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.

𝑣 = 𝑑𝑤

𝑑𝑞

Where v = voltage (V), w = energy (J), q = charge (C)

Current: Rate of charge flow. Measured in amps.

𝑖 = 𝑑𝑞

𝑑𝑡

Where i = current (A), q = charge (C), t = time (s)

Power: Time rate of expending or absorbing energy. Measured in watts.

𝑝 =𝑑𝑤

𝑑𝑡

Where p = power (W), w = energy (J), t = time (s)

Conservation of Energy: Total power delivered = total power absorbed. Energy can only be

transferred.

∑𝑝 = 0

Where p = power (W)

1.2 Basic Circuit Elements Resistor: A device having a designed resistance that electric current can flow through.

𝑣 = 𝑖 ∙ 𝑅 𝑝 = 𝑣 ∙ 𝑖

R = Resistance, measured in

Ohms (Ω)

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 Kirchhoff’s Laws Kirchhoff’s Current Law: The sum of the currents entering a node is zeros. (The incoming

currents must equal the outgoing currents).

For all general cases:

∑𝑖𝑛𝑜𝑑𝑒 = 0

For the diagram to the left:

𝑖𝑎 + 𝑖𝑏 − 𝑖𝑐 − 𝑖𝑑 = 0

Kirchhoff’s Voltage Law: The sum of voltage rises in a loop is zero.

For all general cases:

∑𝑣𝑑𝑟𝑜𝑝𝑠 𝑖𝑛 𝑙𝑜𝑜𝑝 = 0

For the diagram to the

left:

−𝑣𝑎 + 𝑣𝑏 − 𝑣𝑐 = 0

1.4 Basic Resistor Simplifications Resistors in Series:

𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4

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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𝑅𝑒𝑞 = 𝑅1 ∙ 𝑅2

𝑅1 + 𝑅2

1.5 Divider Circuits Voltage Divider Circuit: Voltages are proportional to their corresponding resistance

𝑉1 = 𝑉𝑠 ∙ 𝑅1

𝑅1 + 𝑅2

𝑉2 = 𝑉𝑠 ∙ 𝑅2

𝑅1 + 𝑅2

Current Divider Circuit: Currents are reciprocal to their corresponding resistances

𝑖1 = 𝑖𝑠 ∙ 𝑅2

𝑅1 + 𝑅2

𝑖2 = 𝑖𝑠 ∙ 𝑅1

𝑅1 + 𝑅2

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 Kirchhoff’s 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 Kirchhoff’s

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 (∆) 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 ∆. Use the

right side if going from ∆ to Y.

𝑅1 = 𝑅𝑏𝑅𝑐

𝑅𝑎 + 𝑅𝑏 + 𝑅𝑐

𝑅2 = 𝑅𝑐𝑅𝑎


𝑅3 = 𝑅𝑎𝑅𝑏


𝑅𝑎 = 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅3𝑅1

𝑅1

𝑅𝑏 = 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅3𝑅1

𝑅2

𝑅𝑐 = 𝑅1𝑅2 + 𝑅2𝑅3 + 𝑅3𝑅1

𝑅3

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 𝑖𝑠 = 𝑉𝑠/𝑅

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.

𝑃𝑙 𝑚𝑎𝑥 = 𝑉𝑡ℎ

2

4𝑅𝐿

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 + … + Outputn

To turn off the other sources:

• Current Sources: Open

• Voltage Sources: Short

* Superposition can not be used to calculate power as power is a non-linear function*

Find the current going through the 3KΩ 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.

𝑣𝑙 = 𝐿 𝑑𝑖

𝑑𝑡

* 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.

𝑖𝑐 = 𝐶 𝑑𝑣

𝑑𝑡

* 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: 𝜏 = 𝐿

𝑅= 𝑅𝐶

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, 𝜏 = 𝐿

𝑅

iii) Use the equation 𝑖(𝑡) = 𝑖𝑓𝑖𝑛𝑎𝑙 + (𝑖𝑜 − 𝑖𝑓𝑖𝑛𝑎𝑙)𝑒−𝑡

𝜏⁄ , to get i(t) of the inductor

Switch has been closed for a long time, opens at time 0. Find iL(t) for t ≥ 0, i0(t) for t ≥ 0+, vo(t)

for t ≥ 0+.


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RC Circuit:

i) Find initial voltage through the capacitor, Vo

ii) Find the time constant, 𝜏 = 𝑅𝐶

iii) Use the equation 𝑣(𝑡) = 𝑣𝑓𝑖𝑛𝑎𝑙 + (𝑣0 − 𝑣𝑓𝑖𝑛𝑎𝑙)𝑒−𝑡/𝜏, 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 ≥ 0, vo(t) for t ≥ 0+, io(t) for t ≥ 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.

𝑖(𝑡) = 𝑖𝐹𝑖𝑛𝑎𝑙 + (𝑖0 − 𝑖𝐹𝑖𝑛𝑎𝑙)𝑒−(

𝑅𝐿

)𝑡

(same equation as we used for natural response… think about what I approaches as t goes to ∞)

Ex: Find the expression for il(t) for t≥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.

𝑣𝑐(𝑡) = 𝑣𝐹𝑖𝑛𝑎𝑙 + (𝑣𝑜 − 𝑣𝐹𝑖𝑛𝑎𝑙)𝑒−𝑡/𝑅𝐶


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Ex: Find vc(t) t≥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 ≤ t ≤ 35ms, iL(t) for

t ≥ 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)

𝑑

𝑑𝑡∑ 𝑎𝑙𝑙 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠 =

𝑑

𝑑𝑡(𝑖𝑐 + 𝑖𝑅 + 𝑖𝐿)

=𝑑

𝑑𝑡(

𝑑𝑣

𝑑𝑡𝐶 +

𝑉

𝑅+ ∫

𝑉

𝐿𝑑𝑡 )

=𝑑2𝑣

𝑑𝑣2𝐶 +

𝑑𝑣

𝑑𝑡

1

𝑅+

𝑣

𝐿= 0

𝑑2𝑣

𝑑𝑣2+

𝑑𝑣

𝑑𝑡

1

𝑅𝐶+

𝑣

𝐿𝐶= 0

Parallel RLC has the following characteristic equation (based on the differential equation for

current), neper frequency and resonant radian frequency.

𝑠2 + 𝑠

𝑅𝐶+

1

𝐿𝐶 = 0

𝑠 =−

1𝑅𝐶 ± √ 1

𝑅𝐶

2

−4

𝐿𝐶2

= −α ± √𝛼2 − 𝜔02

∝ = 1

2𝑅𝐶 𝜔 = √

1

𝐿𝐶

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.

𝑣 = 𝐴1𝑒𝑠1𝑡 + 𝐴2𝑒𝑠2𝑡 = Overdamped Response [α2 > ω2, s is real]

𝑣 = 𝐵1𝑒−∝𝑡 cos(𝑤𝑑𝑡) + 𝐵2𝑒−∝𝑡 sin(𝑤𝑑𝑡) = Underdamped Response [α2 < ω2, s is complex]


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𝑣 = 𝐷1𝑡𝑒−∝𝑡 + 𝐷2𝑡𝑒−∝𝑡 = Critically Damped Response [if α2 = ω2, 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 ≥ 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 Kirchoff’s 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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𝑑

𝑑𝑡∑ 𝑎𝑙𝑙 𝑣𝑜𝑙𝑡𝑎𝑔𝑒𝑠 =

𝑑

𝑑𝑡(𝑉𝐿 + 𝑉𝑅 + 𝑉𝑐)

=𝑑

𝑑𝑡(

𝑑𝑖

𝑑𝑡𝐿 + 𝑖𝑅 + ∫

𝑖

𝐶𝑑𝑡 )

=𝑑2𝑖

𝑑𝑡2𝐿 +

𝑑𝑖

𝑑𝑡𝑅 +

𝑖

𝐶= 0

𝑠2 + 𝑅

𝐿𝑠 +

1

𝐿𝐶= 0

𝑠 =−

𝑅𝐿

± √𝑅𝐿

2

−4

𝐿𝐶2

= −α ± √𝛼2 − 𝜔02

∝ = 𝑅

2𝐿 𝜔 =

1

√𝐿𝐶

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 ≥ 0. Find vc(t) for t ≥ 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 𝑗 = √−1. 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.

||𝑛|| = √𝑎2 + 𝑏2

Polar Form: n = c𝑒𝑗𝜃 or 𝑛 = 𝑐⦤𝜃, c is the magnitude, 𝜃 is the phase angle measured CCW from

axis.

A very important concept related to complex numbers is Euler’s identity.

Euler’s Identity: 𝑒±𝑗𝜃 = 𝑐𝑜𝑠𝜃 + 𝑗𝑠𝑖𝑛𝜃


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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 + 60°).

Phasor transform is 𝐕 = 169.7ej60 = 169.7⦤60°

Complex Conjugate: Denoted with a * (ie: 𝑉𝑒𝑓𝑓∗ ). 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.

𝑣 = 𝑉𝑚 cos (𝑤𝑡 + ∅)

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 √2 .

𝑉𝑅𝑀𝑆 = 𝑉𝑚

√2

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.

𝑍 = 𝑅 + 𝑗𝑋

The equations below shows the impedance for the three basic circuit elements. Note that 𝑤 =

2𝜋𝑓, thus your impedance is directly proportional to the impedance of the circuit.

𝑅𝑒𝑠𝑖𝑠𝑡𝑜𝑟: 𝑍 = 𝑅

𝐼𝑛𝑑𝑢𝑐𝑡𝑜𝑟: 𝑍 = 𝑗𝑤𝐿


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𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑜𝑟: 𝑍 = −𝑗

𝑤𝑐= 1/𝑗𝑤𝑐

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 Kirchoff’s 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: 𝑃 = 𝑉𝑚𝐼𝑚

2cos (𝜃𝑣 − 𝜃𝑖)

The reactive power is defined as: 𝑄 = 𝑉𝑚𝐼𝑚

2sin (𝜃𝑣 − 𝜃𝑖)

The instantaneous power is defined as: 𝑝 = 𝑃 + 𝑃𝑐𝑜𝑠(2𝜔𝑡) − 𝑄𝑠𝑖𝑛(2𝜔𝑡)

For the above equations Vm and Im are the magnitudes of the voltage and current phasors. The

angles 𝜃𝑣 and 𝜃𝑖 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 circuit’s 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.

𝑝 = 𝑃 + 𝑃𝑐𝑜𝑠(2𝜔𝑡)

Purely Inductive:

The circuit is out of phase by +90°, thus 𝜃𝑣 − 𝜃𝑖 = 90°. Thus has instantaneous power of

𝑝 = −𝑄𝑠𝑖𝑛(2𝜔𝑡). This means that there is no transfer of real power in purely inductive

cases.

Purely Capacitive:

The circuit is out of phase by -90°, thus 𝜃𝑣 − 𝜃𝑖 = -90°. Thus has instantaneous power of

𝑝 = −𝑄𝑠𝑖𝑛(2𝜔𝑡). This means that there is no transfer of real power in purely capacitive

cases.

Complex Power:

Sum of the real and reactive power, 𝑆 = 𝑃 + 𝑗𝑄. The magnitude of the complex power is the

apparent power. |𝑆| = √𝑃2 + 𝑄2. If working with phasors we have 𝑆 = 𝑉𝑒𝑓𝑓𝐼𝑒𝑓𝑓∗ .

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 – Current lags voltage Inductive load

Leading Power Factor – Current leads voltage Capacitive Load

𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 = 𝑝𝑓 = cos(𝜃𝑣 − 𝜃𝑖)


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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.

elec 221 workbook solutions

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