A Guide for Principles of EE I at Rutgers University

September 3, 2015


Here we are - a new study guide - for Principles of Electrical Engineering 1 at Rutgers University in the Fall of 2015.

I will try to keep this as up to date as possible as the class goes on throughout the semester. Hopefully addressing all topics in the syllabus. The aim is for this to hopefully be able to serve as a study-guide by the time the end of the semester comes around. I hope that this guide will be able to cover all of the topics that are in this course, as well as provide examples of circuits and step-by-step solutions to different problems.


September 3rd 2015

Definition: Circuit

An interconnection of components or elements that form a closed path or many closed paths.

Basic Circuit Components

More Definitions

!Important! A network with \(b\) branches, \(n\) nodes and a total of \(l\) indenedent loops will satisfy the following:

\[b = l + n - 1\]

See the image below for an explanation Circuit and definitions slide

September 6th 2015

Circuit Variables


How does this help us???

Electrical current is actually defined as the flow of positive charge.

It is very important to note that current flow is usually defined as the flow with POSITIVE charge.

Electron Drift

Electrical Current

\[i(t) = \frac{dq(t)}{dt}\]

Voltage or Electric Potential Difference

\[v = \frac{dw}{dq}\]

Direction for Current and Voltage

Image of basic circuit element with flow


Power, \(p\) is defined as \(\frac{dw}{dt} = iv\)

Whenever the current flows in a direction from high potential to low potential (i.e. there is a voltage drop across the element), then the element is consuming the power or energy being delivered to it. (Positive Value

Whenever the current flows in a direction from low potential to high potential (i.e. there is a voltage rise across the element), then the element is generating the power or energy (Negative Value)

One way to think of it - is that whichever terminal (positive or negative) the current is flowing into - that is the going to be the sign of the power equation.


See below for more explanation

Passive Sign Convention

Passive Sign Convention Example

Finding the power consumed or delivered by an Element

If we have multiple elements in a circuit, we know the voltage drop/rise across each, and the current for all but one element, then we can find the power that is being delivered across the final element. We follow just a few simple steps

  1. For each element, find if the power is consumed or delivered.
  2. For each element calculate the power the is being consumed or delivered.
  3. Add all of the power across each element up. Whatever the remaining difference is, the opposite sign (positive/negative) is the amount of power delivered across the last element.

September 10th 2015

Unit 2 Circuit Components

Topics covered in this Unit:

Circuit Components Symbols: Ideal, Independent, Voltage Source

Independent Voltage Source: Generates a voltage drop across its terminals without relying on other voltages in the circuit.

Voltage Source Component

An ideal voltage source supplies the same voltage regardless of current.

Independent Current Source: Generates a current through a branch in a circuit without relying on other currents in the circuit.

Current Source Component

An ideal current source supplies the same current regardless of voltage.

DC (Direct Current) & AC (Alternating Current) Sources

There are two types current sources that are present in electric current. AC (Alternating current), and DC (direct current). In the images below you can see the difference in these two types of currents.

Direct Current

DC Current Image

Alternating Current

AC Current Image

Additional Circuit Sources

Dependent Sources Generate a voltage whose value depends on the value of another voltage or current in the circuit. They are usually represented by a diamond.

Voltage Controlled Voltage Source (VCVS): \(v_s = \alpha v_x\) Current Controlled Voltage Source (CCVS): \(v_s = ri_x\)

Voltage Controlled Current Source (VCCS): \(i_s = gv_x\) Current Controlled Current Source (CCCS): \(v_s = \beta i_x\)

Interconnection of Sources

Voltage Sources

Voltage sources in parallel must be of the same value.

Current Sources

Current sources connected in series must be of the same value.

Passive Circuit Components Symbols

Passive Circuit Component: A device that cannot generate electric energy and does not require external power sources to operate.

There are three types of passive circuit components:

Equation: \(v_R = Ri_R\)

Equation: \(i_C = C\frac{dv_C}{dt}\)

Equation: \(v_L = L\frac{di_L}{dt}\)

Passive Circuit Component


Resistance: The physical property of an element that impedes the flow of current (electron drift)

Represented by \(R\) and is defined by:

\[R = \frac{\rho \cdot l}{A}\]

The unit of resistance is the Ohm –> \(\Omega\)

Ohm’s Law: Linear Model

\(v= i \cdot R\) or \(i = \frac{v}{R}\)

When the voltage is positive and current flows into the positive terminal, then the current drops.

The voltage, \(v\) across and the current, \(i\) through a resistance \(R\) that flows down the potential hill

Power delivered to a resistor: \(p = i^2R=\frac{v^2}{R}\)

Note! Power is always consumed by a resistor (positive).

Hold up! hold up! I just said that power consumed by a resistor is always positive. How is that possible? What if we switch the positive and negative terminals?

Resistors don’t actually have positive and negative terminals. You can just assume - no matter the orientation or current flow - that they will be consuming power.

September 14 2015

Unit 2, Lecture 2

Calculating Power From Ohm’s Law

We know now that \(p = iv\), right? We also know that from Ohm’s law, \(v = iR\). This can then give us a new equation to calculate power based on resistance.

Another form is:

Open and Short Circuit Symbols

Open and Short Ciruit Symbols

The Switch Symbol

Switch Symbol

If a current can take a path through a circuit that eliminates flow through a resistor, then the current will not flow through to that part of the circuit (or resistor) at all.

Kirchoff’s Laws

Arguably some of the most important equations in solving circuits come from Kirchoff’s laws. They help us gain information about a circuit to be able to obtain and solve for different component’s quantities.

Let’s start off with current, \(i\).

Kirchoff’s Current Law

The algebraic sum of all the currents at any node in a circuit equals zero.

What does this mean?

\[\Sigma_j i_j = 0\]

KCL Example 1

Now! We are going to see a similar law, but this time it will apply to voltage.

Kirchoff’s Voltage Law

The algebraic sum of all the voltages around any closed path in a circuit equals zero.

So what exactly does this mean? It means that we can pick any closed loop in the circuit. When we follow that loop we must add (or subtract) the voltages over different elements together to give us a final sum of zero.

\[\Sigma_j v_j = 0\]

KVL Example 1

Example Problem 1

Mastering these problems takes lots of practice - so it could be beneficial to find more problems online or in a textbook to practice calculating and solving for circuits.

September 21st 2015

How many KCL and KVL Independent Equations?

\(b-n+1\) independent loop equations OR \(b_e - n_e + 1\) Is the number of essential branches and \(n_e\) is the number of essential nodes

\(n - 1\) independent node equations OR \(n_e - 1\) if every essential branch has a single current associated with it.

Steps to Solve for a Circuit
  1. Mark all nodes, resistors, currents, and voltages including direction and polarity.

  2. Choose Nodes for KCL

  3. Choose Loops for KVL

  4. Write KCL equations –> \(n_e - 1 = 3\)

Unit 3 - Simple Resistive Circuits

In this unit we’ll address creating circuits that contain resistors in series and parallel - and then finding methods for learning about how to simplify said circuits and then solve for them.

Resistors In Series

Resistors are an element within a circuit - and sometimes these elements are difficult to solve for indicidually. But what we can do is simplify these circuits with multiple of similar elements to make solving these circuits easier.

Resistors in series mean that the resistors are in a “line” with each other. There are no essential nodes between the two resistors. In this case (where there are no essential nodes) we can simply add the resistances of each resistor together

\[R_{eq} = R_1 + R_2\]

From here on our circuit diagrams we can simply just combine the two resistors into one, where the single resistor now has the resistance of \(R_{eq}\).

September 24th 2015

Simplifying Circuits with Resistors in Series and Parallel

Resistors in Series

We can simplify resistors that share a single essential node. We know that they are in series if they share an essential node and there are no other elements in between them. We can simply add their resistances.

So for resistors in series

\[R_{eq} = R_1 + R_2\]

Resistors in Series

Some notes about simplifying resistors in series

Resistors In Parallel

Resistors in parallel is where two resistors share a node with another branch from an element. To find the equivalent resistance we must use the reciprocal of the resistances.

So for resistors in parallel:

\[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\]


\[R_{eq} = \frac{R_1R_2}{R_1 + R_2}\]

Resistors in Parallel

The equivalent resitance of resistors in parallel will ALWAYS be smaller than the minimum resistance in the set of parallel resistors.

The Delta-Wye Connection

In the triangular (Delta) configuration - We name the resistors in a very systematic way.

In the Wye (Y) configuration

Resistors in Parallel

This transformation to and from the Delta and Wye (Y) configurations can be useful in analyzing more complex circuits.

September 28th 2015

Voltage Divider Circuit

So the first thing we need to know is that if we want to measure the voltage over a resistor, then we need to put the voltmeter in parallel with the resistor.

Now in a voltage divider circuit, we take that principle and expand on it to divide the voltage by a certain amount by putting multiple resistors in series with one another.

The voltage divider contains a voltage source of value \(V_x\) and resistors \(R_1\) and \(R_2\) in series with one another and the voltage source.

We know that from KVL the equation for the voltage must be that

\[V_x = V_1 + V_2\]

From KCL we can find that

\[i_x = i_1 = i_2\]

and now from KCL and Ohm’s we have

\[i_x = \frac{V_1}{R_1} = \frac{V_2}{R_2}\]

we want to now find \(V_1\) in terms of \(V_x\).

We can then find that \(V_x = V_1 + \frac{R_2}{R_1}V_1 = (1 + \frac{R_2}{R_1})V_1\)

So then solving for \(V_1\) we find:

\[V_1 = \frac{R_1}{R_1 + R_2}V_x\]

Similarly, if we solve for \(V_2\), then

\[V_2 = \frac{R_2}{R_1 + R_2}V_x\]

Current Divider Circuits

In a current divider circuit we can divide a voltage source’s current by an amount by placing resistors in parallel and putting an ammeter in series with one of the resistors.

So if we have two resistors in series, then through KVL we have:

\[V_x = V_1 = V_2\]

and through KCL we have that

\[i_x = i_1 + i_2\]

And with Ohm’s law included, we can get the equation

\[i_1 = \frac{R_2}{R_1 + R_2} i_x\]

So then if we solve for \(i_2\) we find that

\[i_2 = \frac{}{}i_x\]

October 1st 2015

The Wheatstone Bridge

In a Wheatstone bridge, the \(R_x\), (resistor with unkown resistance) is equal to:

Unit 4a Node-Voltage, Mesh-Currnet, & Source Transformation

The Node-Voltage Method

In the node voltage method, you must pick a node as a reference node. At that node we say that it is \(V_g = 0 V\). From here we can create equations that represent the voltage drops over multiple elements.

You can add the voltage drops over multiple elements to remove the amount of variables involved in the cicuit. From here we can represent each element with a difference between two different node voltages.

We can now convert some of these node voltages into current equations. The equation for current between two nodes with a resistance R is equal to:

\[i = \frac{V_b - V_a}{R}\]

Note that the value \(V_a\) is the value of the node connected to the resistor on one end, while \(V_b\) is equal to the voltage of the node at the other. \(V_b - V_a\) represents the voltage drop across the resistor.

The steps for Vode-Voltage Analysis:

  1. Mark all nodes, node voltages, and choose a reference (ground) node.
  2. Write the node voltage equations
  3. Calculate node voltages, elements, and current voltages.

October 5th 2015

Reviewing Node-Voltage Analysis

For every essential node we assign it a voltage V.

In essence, the nodal-voltage analysis is directly related to kirchoff’s current law. We calculate the voltages by the adding the currents that end up in each essential node.

The equations usually end up in a formal similar to the following:

If a voltage source is the only element between two essential nodes where one of the nodes is the reference node, the number of unknown node voltages is reduced.

Constraint Equations

A constraint equation is an equation in our node voltage (or mesh current) analysis where based on a current source or voltage source that determines when a specified potential difference MUST be a specific value between two nodes due to one of these sources.


When a voltage source is between two essential nodes (other than the reference node) we can combine those to form a supernode.

The Mesh Current Method

The mesh current method mimics the node-voltage method in the idea that it is based off of one of Kirchoff’s laws. In this case, it is based off of Kirchoff’s voltage law (KVL).

In the mesh current method we assume that we always hit resistors in the positive terminal. But when we hit others, we assume that if we hit the negative terminal, it will be negative, and positive terminal, positive.

October 15th 2015

Current Sources: Shared by two mesh currents


Slide with Supermesh

Some notes on a Super Mesh

Case 1:

Case 2:

Node Voltage or Mesh Current?

Deciding which approach to take in a particular circuit usually boils down to determining which method leads to easier math - the fewest numer of simultaneous equations

Source Transformations

How can we change a voltage source or current source, with its opposite type, while still keeping the original behavior of the circuit?

Source Transformations

We want to keep the same behavior of the circuit, so we must use equivalent resistance and collapse parts of the circuit to transform the sources. We need to make sure that current and voltage are preserved.

Source Transformations

Thevenin Equivalent

Suppose we have a section of a circuit name \(\alpha\). This \(\alpha\) is connected to another section \(\beta\). Now \(\alpha\) is a part that we always know will be constant. But what if \(\beta\) can change? How can we represent \(\alpha\) to simplify our calculations with different loads, \(\beta\)?

Thevenin Theorem

If \(\alpha\) is a linear circuit with passive or active elements, with all controlled and controlling branches contained within \(\alpha\), then we can replace entire network \(\alpha\) by an equivalent circuit that contains one independent voltage source in series with a single resistor (impedance) in series with it, such that the current-voltage relationship at \(\beta\) is unchanged.

Norton Equivalent -> A source transformation on Thevenin

Norton’s Theorem

Identical to thevenin theorem only that the equivalent circuit is one independent current source in parallel with a single resistor (impedance)

How to Find the Thevenin/Norton Equivalent

  1. Remove all elements that do not belong to th \(\alpha\) circuit - Usually refers to removing the output load
  2. Do TWO of the following

-The short circuit current is calculated by making a short between terminals a and b, and finding the current flowing through this short from the positive terminal to the negative one.




Equivalents Examples 1 Equivalents Examples 1

October 19th 2015

Thevenin and Norton Equivalents with Independent Sources

Method for direct calculation of \(R_{TH}=R_{EQ}\).

  1. Deactivate all independent voltage sources –> Set them equal to \(0V\).
    • Short Circuit.
  2. Deactivate all independent current sources –> Set to\(0A\).
  3. Find the equivalent resistance between the \(a\) and \(b\) terminals

(This only works if there are no dependent sources.)

However, we can modify this method slightly so that it is possible to calculate the \(R_{TH}\) even if there are dependent sources.

Equivalent resistance calculation

So after we calculate our equivalent resistance. We then need to find the voltage drop (or total current) through that branch of the circuit.

After finding the current through it, then we can replace the entire circuit (up to the terminals) with our thevenin or Norton equivalents.

Equivalents Examples 2

Thevenin and Norton Equivalents with Dependent Sources

Okay, so now we know how to calculate the Thevenin and Norton Equivalents, but what if there are dependent sources?

We need to do the following:

  1. Deactivate all independent voltage sources –> Set to \(0V\).
  2. Deactivate all independent current source –> set to \(0A\).
  3. Add a test source, \(V_T\), between the terminals \(a\) and \(b\).
  4. The Thevenin resistance is then \(R_{TH} = \frac{V_T}{i_T}\).

For example, if we find \(V_T = 1V\), then we just need to find the \(i_T\) from \(R_{TH} = \frac{1}{i_T}\).

While this method does introduce an extra variable, it still makes it relatively easy to calculate the equivalents with the node-voltage or mesh-current methods.

Notes on Supermeshes

Case 1 All mesh currents in a super mesh follow the same directions:

Case 2 Not all mesh current in a super mesh follow the same direction:

Maximum Power Transfer

Maximum Power Transfer 1

To find the maximum power, we need to take the derivative of the power and find the poitn where the derivative equals 0, because that will maximize (or minimize) the value for power.

So if we take the equation \(P = i^2(R_TH + R_L)\) or \(P = \frac{v^2R_L}{(R_{TH} + R_L)^2}\).

Now if we take the derivative with respect to \(R_L\), then we find that:

\[\frac{dP}{dR_L} = [\frac{v^2R_L}{(R_{TH} + R_L)^2}] = 0\]

So then from there we can find the P_{max} is when \(R_L = R_{TH}\).

Then our \(P_{max} = \frac{V^2_TH}{4R_L}\)

Finding the Thevenin/Norton Equivalents for a larger Circuit.

See The following problem to see how to calculate the Thevenin equivalent for a whole circuit

Thevenin Problem 1 Thevenin Problem 1 Thevenin Problem 1

October 22nd 2015

The Superposition Principle



At each stage, set ALL sources ecept for ONE to zero and find the contribution of this indicidual source to the current through an element or voltage across an element

Operational Amplifiers

Operational Amplifier

For our purposes, operational Amplifiers are a 5 terminal element.

There are 5 terminals:

Operational Amplifier 2

An operational amplifier amplifies the difference between the inverting input and the non-inververting input.

Operational Amplifier 3

We can simplify the behavior for an op. amp. so long as the voltage remains in the linear region of the amplifier’s operation. This allows us to simpligy the model.

Operational Amplifier 4

A good Op-Amp circuit model has 3 important characteristics

Ideal Operational Amplifier

Operational Amplifier Feedback Loop

Operational Amplifier Feedback 1

We can simplify the model of the op-amp so that we can solve for the output voltage as a function of the input voltage.

Operational Amplifier Simplified 1

October 26th 2015

Continuing Non-Inverting Amplifiers

In short: The voltage output from an operational amplifier will stay between the two input voltages of \(-V_{cc}\) and \(+V_{cc}\).

The slope of the range then depends on our Amplifier’s gain, \(V_{cc}/A\).

\[V_o = \frac{A}{1 + FA}\]

Op Amp Diagram

\[V_o = ( 1 + \frac{R_1}{R_2})V_{IN}\]

Inverting Operational Amplifiers

Now we’re going to take a look at inverting amplifiers

Op Amp Diagram

From this we focus on the 4 different types of amplifier circuits. The 4 types we focus on are:

  1. Inverting
  2. Non-Inverting
  3. Summing
  4. Difference

Inverting Op Amp Circuit

Inverting Op Amp Circuit

In an inverting amplifier our gain values \(K\) will be negative because we want the output to be the inverted from the input.

The output voltage is calculated as:

\[v_0 = \frac{-R_f}{R_s}v_s\]

It might also be useful to note that

Non-Inverting Op Amp Circuit

Non-Inverting Op Amp Circuit

In an inverting amplifier our gain values \(K\) will not be negative because we want the output to be the same sign as the input.

The output voltage is calculated as:

\[v_0 = \frac{R_f + R_s}{R_s}v_g\]

Operation in the linear range requires that:

\[\frac{R_s + R_f}{R_s} < \left|\frac{V_{CC}}{v_g}\right|\]

Summing Op Amp Circuit

Summing Op Amp Circuit

In a summing amplifier circuit, the output voltage can be related by:

\[v_0 = -(\frac{R_f}{R_a}v_a + \frac{R_f}{R_b}v_b + \frac{R_f}{R_c}v_c)\]

Now if \(R_s = R_a = R_b = R_c\) (all resistors equal) then we can say the output voltage is

Then if we find that \(R_s = R_f\), the summing amplifier’s output is simply

Difference Op Amp Circuit

Difference Op Amp Circuit

Difference amplifier circuits are slighly different from the previous ones. However we can still simplify the equation to obtain our output voltage. We say for the difference circuit that

\[v_0 = \frac{R_d(R_a + R_b)}{R_a(R_c + R_d)}v_b - \frac{R_b}{R_a}v_a\]

But then if we find that the circuit variables satisfies the equation: \(\frac{R_a}{R_b} = \frac{R_c}{R_d}\), then we can represent this as

\[v_0 = \frac{R_b}{R_s}(v_b - v_a)\]

November 2nd 2015

Unit 6 - Inductance and Capacitance

Up until now we assumed that everything in a circuit happened instantaneously.

capacitors and inductors have inputs and outputs that depends on time!

These elements store energy at different times and they may either produce or absorb energy.

Capacitors and Inductors

Inductor and Capacitors

Capacitors and Inductors: Energy Storage


Capacitance is a measure of the ability of a device to store energy in the form of a separated charge or an electric field

\[C = \frac{\epsilon A}{d}\]
\[q = C \cdot v(t)\]

It should be noted that if there is no change in voltage, current DOES NOT flow through a capacitor.

Capacitor - Current, Charge, Power, and Energy

We can calculate the energy stored in a capacitor over time using the following formulas:

So then using the integration from \(t_0\) to \(t\) for charge, then we can find the work required which we find as

\[W = \frac{1}{2}Cv^2(t) = \frac{1}{2}\frac{q^2(t)}{C} = \frac{1}{2}q(t)v(t)\]

Finding the Voltage and Current from a graph of capacitor states.

We can find the graph for voltage or current for a capacitor given that we have at least one of the graphs.

We can also solve for the total charge on a place capacitor knowing that \(v(t) = \frac{q(t)}{C}\)

Energy From a Capacitor over Time

The work from a capacitor is the integral of the power over a period of time.

Simplifying Capacitors

It’s very easy to simplify capacitors. They are similar to resistors in how we simplified them. The only difference is that the way we add them together is flipped.

That is, in parallel

\[C_{eq} = C_1 + C_2 + ... + C_n\]

and in series:

\[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}\]

November 5th 2015

Energy Stored in a Capacitor

\[i(t) = C\frac{dv(t)}{dt}\] \[w = \frac{1}{2}Cv^2(t)\]


To calculate the inductance of a inductor we can use the following equation:

\[L = \frac{\mu N^2A}{l}\]

And the voltage from an inductor

\[v(t) = L\frac{di(t)}{dt}\]

Inductor Voltage

The factors that affect inductance:

The work done by an inductor is proportional to the current flowing through it. We can say that:

\[w(t) = \frac{1}{2}Li^2(t)\]

Inductors in Series and in Parallel

Inductors simplify similar to resistors.

In Series

\[L_{eq} = L_1 + L_2 + L_3 + ... + L_n\]

In Parallel:

\[\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + ... + \frac{1}{L_n}\]

Inductor and Capacitors - A Summary

Initial Conditions of DC Switched Circuits

If a DC circuit is steady state then,

November 9th 2015

Unit 9: DC & Sinusoidal Transient and Steady State Response Phasor (frequency) Domain Analysis

A Sinusoidal Signal - AC (Alternating Current)

In AC current the voltage level alternates with respect to a trigonometric sine function. The general equation is

\[v(t) = V_mcos(\omega t + \phi)\]

And our variables:

So when our voltage shifts up and down, we also find that we have a current alternating as well.

However sometimes these are not always in phase, that is the peaks and troughs of each function will not line up perfectly.

So we can say that if voltage is lagging then the current is going to be leading.

Or conversely, if the voltage reaches its peak before the current does, we say the voltage leads and that the current lags.

Lagging and Leading AC Equations

Charging a Capacitor from a DC source through a Resistance

We can analyze the circuit using KVL.

First Order RC Circuit

From this we have a first order differential equation. Generally this equation shouldn’t be too difficult to solve.

After solving this differential equation for \(v_0\) we find that

\[v_0(t) = (v_0(0) - E)e^{\frac{-t}{RC}} + E\]

RC Circuit Charging

Capacitor Charging for an AC Source

So now what happens if our source \(V_{in}\) is an AC source where the function of the output is \(Acos(\omega t)\)?

After quite a longer derivation we find that

\[v_o(t) = (v_o(0) - \frac{A}{1 + \omega^2R^2C^2})e^{\frac{-t}{RC}} + \frac{A}{1 + \omega^2R^2C^2}(cos(\omega t) + \omega RCsin(\omega t))\]

So what can we tell from this equation?

For a sinusoidal input voltage, the steady state output voltage is a sinusoidal signal of the same angular frquency \(\omega\) but could have **different ampligtude and phase angle.

Currents will also be sinusoidal signals with the same angular frequency \(\omega\) and a different amplitude and phase angles.

If we simplify the analysis by using only the amplitude and phase information we get Phasor Domain Analysis

When we use \(cos(\omega t)\) as the reference sinusoid, we have the following transformation from the time domain to phasor domain:

\[Acos(\omega t + \theta) ==> Ae^{j\theta} = \frac{A}{\theta}\]

When we use \(cos(\omega t)\) as the reference sinusoid, we have the following inverse tranformation from phasor domain to time domain.

November 12th 2015

Complex Numbers

A complex number is in the form

\[a + jb\]

The simples complex value is \(i = j = \sqrt{-1}\)

Recall Euler’s Formula is

\[e^{jx} = cos(x) + jsinx\]

Where \(a = cos(x)\) and \(b = sin(x)\).

\(a\) and \(b\) are both real numbers.

Plotting Imaginary Numbers

Imagine our axes in a 2-dimensional graph. Normally in 2D we define the horizontal axis as the x axis, and the vertical axis as the y-axis.

However in dealing with imaginary numbers we change the axes slightly.

When plotting imaginary numbers:

What about on a polar coordinate system?

We can simply represent \(a + jb\) as

\[a + jb = M(cos(\theta) + jsin(\theta))\]

This allows us to express the complex numbers in a polar coordinate system on a graph as well.

So now we have two ways of writing complex numbers

  1. \[a + jb\]
  2. \[M(cos(\theta) + jsin(\theta))\]

But wait!

Do you remember the Euler formula from earlier?

We can manipulate the 2 representation above to actually make our 3rd representation


Converting from Rectangular to Polar Coordinates

So we can convert from rectangular coordinates to polar by simply using the relationships:

Adding and Subtracting Complex Numbers

When adding or subtracting these types of numbers, we simply only add or subtract only the real parts to obtain the sum or difference of real answers. Then we we simply add or subtract all imaginary parts to obtain the sum or difference of imaginary parts.

For example, given the following:

\[(a + jb) + (c + jd)\]

The result is

\[(a+c) + j(b + d)\]

But what happens if we have items in polar coordinates?

Well, the simplest solution is to actually just convert the numbers to rectangular

Multiplication of Complex Numbers

Given the numbers:

Then the multiplication result of these two is

\[ac + jad + jbc + j^2bd\]

Simplified this yields

\[(ac - bd) + j(ad + bc)\]

We subtract \(bd\) because when we have \(j^2\), we find it is equal to \(-1\) given the definition of \(j\)

For polar coordinates:

\[M_1M_2e^{j(\theta_1+\theta_2)} = M_1M_2, \theta_1+\theta_2\]

Complex Conjugate

The complex conjugate of a number \(z = a + jb\), we find that the complex conjugate denoted \(z^*\), we find is equal to

\[z = a + jb, z^* = a - bj\]

Properties of Complex Conjugates

If we multiply two complex conjugates together, we find that in the expansion:

\[a^2 + jab - jab -j^2b\]

which is equal to the square magnitude:

\[a^2 + b^2\]

Division of Complex Numbers

The division of complex number using the rectangular coordinate system is difficult and requires a fair amount of algebra. However, converted to polar coordinates it is much easier

\[\frac{M_1e^{j\theta_1}}{M_2e^{j\theta_2}} = M_1M_2, \theta_1-\theta_2\]

November 19th 2015

Conversions from the Time Domain to Phasor Domain

Conversions from the time domain to a phasor domain make our evaluations and analysis of AC response circuits much easier.

For every item in the time domain, we have an equivalent in the phasor domain. Remember that \(j=\sqrt{-1}\).

For a general overview of the Time-Domain to Phasor-Domain conversions see the table below.

Time Domain Phasor (Frequency) Domain
\(f(t)=Acos(\omega t + \theta)\) \(Ae^{j\theta}=A\angle\theta\)
\(v(t)=V_mcos(\omega t+\theta_v)\) \(V=V_me^{j\theta_v}=V_m\angle\theta_v\)
\(i(t)=I_mcos(\omega t+\theta_i)\) \(I=I_me^{j\theta_i}=I_m\angle\theta_i\)
\(v_\Delta (t)=\frac{dv(t)}{dt}=V_m\omega cos(\omega +\theta_v +90)\) \(V_\Delta = j\omega V\)
\(v_{\int}(t)=\int v(t) = \frac{V_m}{\omega}cos(\omega t+\theta_v -90)\) \(V_{\int}=\frac{1}{j\omega}V\)

Now when dealing with RC and RLC circuits we normally would calculate the voltage/current/resistance by using Ohm’s law (\(V=IR\)). When dealing with phasors our calculations are simplified.


Because our voltage is a function of time we need to deal with functions when we calculate our values. But in the phasor domain it is simply V=IR (When \(V\) and \(I\) are in phase!!).

Time Domain: \(i(t)R = v(t)\)

Phasor Domain: \(\frac{V}{I} = R\)


Time Domain: \(L\frac{di(t)}{dt}=v(t)\)

Phasor Domain: \(\frac{V}{I} = j\omega L\)


Time Domain: \(C\frac{dv(t)}{dt}=i(t)\)

Phasor Domain: \(\frac{V}{I} = -j\frac{1}{\omega C}\)


Impedance is in the units of Ohms.

\[Z = \frac{V}{I} = R + jX\]


Admittance is in the units of Siemens (\(\frac{1}{\Omega})\).

\[Y = \frac{1}{Z} = \frac{I}{V} = G + jB\]

RLC in Time vs Phasor

Time Domain Phasor Domain
Resistance \(R\) Impedance \(Z=R\)
Inductance \(L\) Impedance \(Z=j\omega L\)
Capacitance \(C\) Impedance \(Z=-j\frac{1}{\omega C}\)

November 23rd 2015

AC State Analysis

In solving these circuits we need to first move from the time domain to the phasor domain before analyzing the circuit. After which we then analyze the circuit, and then finally convert back to the time domain

Maximum Power Transfer

This occurs when the \(Z_{load} = Z_{TH}\).

November 30th 2015

Analyzing AC Circuits in the Phasor Domain

When analyzing circuits in the phasor domain we need to convert all of the elements in the circuit to have impedance values for the phasor domain.

Remember Impedances are in the units of Ohms (\(\Omega\)).

Also remember that to convert different elements to the phasor domain they each have their own respective equations.

After converting all elements to the phasor domain we can then solve for our circuit using node voltage or mesh current method using the impedances for each.

However you must keep imaginary and real values separated. Remember that the value \(j\) represents \(\sqrt{-1}\). So we cannot simply add the impedance values from resistors and then capacitors (or inductors) because we must separate the real and imaginary values.

This means that it is possible for solutions (voltage, current) to a circuit to be in terms of a real and imaginary values when solving for the entire circuit in the phasor domain.

How is that possible?

Remember that we’re dealing in the phasor domain because we have an AC current/voltage source. The equation that represents the source has the form \(V(t) = Acos(\omega t + \phi)\).

If we recall from earlier we represented our complex terms as a real portion and an angle.

The imaginary part of a solution will represent the tangent value of the angle, \(\phi\). Then the real part will represent the amplitude, \(A\).

December 3rd 2015

Chapter 10a - Power and Phasor Domain Analysis

The root mean square (RMS) value of a periodic function \(x(t)\) is defined as:

\[X_{rms} = \sqrt{\frac{1}{T}\int^{t_o+T}_{t_o} x^2(\tau)d\tau}\]

This RMS value is equivalent to the **maximum (peak) voltage divided by \(\sqrt{2}\).

\[V_{rms} = \frac{V_m}{\sqrt{2}}\]

The same thing applies for the current in AC circuits.

\[I_{rms} = \frac{I_m}{\sqrt{2}}\]

The rms value of any periodic voltage or current delivers the same average power to a resistor as a DC voltage with the same value.

So for the average power delivered to a circuit, we find that given a simple AC source with function \(V_mcos(\omega t + \theta_v)\), then the average power delivered to the resistor is:

\[P_{avg} = \frac{V_{rms}^2}{R}\]

So then we know how to calculate our power in the time domain…but what about the phasor domain?

The instantaneous power in an AC system in the time domain is

\[P = -\frac{V_mI_m}{2}sin(\theta_v-\theta_i)sin(2\omega t)\]

Relation between voltage and Current on Capacitors and Inductors

ELI the ICE man

This can help us rememebr when voltage lags or leads in a capacitor and how we can estimate the phase angles.

Another way that you could remember this is that capacitors have a negative imaginary impedance value. Because it is negative, this corresponds to a -90 degrees - LAG.

We can also say the same for inductors where inductors have a positive imaginary impedance value, so this positive imaginary value corresponds to a +90 degree phase - LEAD.

A Guide for Principles of EE I at Rutgers University - September 3, 2015 - zac blanco