ELEC 221 Workbook Solutions
Transcript of ELEC 221 Workbook Solutions
ELEC 221
Workbook Solutions
ELEC 221 Workbook Solutions
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1.1 Definitions and Equations Voltage: Amount of potential energy difference between two points. Energy required per unit
charge for separation. Measured in volts.
π£ = ππ€
ππ
Where v = voltage (V), w = energy (J), q = charge (C)
Current: Rate of charge flow. Measured in amps.
π = ππ
ππ‘
Where i = current (A), q = charge (C), t = time (s)
Power: Time rate of expending or absorbing energy. Measured in watts.
π =ππ€
ππ‘
Where p = power (W), w = energy (J), t = time (s)
Conservation of Energy: Total power delivered = total power absorbed. Energy can only be
transferred.
βπ = 0
Where p = power (W)
1.2 Basic Circuit Elements Resistor: A device having a designed resistance that electric current can flow through.
π£ = π β π π = π£ β π
R = Resistance, measured in
Ohms (Ξ©)
Switches: Two positions open and closed. Current cannot flow through an open circuit as there
is no path.
Open Circuit: i = 0, no current flow
Short Circuit: Can have current flow, v = 0 due
to no resistance
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Independent Sources: An idealized circuit component that fixes the voltage or current in a
branch.
Ideal Voltage Source
Ideal Current Source
Dependent Sources: A voltage or current source whose value depends on a voltage or current
somewhere else in the network. Usually represented by a diamond shape.
Dependent Voltage Source
Dependent Current Source
1.3 Kirchhoffβs Laws Kirchhoffβs Current Law: The sum of the currents entering a node is zeros. (The incoming
currents must equal the outgoing currents).
For all general cases:
βπππππ = 0
For the diagram to the left:
ππ + ππ β ππ β ππ = 0
Kirchhoffβs Voltage Law: The sum of voltage rises in a loop is zero.
For all general cases:
βπ£πππππ ππ ππππ = 0
For the diagram to the
left:
βπ£π + π£π β π£π = 0
1.4 Basic Resistor Simplifications Resistors in Series:
π ππ = π 1 + π 2 + π 3 + π 4
Resistors in Parallel: Any two elements that connect to form a loop are said to be in parallel.
Parallel elements have the same voltage.
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π ππ = π 1 β π 2
π 1 + π 2
1.5 Divider Circuits Voltage Divider Circuit: Voltages are proportional to their corresponding resistance
π1 = ππ β π 1
π 1 + π 2
π2 = ππ β π 2
π 1 + π 2
Current Divider Circuit: Currents are reciprocal to their corresponding resistances
π1 = ππ β π 2
π 1 + π 2
π2 = ππ β π 1
π 1 + π 2
Example: Find the output voltage (v0)
2.1 Node-Voltage Analysis First make sure that the circuit schematic is neat and has no branches crossing over each other.
After label, all the essential nodes on the circuit. Next choose an essential node as ground, this
will be your reference node. This choice is arbitrary; however, it is recommended to ground the
node with the most branches connected to it.
Each node equation is generated by applying Kirchhoffβs current law at each node. Thus, the
current entering each node minus the current leaving each node must equal zero. After all the
nodes have their respective equations, it just comes down to re-arranging equations.
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i) Standard Example: Solve for the node voltages
ii) With Dependent Source
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iii) Supernode Cutset
A Supernode is when there is a voltage source between two essential nodes. The two nodes can
be combined to form a Supernode. The voltage difference between the two nodes is equal to
the voltage source between them, leading to easier calculations.
2.2 Mesh-Current Analysis A mesh is any loop without a smaller loop inside of it. Mesh-Current analysis applies Kirchhoffβs
laws to write equations for each mesh. This method is okay, however for it to work you must
know the voltage drop across every circuit element you pass through, thus it raises difficulties
when going through current sources or dependent sources.
i) Standard Example: Find mesh currents
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ii) Special Case 1: Current Source Between Two Meshes
To create a Supermesh, the current source between the two meshes is simply avoided when
writing current equations. Can then use this current source value to relate some mesh currents.
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3.1 Delta-Wye Conversion This conversion is used to establish equivalent circuits between a delta (β) and wye (Y)
arrangement. This conversion can be used when three circuit elements are not sources and are
connected in one of the proper arrangements seen below.
Can convert between these two arrangement with the conversions seen below.
Equations for the resistors can be seen below. Use the left side if going from Y to β. Use the
right side if going from β to Y.
π 1 = π ππ π
π π + π π + π π
π 2 = π ππ π
π π + π π + π π
π 3 = π ππ π
π π + π π + π π
π π = π 1π 2 + π 2π 3 + π 3π 1
π 1
π π = π 1π 2 + π 2π 3 + π 3π 1
π 2
π π = π 1π 2 + π 2π 3 + π 3π 1
π 3
3.2 Source Transformations Replace a voltage source in series with a resistor with a current source in parallel with a resistor.
Can also do the reverse of the previous statement. Very useful in simplifying circuits, should be
used immediately if one of the above situations is present in your circuit.
where ππ = ππ /π
Note that the resistor value (R) stays the same for the source transformation.
Example: Find the power from the 12V source.
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4.1 Thevenin and Norton Equivalent Circuits Thevenin: Network can be replaced by equivalent circuit with an independent voltage source
called Vth , in series with a resistor. The Thevenin equivalent is obtained by looking into a circuit
from some terminals. Much more common than Norton and will be used in future courses. A
Thevenin diagram can be seen below.
Norton: Network can be replaced by equivalent circuit with an independent current source in
parallel with a resistor.
i) For a network with only independent sources:
The first way to find a Thevenin is by simplifying the circuit to a voltage source in series with a
resistor using source transformations seen in the previous example. The voltage source is the
Vth, and the resistor is the Rth.
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However, for a network with only dependent sources, it may be more efficient to remove the
sources. Simply short the voltage sources and open current sources. Then combine the resistors
to find Rth. Find the Thenvein resistor for the question below by removing the sources.
Redraw the circuit with the voltage source shorted and the current source open.
Maximum Power Transfer: To maximize the power from a network to a load.
ππ πππ₯ = ππ‘β
2
4π πΏ
5.1 Superposition Principle When a system is excited by several independent sources or excitations, the total output is the
sum of all individual responses. Solve for the desired value from each excitation then sum add
them together.
Total Output = Output1 + Output2 + β¦ + Outputn
To turn off the other sources:
β’ Current Sources: Open
β’ Voltage Sources: Short
* Superposition can not be used to calculate power as power is a non-linear function*
Find the current going through the 3KΞ© resistor below, using superposition.
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6.1 Non-Linear Circuit Elements in DC Conditions Inductor: An element which opposes change in current. Composed of a coil of wire wrapped
around a core.
π£π = πΏ ππ
ππ‘
* INDUCTOR CURRENT CAN NOT JUMP, This is why you short circuit inductors at DC *
Capacitor: An element which opposes change in voltage. Composed of two conductors
separated by an insulator.
ππ = πΆ ππ£
ππ‘
* CAPACITOR VOLTAGE CAN NOT JUMP, This is why you open circuit capacitors at DC *
Adding Inductors and Capacitors: Inductors add the same as resistors, capacitors add
opposite of resistors (series = parallel, parallel = series).
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7.1 First Order RL and RC Circuits, Second Order
RLC Circuits in DC Conditions RL: Sources + Resistors + Inductor
RC: Sources + Resistors + Capacitors
**You assume that the initial condition of the switch has been there for a very long time.**
Time Constant: π = πΏ
π = π πΆ
When the elapsed time exceeds five time constants, the current is less than 1% of the initial
value. Thus considered past this point to be essentially zero.
Natural Response: Response of a capacitive/inductive circuit to the initial conditions (no
impulse).
RL Circuit:
i) Find initial current through the inductor, Io
ii) Find the time constant, π = πΏ
π
iii) Use the equation π(π‘) = ππππππ + (ππ β ππππππ)πβπ‘
πβ , to get i(t) of the inductor
Switch has been closed for a long time, opens at time 0. Find iL(t) for t β₯ 0, i0(t) for t β₯ 0+, vo(t)
for t β₯ 0+.
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RC Circuit:
i) Find initial voltage through the capacitor, Vo
ii) Find the time constant, π = π πΆ
iii) Use the equation π£(π‘) = π£πππππ + (π£0 β π£πππππ)πβπ‘/π, to get v(t) of the capacitor
Switch has been in position x for a long time. At t = 0, switch instantly moves to position y. Find
vc(t) for t β₯ 0, vo(t) for t β₯ 0+, io(t) for t β₯ 0+.
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Step Response: Response of a capacitive/inductive circuit to step inputs (abrupt changes in
voltage or current, think applying a source).
RL Circuit: Given the circuit below the step response current can be found using the equation
below.
π(π‘) = ππΉππππ + (π0 β ππΉππππ)πβ(
π πΏ
)π‘
(same equation as we used for natural responseβ¦ think about what I approaches as t goes to β)
Ex: Find the expression for il(t) for tβ₯0. Find initial voltage in inductor just after switch is
moved to b.
RC Circuit: Given the circuit below the step response voltage can be found using the equation
below.
π£π(π‘) = π£πΉππππ + (π£π β π£πΉππππ)πβπ‘/π πΆ
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Ex: Find vc(t) tβ₯0.
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7.2 Sequential Switching: The circuit switches move more than one time. Time references for all switching can not be at t
= 0. Need to use inductor currents and capacitor voltages to find initial conditions for the
switching as these values can not instantaneously jump.
Ex: The two switches have been closed for a very long time. At t = 0, the first switch is
opened. Then 35ms later the second switch is opened. Find iL(t) for 0 β€ t β€ 35ms, iL(t) for
t β₯ 35ms.
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7.3 RLC Circuits: Circuits with a resistor, inductor and capacitor. Still study the circuits step and natural responses.
Parallel RLC Circuit: Solve for the voltage across the parallel branches, then for each branch
current.
We solve parallel RLC circuits by writing down the differential equation for the sum of the
currents through each branch (must equal 0, per KCL)
π
ππ‘β πππ ππ’πππππ‘π =
π
ππ‘(ππ + ππ + ππΏ)
=π
ππ‘(
ππ£
ππ‘πΆ +
π
π + β«
π
πΏππ‘ )
=π2π£
ππ£2πΆ +
ππ£
ππ‘
1
π +
π£
πΏ= 0
π2π£
ππ£2+
ππ£
ππ‘
1
π πΆ+
π£
πΏπΆ= 0
Parallel RLC has the following characteristic equation (based on the differential equation for
current), neper frequency and resonant radian frequency.
π 2 + π
π πΆ+
1
πΏπΆ = 0
π =β
1π πΆ Β± β 1
π πΆ
2
β4
πΏπΆ2
= βΞ± Β± βπΌ2 β π02
β = 1
2π πΆ π = β
1
πΏπΆ
Essentially, we will just solve the characteristic equation using the quadratic equation. The result
of the quadratic equation will tell us which type of response the circuit is experiencing. Think
back to first order differential equations.
There are three possible solutions for the branch voltages, determined by the result of the
quadratic equation. These equations should look familiar as they are identical to your differential
equations class.
π£ = π΄1ππ 1π‘ + π΄2ππ 2π‘ = Overdamped Response [Ξ±2 > Ο2, s is real]
π£ = π΅1πββπ‘ cos(π€ππ‘) + π΅2πββπ‘ sin(π€ππ‘) = Underdamped Response [Ξ±2 < Ο2, s is complex]
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π£ = π·1π‘πββπ‘ + π·2π‘πββπ‘ = Critically Damped Response [if Ξ±2 = Ο2, 1 real root]
If the circuits have an external source applied to them then the equations above will have an
initial value added to them.
Ex: What is the initial current through the inductor branch? What is the initial value of
diL/dt? What are the roots of the characteristic equation? Find an expression for iL(t)
when t β₯ 0. I = 30mA.
Series RLC Circuit: Essentially the same as parallel RLC Circuits, just has a different characteristic
equation and you solve for the current as it is the common characteristic in series circuits. The
differential equation is derived from Kirchoffβs voltage law, the sum of the voltages around the
loop must be 0. Equations can be seen below for characteristic equation, neper frequency and
resonant radian frequency.
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π
ππ‘β πππ π£πππ‘ππππ =
π
ππ‘(ππΏ + ππ + ππ)
=π
ππ‘(
ππ
ππ‘πΏ + ππ + β«
π
πΆππ‘ )
=π2π
ππ‘2πΏ +
ππ
ππ‘π +
π
πΆ= 0
π 2 + π
πΏπ +
1
πΏπΆ= 0
π =β
π πΏ
Β± βπ πΏ
2
β4
πΏπΆ2
= βΞ± Β± βπΌ2 β π02
β = π
2πΏ π =
1
βπΏπΆ
The equations and methods of solving are the same as the parallel RLC circuits. The conditions
for damping (underdamped, overdamped, and critically damped) are also the same.
Ex: The capacitor is initially charged to 200V. At t = 0 the switch is closed and the
capacitor discharges. Find i(t) for t β₯ 0. Find vc(t) for t β₯ 0.
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8.1 Introduction to Complex Numbers Instead of using the standard Cartesian plane system which has an x and y axis to represent
numbers, use the Complex plane. The complex plane has a real and an imaginary axis. Complex
numbers consist of a real and an imaginary part. Complex values have two forms they can be
represented in. Note that π = ββ1. These techniques will be used in AC analysis of circuits.
Rectangular Form: n = a + bj , the a term is real, the j makes the b term imaginary.
||π|| = βπ2 + π2
Polar Form: n = cπππ or π = π⦀π, c is the magnitude, π is the phase angle measured CCW from
axis.
A very important concept related to complex numbers is Eulerβs identity.
Eulerβs Identity: πΒ±ππ = πππ π + ππ πππ
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The Phasor: A phasor is a complex number that contains the amplitude and phase of a
sinusoidal function. This value is represented in the frequency domain. Generally represented
with a squiggly line above the variable or the variable is bolded.
Can convert a number to a phasor by doing a phasor transform. A sample phasor transform is
shown below.
Ex: Voltage defined by AC outlet as v(t) = 169.7cos(110t + 60Β°).
Phasor transform is π = 169.7ej60 = 169.7⦀60Β°
Complex Conjugate: Denoted with a * (ie: ππππβ ). It is the number with the same real and
imaginary parts in magnitude but opposite sign.
Working in the complex plane leads to simpler calculations. To add complex numbers in
rectangular form, add the real components and then add the imaginary parts. To multiply
complex numbers in polar form, multiply the magnitudes and add the angles. However, your
calculator can do all of this for you.
9.1 Circuits with AC Conditions A sinusoidal source, is either a voltage or current source that varies sinusoidal. Follows the basic
wave equation seen below.
π£ = ππ cos (π€π‘ + β )
Most times when talking about power in AC circuits the voltage and current values are
expressed in RMS. RMS is the root mean square value, and is the magnitude of the source
divided by β2 .
ππ ππ = ππ
β2
When working with AC conditions normally work in the frequency domain, so we use
impedances. Impedances (Z) are the ratios of a circuit elements voltage phasor to its current
phasor. Impedance is analogous to resistance, inductance and capacitance in the time domain.
Impedance can be represented by the formula below. Where the imaginary part of the
impedance is the reactance.
π = π + ππ
The equations below shows the impedance for the three basic circuit elements. Note that π€ =
2ππ, thus your impedance is directly proportional to the impedance of the circuit.
π ππ ππ π‘ππ: π = π
πΌπππ’ππ‘ππ: π = ππ€πΏ
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πΆππππππ‘ππ: π = βπ
π€π= 1/ππ€π
Some other terms related to impedance are admittance and susceptance. Admittance is the
reciprocal of the impedance denoted by Y. Susceptance is then the imaginary part of the
admittance.
10.1 Circuit Analysis in AC Conditions When working with impedances, all of Kirchoffβs Laws still hold. Essentially this means that the
circuit analysis is the same as using time domain elements. Note however that if you have
multiple sources they can only be combined if they are operating at the same frequency. If they
are not operating at the same frequency you must use superposition to solve the circuit.
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Ex: Solve for the voltages denoted by V1 and V2.
11.1 Power in AC Conditions Instantaneous, Average and Reactive Power
The average (real) power is defined as: π = πππΌπ
2cos (ππ£ β ππ)
The reactive power is defined as: π = πππΌπ
2sin (ππ£ β ππ)
The instantaneous power is defined as: π = π + ππππ (2ππ‘) β ππ ππ(2ππ‘)
For the above equations Vm and Im are the magnitudes of the voltage and current phasors. The
angles ππ£ and ππ are the phase shift of the voltage and current phasors.
The average power (P) is also called the real power. This value represents the real portion of the
power, which is converted from electric to non-electric forms of energy. The reactive power (Q)
is the imaginary portion of the power. Depending on what type of circuit you have, the total
instantaneous power differs greatly.
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Purely Resistive:
If the circuit is purely resistive than the circuitβs voltage and phasor currents are in phase. This
means their phase difference is 0. Thus the reactive power term will go to 0, making the
instantaneous power completely real as seen below.
π = π + ππππ (2ππ‘)
Purely Inductive:
The circuit is out of phase by +90Β°, thus ππ£ β ππ = 90Β°. Thus has instantaneous power of
π = βππ ππ(2ππ‘). This means that there is no transfer of real power in purely inductive
cases.
Purely Capacitive:
The circuit is out of phase by -90Β°, thus ππ£ β ππ = -90Β°. Thus has instantaneous power of
π = βππ ππ(2ππ‘). This means that there is no transfer of real power in purely capacitive
cases.
Complex Power:
Sum of the real and reactive power, π = π + ππ. The magnitude of the complex power is the
apparent power. |π| = βπ2 + π2. If working with phasors we have π = πππππΌπππβ .
Power Factor:
The ratio of the real power over the apparent power delivered to the load. To get the power
factor of a circuit you must know the phase angle of both the voltage and current phasor, and if
the load is lagging or leading.
Lagging Power Factor β Current lags voltage Inductive load
Leading Power Factor β Current leads voltage Capacitive Load
πππ€ππ πΉπππ‘ππ = ππ = cos(ππ£ β ππ)
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Ex: Find the load current IL and load voltage VL. Find the average and reactive power
delivered to the load. Find the average and reactive power delivered from the source.