Brief 5AC RL and RC CircuitsElectrical Circuits Lab

Brief 5
AC RL and RC Circuits
Electrical Circuits Lab I
(ENGR 2105)
Dr. Kory Goldammer
Review of Complex Numbers and Transforms
Transforms
The Polar Coordinates / Rectangular Coordinates Transform
The Complex Plane
We can use complex numbers to solve for the phase shift in AC
Circuits
Instead of (x,y) coordinates, we define a point in the Complex
plane by (real, imaginary) coordinates
Real numbers are on the horizontal axis
Imaginary numbers are on the vertical axis

The Complex Plane (cont.)
Imaginary numbers are multiplied by j
By definition,
(Mathematicians use i instead of j, but that would confuse us
since i stands for current in this class)
Imaginary Plane: Rectangular Coordinates
We can identify any point in the 2D plane using (real,
imaginary) coordinates
Complex Plane Using Rectangular Coordinates
Imaginary Plane: Transform to Polar Coordinates
We can identify any point in the complex plane using (r,)
coordinates.
The arrow is called a Phasor.
r is the length of the Phasor, and is the angle between the
Positive Real Axis and the Phasor
is the Phase Angle we want to calculate
r
Imaginary Plane: Transform to Polar Coordinates
(we will discuss the meaning of r later)

Use either the sin or cos term to find :
But we need in radians:
r
=7.07
Complex Math
For addition or subtraction, add or subtract the real and j terms
separately.
(3 + j4) + (2 – j2) = 5 + j2
To multiply or divide a j term by a real number, multiply or
divide the numbers. The answer is still a j term.
5 * j6 = j30
-2 * j3 = -j6
j10 / 2 = j5
Complex Math (cont. 1)
To divide a j term by a j term, divide the j coefficients to
produce a real number; the j factors cancel.
j10 / j2 = 5
-j6 / j3 = -2
To multiply complex numbers, follow the rules of algebra,
noting that j2 = -1

To divide by a complex number: Can’t be done!
The denominator must first be converted to a Real number!
Complex Conjugation
Converting the denominator to a real number without any j term
is called rationalization.
To rationalize the denominator, we need to multiply the
numerator and denominator by the complex conjugate
Complex Number Complex Conjugate
5 + j3 5 – j3
–5 + j3 –5 – j3
5 – j3 5 + j3
–5 – j3 –5 + j3
Multiply the original equation by the complex conjugate divided
by itself (again, j2 = -1):
Phase Shift
Time Domain - ω Domain Transforms
Transforming from the Time (Real World) Domain to the
(Problem Solving Domain

Note that in the ω Domain, Resistance, Inductance and
Capacitance all of units of Ohms!ElementTime Domainω
Domain TransformApplied Sinusoidal AC Voltage
(Volts) (ω=2πf)Vp
(Volts)Series Current(Amps) (ω=2πf)(Amps)ResistanceR
(Ohms)R
(Ohms)InductanceL
(Henry’s)(Ohms) (ω=2πf)CapacitanceC
(Farads)(Ohms) (ω=2πf)
Solving For Current
Impedance - Ohms
So far, we’ve talked about resistance. Ohm’s law tells us:
Resitance limits, or “impedes”, the flow of current.
The more Ohms in the circuit, the less current.
Impedance (Z): The number of Ohms in a circuit.
Resistance has units of Ohms, but so do Capacitance and
Inductance in the ω Domain!
Impedance – Ohms (cont. 1)
The impedance of the Capacitor:
When the frequency is small, ZC is large.
When the frequency is large, ZC is small

The impedance of the Inductor:
When the frequency is large, ZL is large.
When the frequency is small, ZL is small
Impedance – Ohms (cont. 2)
To find the total impedance of a series circuit:
ElementTime Domainω Domain TransformApplied Sinusoidal
AC Voltage
(Volts) (ω=2πf)Vp
(Volts)Series Current(Amps) (ω=2πf)(Amps)ResistanceR
(Ohms)R
(Ohms)InductanceL
(Henry’s)(Ohms) (ω=2πf)CapacitanceC
(Farads)(Ohms) (ω=2πf)
Ohm’s Law Still Works
Ohm’s law is well-named. It applies to anything that has units
of Ohms!
Instead of , it becomes:
Or,
Ohms law works in both the time- and ω-domains!
Strategy
Solving in the time domain is difficult!
Transform v(t) to the ω domain

Calculate Z in the ω domain
Solve for Current in the ω domain:
Transform Current back to the time domain
Calculating current in an R (Resistor Only) Circuit – Nothing
Changes!
Time Domain
ω Domain
Vp
Phase Angle for an R Circuit
Time Domain
0.1
Interpretation: The current in the circuit will be in phase with
the applied voltage.
+j
+
I = 0.1 Amps
θ = 0o

Calculating current in an L (Inductor Only) Circuit
Time Domain
ω Domain
Vp
Phase Angle for an L Circuit
Time Domain
1
Interpretation: The current in the circuit will Lag the voltage
by 90o (i.e. Current reaches it’s peak at a later time)
+j
+
I = -j1 Amps
Calculating current in a C (Capacitor Only) Circuit

Time Domain
ω Domain = 10 Volts
Vp
Phase Angle for a C Circuit
Time Domain
1
Interpretation: The current in the circuit will Lead the voltage
by 90o
+j
+
I = j1 Amps
Calculating current in an RL Series Circuit
Time Domain
ω Domain =10Volts

Vp
Calculating Current for RL Circuit
Rectangular: Amps
Polar:
But we need in radians:
+j
+
θ
)
I =0.5 – j0.5 Amps
Calculating Current for RL Circuit – Time Domain
The Current is out of phase with the applied voltage by 45o

Transform Theta from Radians to Time
In the previous example, we said the two waveforms were out of
phase by
How much is the phase shift in terms of time?
We know in 1 Period, the wave goes through , or 360o. So in
this case, the shift is:
We know 1 period takes
3) The phase shift is of a period, so
Experiment #5 Data Sheet
1. RL Circuit measurements:
2.1.Measured peak (Vp) voltage of the Voltage Source:
__________
2.2.Peak current (A): __________
2.3.Time delta (μsec) between current and voltage peaks:
__________
2.4.Time-domain expression for i(t), based on measures above:
__________

2.5.Calculated expression for i(t), based on measured R and L:
__________
2.6.List your ideas for discrepancies, if any:
_________________________
2. RC Circuit measurements:
2.1 Measured peak voltage of the Voltage Source: __________
2.2 Peak current (A): __________
2.3 Time delta (μsec) between current and voltage peaks:
__________
2.4 Time-domain expression for i(t), based on measures above:
__________
2.5 Calculated expression for i(t), based on measured R and C:
__________
2.6 List your ideas for discrepancies, if any:
_________________________

_____________________________________________________
________
© N. B. Dodge 01/12
ENGR 2105 – Inductors and Capacitors in
AC Circuits and Phase Relationships
1. Introduction and Goal: Capacitors and inductors in AC
circuits are
studied. Impedance and phase relationships of AC voltage and
current are
defined. Frequency-dependence of inductor and capacitor
impedance is
introduced. Phase relationships of AC voltage and current are
defined.
2. Equipment List:
• Multisim
• Scientific Calculator
3. Experimental Theory: Capacitors and inductors change the
voltage-
current relationship in AC circuits. Since most single-frequency
AC circuits

have a sinusoidal voltage and current, exercises in Experiment 5
use
sinusoidal AC voltages. Note that in an RLC AC, current
frequency will be
identical to the voltage, although the current waveform will be
different.
“Imaginary” Numbers, the Complex Plane, and Transforms:
3.1.1 Definition of j: As √−1 is not a real number, Electrical
Engineers define � = +√−1. Physicists and
Mathematicians use � = −√−1 for this same purpose, so
� = −�, but that will not affect our theory.
3.1.2 Electrical Engineering problem solutions often include
imaginary numbers. It is useful to consider real and
imaginary numbers as existing in a two dimensional space,
one axis of which is a real-number axis, and the other of
which is the “imaginary” axis.
3.1.3 In the complex plane (Figure 1), the horizontal axis is the
real axis,
and the vertical axis is the y axis. Real numbers (–7 , 10) lie on
the x-
axis, imaginary numbers (–3j, j42) lie on the y-axis. Complex
numbers lie off axis. For example, 2 + j6 would lie in the first

quadrant, and (–43 – j17) would lie in the third quadrant.
© N. B. Dodge 01/12
3.1.4 Transforms: Transforms allow moving a problem from a
coordinate system or domain where it is difficult to solve to
one where it is easier to solve (Figure 2).
3.1.5 Simpler equations in the transform domain make the
problem easier to solve than in the original domain. We can
transfer sinusoidal, single-frequency AC circuit problems to
a domain where we can use algebra to solve them rather than
calculus. Solving problems in algebra is almost always
easier than calculus!
3.1.6 The catch: We need transforms (formulas) to the new
domain. Then “inverse transforms” are required to return the
solution to the time domain, where it is useful.
3.2 A New Domain: In the phasor or frequency (ω) domain,

sinusoidal AC
circuit problems are easier to solve. Note: � = ��, where f is
the
frequency in Hertz. Recall that Hertz has units of 1/second, or
“per
second”. Thus, ω is in radians/sec.
3.2.1 Transforms: Skipping the derivation (you will do the
derivation if you take ENGR 2305), we simply list
frequency domain transforms. Note that AC voltage is
© N. B. Dodge 01/12
usually expressed as �(�) = ��cos(ωt), where Vp is the peak
voltage.
3.3 Transform from the Time Domain to the Frequency
Domain: Some
circuit elements have a different representation in the
frequency, or ω,
domain.
Element Time Domain ω Domain Transform
Sinusoidal AC Voltage ��cos(ωt) Vp (Volts)
Resistance R (Ohms) R (Ohms)

Inductance L (Henry’s) �� (Ohms)
Capacitance C (Farads) 1 ��⁄ (Ohms)
3.4 Comments:
3.4.1 Resistance: There is no transform for Resistors. We use
the resistance value R (Ohms) in both domains.
3.4.2 Inductance: In the time domain, we use the value of the
Inductor, L (Henry’s). This value transforms to �� in the
ω domain. �� has the same units as Resistance and
represents the amount the Inductor opposes current.
3.4.3 Capacitance: In the time domain, we use the value of the
Capacitor, C (Farads). This value transforms to 1 ��⁄ in
the ω domain. 1 ��⁄ has the same units as Resistance and
represents the amount the Capacitor opposes current.
3.4.4 Voltage: Voltage in the ω domain is just the peak voltage,
Vp. There is no frequency information in the voltage
transform.
3.5 Solving For Currents in the ω Domain: In most Electrical
Engineering
problems, we know voltages and component values: Resistance
(R),
Inductance (L), and Capacitance (C). We generally solve for
circuit
currents, which is easier in the frequency domain. Note: V = IR
in the

time domain; V = IZ in the ω domain, where Z is the circuit
impedance. It is the sum of all the impedance contributions
from
resistors, capacitors and inductors, which all have the unit of
Ohms.
� = � + �� + � ��⁄ .
In both domains, voltage is still in Volts and current is still in
Amperes.
3.5.1 Resistor AC circuit solution in the ω domain: In Figure 3,
we identify time domain voltage (�(�) = ��(��) =
10cos⁡(1000�). From this, we obtain the ω domain voltage
© N. B. Dodge 01/12
transform: Vp = 10 Volts and ω = 1000 radians/second.
Resistance = 100 Ω. Since V=I Z in the ω domain, then
� = �� ⁄ = �� ⁄ = 10 100⁄ = 0.1⁡��
This answer is converted to the time domain later in the lab.
3.5.2 Inductor AC circuit (Figure 4): � = 10cos(1000�), so ω =
1000 rad/s. In the ω domain, 10 mH transforms to jωL =

j(1000)(10 ∗ 10−3) = �10 Ohms. Vp = 10 Volts. So
� = �� ⁄ = �� ωL⁄ = 10 �10⁄ = −�1⁡��
(Time Domain answer below.)
3.5.3 Capacitor AC circuit (Figure 5): ⁡� = 10cos(1000�), so
ω
= 1000 rad/s. In the ω domain, 100 μF transforms to
1 �⁄ ωC = 1 j(1000)(100 ∗ 10−6)⁄ = −�10 Ohms in the ω
domain. Vp = 10 Volts. So
� = �� ⁄ = �� ωL⁄ = 10 −�10⁄ = �1⁡��
(Time Domain answer below.)
3.5.4 RL Circuit: Resistor and Inductor: In Fig.6, R and L
transform to 10 Ω and �ωL = j(1000)(10 ∗ 10−3)=j10 Ω.
Then
© N. B. Dodge 01/12
� = �� ⁄ = 10 (10 + �10)⁄ =
10(10 − �10) (100 + 100)⁄ = 0.5 − �0.5 Amps

3.6 Inverse Transforms: To make solutions from the ω domain
useful, we
must do the reverse (or inverse) transform to the time domain.
3.6.1 The answers above are in Cartesian coordinates (X ± jY).
These X ± jY results would be more useful in polar
coordinates. We want to find r (r is actually the peak
current, Ip) and θ.
� = �� =
√(|"��"⁡��|)2 +
(|"��"⁡��|)2
� = tan−1 (|"��⁡��"|)
|"��"⁡��|⁄
3.6.2 The time domain current is then:
�(�) = ��cos⁡(ωt + θ)
3.6.3 Resistor Circuit: The current had only a Real component.
I
= 0.1 Amps.
� = �� = √(0.1)
2 + (0)2 = 0.1⁡��
� = tan−1 0 0.1⁄ = 0⁡��

�(�) = �� (�� + �) = �.��(��)
The voltage in the time domain is � = 10cos(1000�).
Comparing this
to the current, we see that V and I have different amplitudes,
but they will rise
and fall “in phase” with each other, as shown in Figure 7.
© N. B. Dodge 01/12
Fi
3.6.4 Inductor Circuit: The current had only an Imaginary
component. I = -j 1 Amps.
� = �� = √(0)
2 + (1)2 = 1⁡��
� = tan−1 −1 0⁄ = −90⁡�� = −� 2⁄
�(�) = ��⁡(�� + �) = ��(�� − � �⁄
)⁡��
The result is sinusoidal current with a peak value of 1
ampere, with an associated phase angle. That is, it oscillates

at the same radian frequency of 1000 rad/s, but is not in
lock-step with the voltage. Its oscillation is 90 degrees
behind the voltage, as shown in the graph below (Figure 8).
This “phase angle” is constant. Current is always exactly 90
degrees behind the voltage, a significant characteristic of
inductors in sinusoidal AC circuits
3.6.5 Capacitor Circuit: The current had only an Imaginary
component. I = j 1 Amps.
� = √(0)2 + (1)2 = 1⁡��
� = tan−1 1 0⁄ = +90⁡�� = + � 2⁄
�(�) = ��⁡(�� + �) = ��(�� + � �⁄ ) Amps
© N. B. Dodge 01/12
The result is sinusoidal current with a peak value of 1
ampere, with an associated phase angle. That is, it oscillates
at the same radian frequency of 1000 rad/sec, but is not in

lock-step with the voltage. Its oscillation is 90 degrees ahead
of the voltage, as shown in the graph below (Figure 9). This
“phase angle” is constant. Current is always exactly 90
degrees ahead of the voltage, a significant characteristic of
Capacitors in sinusoidal AC circuits
3.6.6 RL Circuit: The current had both a Real and an Imaginary
component. I = 0.5 - j 0.5 Amps.
� = √(0.5)2 + (0.5)2 = 0.707⁡��
� = tan−1 −0.5 0.5⁄ = −45⁡�� = − � 4⁄
�(�) = ��⁡(�� + �) = (�.��)��(�� − � �⁄
)⁡��
The result is sinusoidal current with a peak value of 0.707
Amperes, with an associated phase angle. That is, it
oscillates at the same radian frequency of 1000 rad/s, but is
not in lock-step with the voltage. Its oscillation is 45 degrees
behind the voltage. This “phase angle” is constant. Current
is always exactly 45 degrees behind the voltage.

4. Pre-Work: Prior to lab, review the experimental theory and
experimental
procedure and complete the Worksheet.
5. Experimental Procedure:
5.1 V-I We first study an RL circuit. For the following, use �
=
1000
��
�
⁡
5.1.1 Construct a series RL circuit as in Fig. 6, using a 10 mH
inductor and 16 Ω resistor. Recall that the frequenc y can be
calculated as � =
�
2�
. Connect ground between the inductor
and the AC Voltage source. Connect a combined voltage
and current probe between the AC source and the resistor
© N. B. Dodge 01/12
(This setup will simultaneously measure the output of the
AC Voltage Source and the total current in the circuit).

5.1.2 Using time cursors, measure time difference, ∆ �, between
Ip
(on the “Current” Curve) and Vp (on the “Voltage” curve).
This ∆� will be used to determine the phase angle.
5.1.3 Take a screen shot of your Multisim setup for this
measurement and include it in your Lab Deliverables.
5.2 Examining V-I Relationship in an AC RC Circuit: Replace
the
inductor with a 10 μF capacitor to create an RC circuit. Leave
the 16
Ω resistor in place and continue using � = 1000
��
�
.
5.2.1 Using the cursors, measure “∆�” between current and
voltage peaks.
5.2.2 Take a screen shot of your Multisim setup for this
measurement and include it in your Lab Report.
5.2.3 ∆� is used to calculate the phase angle. We first find the
ratio of ∆� to the total amount of time required to complete
one period, T. Where � = 1 �⁄ , and � = 1000 Hz for this
experiment.
�� = ∆� �⁄ = ∆� (1 �⁄ )⁄
5.2.4 This ratio must also represent the ratio of the phase shift,

θ,
to 360 degrees:
�� = � 360�⁄
5.2.5 Equating the two terms, ∆� �⁄ = � 360�⁄ . Rearranging,
we
can solve for the phase shift:
�ℎ ��⁡�ℎ ��⁡� = (∆� �⁄ ) ∗ 360�
6. Cleanup: Return parts and cables to their respective homes.
Make sure that
your work area is clean.
7. Writing the Lab Report: For your lab deliverables, do the
following:
7.1 Since �(�) = ��cos(�� + �), construct expressions for
i(t) in the RL
and RC circuits using the given circuit values. Do this by
transforming to the �-domain and calculating �� and �, and
then
transform back to the time domain.
7.2 From your measurements, write an expression for i(t) in
each case.
7.3 Compare the i(t) expressions developed in 7.1 and 7.2.
Discuss any
discrepancies.

© N. B. Dodge 01/12
Lab 5 Worksheet
Note: Lab 5 is generally the most challenging exercise in ENGR
2105. Please
watch the videos and read Lab 5 carefully at least twice and
then take your time on
the exercises below to make sure that you understand the
theoretical material.
1. In what quadrant of the complex plane are these numbers
located?
−12+j7 __________________ −10−j50 __________________
8−j2 __________________ 1+j100 __________________
2. Use the complex conjugate to convert the expressions to a
real term and an
imaginary term.
26 (6 − �4)⁄ __________________ (8 − �8) (2 + �2)⁄
__________________
3. Inductor and capacitor impedances are given as: �� = ��
and �� =
1 ��⁄ . Assume you have a 10μF capacitor and a 10mH

inductor. Calculate
the impedance of these components at the following frequencies
and list in
the space provided:
1 MHz (1,000,000 Hz): �� = _______________Ω 1 ��⁄ =
_________ Ω
50KHz (50,000 Hz): �� = ______________ Ω 1 ��⁄ =
___________ Ω
0Hz: �� = _______________ Ω 1 ��⁄ =
_______________ Ω
4. Different items in the time domain transform in different
ways to the ω
domain:
Element Time Domain ω Domain Transform
Applied Sinusoidal
AC Voltage
�� (��)
(Volts)
Vp
(Volts)
Series Current �� (�� + �)

(Amps)
��
(Amps)
Resistance R
(Ohms)
R
(Ohms)
Inductance L
(Henry’s)
��
(Ohms)
Capacitance C
(Farads)
� ��⁄
(Ohms)
© N. B. Dodge 01/12
Given a circuit with a Time domain values:

�(�) = 10 cos(1000�)
� = 100 Ω
L = 10 mH
C = 10 μF
Calculate the values in the ω domain for:
Applied Voltage __________Volts
Resistance __________Ω
Impedance from Inductor __________ Ω
Impedance from Capacitor __________ Ω
5. After transforming voltage and circuit to the ω domain, find
the current by
dividing voltage by impedance. This usually results in a
complex number. To
convert back to the time-domain, which is the answer sought, do
four things:
• Rationalize using complex conjugate; the result is an X ± jY
representation.

• Convert complex number to polar-coordinates:
• The radial distance is the peak current. The angle is the phase
angle.
6. Based on the procedure in 5 above, convert the following ω
domain currents
back into the time domain (assume ω = 1000):
I = 10+j10 _____________ I = −8+j4 _____________
7. If �(�) = 10 cos(1000�), R = 100 Ω, C = 100 μF , determine
the expression
for i(t) .
https://www.youtube.com/watch?v=o5HWQZhane4&ab_channel
=KoryGoldammer
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KoryGoldammer
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KoryGoldammer

Brief 5AC RL and RC CircuitsElectrical Circuits Lab

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