). The total
voltage across the two elements must (Kirchoff’s loop rule) equal the applied sinusoidal voltage,
so the diagonal represents this sum and incorporates the fact that the two voltages are out of
phase. Notice that the total voltage V leads the current I by a phase angle  . It is convenient to
define a pseudo-resistance to the circuit which we call Z and define as the impedance of the
circuit. Z is a very important quantity that works in a way that R works for a resistor, that is, we
can have a pseudo-Ohm’s Law given by:
V=IZ
------------(1)
From the diagram above, using the Pythagorean theorem, we find that
V 2  VL2  VR2
Using VL
 IX L , VR  IR
and equation (1),
I 2 Z 2  I 2 X L2  I 2 R 2
Solving this equation for Z, we get
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Z  R 2  X L2  R 2   2 L2
This last relationship will form the basis of the RL experiment. Using the specified values for R
and L, calculate the value of Z for the following frequencies (f):

f (Hz)
100
500
1000
5000
10000
Z
Set up the experiment with an applied voltage of about 2 volts. Implement the “scope” and drag
the applied voltage and the current ((voltage measured across the resistor)/R). The scope should
show a “trace” like the following:


Determine the following values for an input voltage of 2 V:
Fequency (Hz)
100
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Current
360 degrees
Phase angle
Impedence
500
1000
5000
10000
Use this space for calculations. Show the equations that you use (ask for help if you need it).
Collect the following data for frequencies from 500 Hz to a maximum of 8,000 Hz. Note that
 L 
 so you can check the consistency of all of the
form the graph you can use   tan 1 
 Rtotal 
proceeding by calculating L over the range of the data and seeing how consistent it is. Note that
as the angle gets close to 90º there may be some large errors in the calculation.
f
500
1000
2000
4000
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measured
tan 
L - Calculated
6000
8000
Discuss your results here:
AC RESONANCE
We will now add a capacitor to the mix. The equations for the capacitor (see class notes and the
textbook) are:
1
I sin(t )
C
1
Xc 
(ohms)
C
I

vC 
cos(t  )
C
2
vc 
The capacitor voltage, in a similar argument to the inductance, is found to lag the current by 90°
and for a capacitor, things are pretty much the same except that the capacitive reactance is given
by 1/C.
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Putting everything together here, we can look at the combined phasor diagram for an RLC circuit
and discover some interesting facts.
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1. The voltages VL and VC are 90° ahead and behind the current respectively. Since they
form a “straight line” in the diagram, we can add them like vectors to determine the net
reactance.
2. In the left diagram the inductive reactance dominates and the current lags the voltage (VL
- VC).
3. In the right, the capacitive reactance dominates and the current leads the voltage (VL VC).
4. Following the previous discussion for the inductor alone (and reading the textbook!)
2
2
leads to the expression Z  R  ( X L  X C ) .
5. Notice that when XL=XC, Z is a minimum and is equal to R. This produces the maximum
current (I = V/R). This condition is known as resonance.
We now proceed with a look at an RLC circuit’s behavior as a function of frequency. We
basically do the same experiment after adding a capacitor in series.
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Setting the capacitive and inductive reactances equal, derive an expression for the resonant
frequency f0.
For a series RLC circuit, using R= 10 Ω, L = 8.2 mH and C = 100 µF,
(a) determine the resonance frequency.
(b) Fill in the following table with experimental values. If the data looks “ratty”, you may
need to increase the sampling rate in the software. Ask an instructor about this.
Frequency (Hz)
Discuss your results:
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phase angle
impedance
current
Plot the current vs. the frequency (a resonance plot). On the same graph plot the phase shift,
keeping track of the sign.
i
Phase
Angle
frequency
Discuss your results:
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
Solve the following problem:
An AC source with Vmax = 155 V and f = 45.0 Hz is connected between points a and d in the
figure.
Calculate the maximum voltages between the following points:
(a) a and b
(b) b and c
(c) c and d
(d) b and d
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Parallel resonance:
A parallel RLC circuit has resonance behavior similar to that of the series RLC circuit, but the
roles of voltage and current are reversed. For this circuit, the instantaneous potential difference
is the same for all three circuit elements and is equal to the source voltage.
At one particular frequency ωo, the two reactances XL and XC are equal and this frequency is
called resonance frequency.
(a) Derive the equation for the resonane frequency.
(b) Draw the phasor diagram for the parallel RLC circuit.
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(c) Using the phasor diagram, derive the equation for impedance Z of the circuit.
(d) Draw the phasor diagram for the parallel RLC circuit at resonance.
(e) For the circuit in the following figure, R = 200 Ω, L = 0.800 H, and C = 5.00 μF. When the
source is operated at the resonance frequency, the current amplitude in the inductor is 0.400
A. Determine the current amplitude (i) in the branch containing the capacitor and (ii)
through the resistor.
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