## What is Resistance & Resistivity?

### What are resistors?

Resistors are the first of the three passive circuit elements you will be studying in this course.  Knowing how resistance is related to current and voltage is fundamental to analyzing a circuit.  Fortunately, it’s also one of the easiest concepts you will learn.

### What is resistance?

Resistance (R) is a measure of how much a material prevents the flow of current and is measured in Ohms (Ω).  The resistance of a material depends largely on what the material is, but length and cross-sectional area also play a significant role.  These three factors are related to a material’s resistance by the following equation.

### $R = \frac{\rho\ell}{A}$

Where ρ is the resistivity of the material (in Ω·m), ℓ is its length (in m), and A is its cross-sectional area (in m2).  From this equation, a wire’s resistance is doubled when its length is doubled.  In contrast, doubling the wire’s radius effectively divides its resistance by four.

### What is resistivity?

Resistivity is a property of a material (not something that can be changed unless the material itself is changed).  Therefore, it plays the largest role in determining whether or not the material is a good conductor.  Silver, for example, has a resistivity of 1.6×10-8 Ω·m.  Meanwhile, wood has a resistivity of around 1014 Ω·m.  The disparity between these two resistivities is so great that length and cross-sectional area hardly matter; wood can never be a good conductor.  Note: While silver is a better conductor than copper, copper is used in wires because it’s much cheaper.  Written by Ryan Eatinger (reatinge@ksu.edu).  Thank you!

## Kirchoff’s Voltage Law

### What does Kirchoff’s Voltage Law state?

KVL states that the voltage rises around a loop is equal to the voltage drops.

$\sum_{}^{} V_{rises} = \sum_{}^{} V_{drops}$

Consider Figure 3.  The voltage source supplies 12 V to the two resistors.  This is a voltage rise.  The resistors, therefore, must drop the voltage 12 V to match the 12 V produced by the supply.  When determining KVL loop equations,

1)     Determine a starting point (a voltage source is typically a good place).

2)    Choose a direction (clockwise or counterclockwise).

3)    Label each component with a plus on one side and a minus on the other.

Obviously, below the voltage source, the voltage is 0 V and above it the voltage is 12 V.  The 12 V can be seen as a voltage rise or drop depending on which way you move around the loop.  The source is interpreted as a voltage rise when moving clockwise (minus to plus) and a voltage drop when moving counterclockwise (plus to minus).

### A few Kirchoff’s Voltage Law examples

Example: Find the voltage across resistor R1.

Solution: To find this voltage, we will start at the voltage source and move clockwise.  When moving clockwise, the voltage source is the only voltage rise (minus to plus) and the two resistors are both voltage drops (plus to minus).  Therefore the equation can be written as

$\sum_{}^{} V_{rises} = \sum_{}^{} V_{drops}$

$12 = V_{R1} + 3$

$V_{R1} = 9V$

It does not matter whether you go clockwise or counterclockwise.  If we had gone counterclockwise, the voltage source would have become the voltage drop and the voltage across the two resistors the voltages rises.  Be careful of the orientation of the voltage across components.  If VR1 was labeled with the plus on the right and the minus on the left, the voltage calculated would be -9 V.

Example: Find VR1

Solution: In nearly all the problems you’ll see, the orientation of the voltages is already labeled for you (orientation, as in which side has the plus and which has the minus).  You only need to worry about using the given orientation properly within your KVL equation.

In this case we will start from V1 and move clockwise.  V1 and V3 are the only voltage rises.  V2 is opposing the other two sources so it is a voltage drop, as are VR1 and VR2.  Our equation becomes:

$\sum_{}^{} V_{rises} = \sum_{}^{} V_{drops}$ $V_{1} + V_{3} = V_{R1} + V_{2} + V_{R2}$ $20 + 20 = V_{R1} + 30 + 3$

$V_{R1} = 7 V$ (Always label your values!)

So far, we have only worked with single loops, but you should recognize that there can be multiple loops in one circuit.  There can even be loops within loops.  For every loop, Kirchoff’s voltage law still applies.  Three loops are shown in Figure 7, but there are others that aren’t marked.  As you work with more complicated circuits, you should learn to identify any loops in a circuit and be able to apply KVL to each loop.  Written by Ryan Eatinger (reatinge@ksu.edu).  Thank you!

## Kirchoff’s Current Law

While analyzing a circuit may seem like a daunting task at first, circuit theory is simply a matter of breaking down larger circuits into smaller, more manageable sections and applying a few fundamental principles to these sections. One of these basics is known as Kirchoff’s Current Law.

### Kirchoff’s Current Law (KCL)

KCL states that the sum of the currents entering a node is equal to the sum of the currents leaving.

$\sum_{}^{} i_{entering} = \sum_{}^{} i_{leaving}$

In Figure 1, node 1 has three currents entering (a, c, and d) and two currents leaving (b and e).  Using KCL, all five currents are related to each other by the equation.

$\sum_{}^{} i_{entering} = \sum_{}^{} i_{leaving}$ $a + c + d = b + e$

It is important to understand the definition of a node.  A node is defined as a point where the voltage is the same.  While it is easy to identify the node in the example above as the point where all of the wires meet, that is not always the case.

Consider Figure 2.  When analyzing a schematic, do not assume a node is merely the intersection of wires.  For the sake of appearance, schematic diagrams cannot bring every wire to the same point as I have in Figure 1.  There are several intersecting wires in Figure 2, yet there are only three nodes in the circuit or, in other words, three different voltages.  Everywhere along the red wire, the voltage is 12 V.  Everywhere along the black wires, the voltage is 0 V.  After taking this class, you should be able to find the voltage at the blue node.

### Labeling Voltage on a Schematic

Before moving on to KVL, you should know how to read voltages on a schematic.  Voltage sources and other components often have plusses and minuses labeled on either side of them.  Keep in mind that voltage is the electric potential difference between two points.  This implies that voltage is a relative measure; one voltage is taken with respect to another.  The plusses and minuses are needed to signify the reference point for a measured voltage.  Finding the voltage across a component is a matter of finding the voltage at the ‘plus’ node and the ‘minus’ node and then subtracting the latter from the former.  For the supply on the left, Vs can be found from the following equation.

$V_{s} = V_{a} - V_{b}$

## Complex Numbers

A complex number can be represented graphically in the complex plane.

This point can be described in several different ways:

$x+jy = r e^{\j \theta} = r cos\theta + r j sin\theta$

Note that the third relationship is derived when substituting these trigonometric functions into the first relationship:

$x = r cos\theta \quad \& \quad y = r sin\theta$

Also:

$(x ^ {2} + y ^{2}) ^ { \frac{1}{2}} = r \quad \& \quad \theta = tan ^ {-1}( \frac{y}{x})$

Rectangular: $z_{1} = x_{1} + j y_{1} \quad \& \quad z_{2} = x_{2} + jy_{2}$

$z_{1} + z_{2} = (x_{1} + x_{2}) + j( y_{1} + y_{2})$

$z_{1} - z_{2} = (x_{1} - x_{2}) + j( y_{1} - y_{2})$

Polar: There is no simple way of doing this. The best method would be to convert to rectangular form using the formula: $r cos\theta + r j sin\theta$

### Multiplication/Division of Complex Numbers:

Rectangular: $z_{1} = x_{1} + j y_{1} \quad \& \quad z_{2} = x_{2} + jy_{2}$

$z_{1} \cdot z_{2} = (x_{1} + j y_{1})(x_{2} + j y_{2}) = x_{1}x_{2} + j(x_{2}y_{1} + x_{1}y_{2}) - y_{1}y_{2}$

$\frac{z_{1}}{z_{2}} = \frac{x_{1} + j y_{1}}{x_{2} + jy_{2}}(\frac{x_{2} - j y_{2}}{x_{2} - jy_{2}}) = \frac{x_{1}x_{2} + j(x_{2}y_{1} - x_{1}y_{2}) + y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}$

When multiplying two complex numbers in rectangular form, the FOIL method must be used. When dividing complex numbers in rectangular form, multiply the numerator and denominator by the complex conjugate of the denominator.

Polar: $z_{1} = c_{1} e ^ {j \theta_{1}} \quad \& \quad z_{2} = c_{2} e ^ {j \theta_{2}}$

$z_{1} \cdot z_{2} = (c_{1} e ^ {j \theta_{1}})(c_{2} e ^ {j \theta_{2}}) = c_{1} c_{2} e ^ {j(\theta_{1} + \theta_{2})}$ $\frac{z_{1}}{z_{2}} = (\frac{c_{1}}{c_{2}}) e ^ {j(\theta_{1} - \theta_{2})}$

## Basic Electrical Engineering Concepts

Before you start working with circuits, you need to understand the main concepts upon which the core of electrical engineering lies.  Understanding the basics will help you keep up with the material and reduce the number of errors you make in the future.  While there are some very important equations that you need to know, circuit analysis is not simply a matter of plugging numbers into an equation.  You need to understand what voltage, current, and resistance is, and how they relate to each other in order to take advantage of those equations and understand what’s really going on in the circuit.  This article will introduce you to the most basic concepts.

### Engineering Notation

In this course and many other courses in the department, we want you to work with engineering notation.  Scientific notation is useful for very small or very large numbers.  However, you should use and familiarize yourself with engineering notation for all other numbers.  Some common engineering prefixes are shown in the table below.

### Electric Charge

You know that an atom, in its neutral state, has a charge of zero.  You also know that if a neutral atom gains an electron, it becomes an ion with a charge of 1-.  This definition of charge works fine when talking about one atom, but when working with large numbers of atoms, a more practical definition of charge is needed, i.e. electric charge.  The unit for electric charge (denoted by the letter ‘q’) is the Coulomb (C).  When measured in Coulombs, an electron has a charge of approximately -1.60 × 10−19 C.  A proton, then, would have a charge of +1.60 × 10−19 C.

Example: Find the charge of 5.1×1018 ions of copper (Cu).  Each copper ion has a charge of 2+.

Solution: The copper ions have a surplus of protons, which means that the copper will have a positive charge.  Multiply the fundamental charge of each ion times the number of ions.  This gives the number of extra protons.  Now multiply the number of extra protons by the charge, in Coulombs, of a proton.

q = 2 x 5.1 x $10^{18}$ x 1.60x$10^{-19}$ = 1.63 C

### Coulomb’s Law

Coulomb’s law defines the magnitude of the force between two charges as:

$F = \frac{q1q2}{4\pi \epsilon r^2}$

where q1 and q2 are the two charges in Coulombs, r is the distance between the two points in meters, and  the permittivity constant of free space is equal to 8.85 × 10−12 F/m (Farads/meter).  If the force is negative, the two charges attract each other while a positive force means the two charges repel each other.  The constant $\frac{1}{4\pi \epsilon}$ is known as the electrostatic constant Kc, where

$Kc = \frac{1}{4\pi \epsilon} = 8.99 x 10^9$  $\frac{N*m^2}{C^2}$

Note: Newtons (N) are a unit of force.

If you’ve ever experimented with magnets, then you have witnessed this law before.  Two magnets repel and attract each other depending on the orientation of their poles.  This equation also shows that the force between two charges grows exponentially as they move together because of the r2 in the denominator.  You may have noticed this phenomenon as well.  Two magnets attract or repel each other when placed very close together, but the force between them dies off rather quickly as they’re pulled apart.

### Electric Current

Current is the flow of charge per unit time and is measured in amperes (A).  Current is represented by the letter ‘I’.

$I = \frac{Q}{t}$  –> $1A = \frac{1C}{1s}$

As the formula above indicates, one amp of current is equal to the flow of one Coulomb of charge per one second.  In other words, a wire carrying one amp of current moves one Coulomb of charge through the wire every second.  When you are working with electricity, keep in mind that one amp is a large amount of current; less than 100 mA of current can kill you!

### Voltage

Voltage is the difference in electric potential between two points and is measured in Volts (V).  If you’ve taken physics, electric potential is similar to the concept of potential energy, except in this instance, electric potential is equal to the potential energy per unit charge.  Voltage can be seen as the electric pressure, or driving force, that causes current to flow.  You may also see voltage referred to as electromotive force.

An important concept to understand when working with voltage and current is that there can be voltage without current flow, but there cannot be current flow without a voltage.  For example, think of two people on opposite sides of a box.  If both of them apply the same amount of force, the cart will not move.  However, if one of them applies more force, the box will move.  In both instances, a force (voltage) was applied to the cart (electrons), but only when there was a difference in force did we witness the cart move (current).

### Conventional Current Flow

In the example above, the cart moves away from the person applying more force toward the person applying less force.  Current and voltage interact in a similar manner in that current flows from higher voltages to lower voltages.  In other words, current is said to flow from the positive terminal of a battery to the negative terminal.  This is because current is described as the flow of positive charges.  The electrons that actually carry the charge through the wire have a negative charge.  Therefore, by definition, current flows in the opposite direction of the flow of electrons.

### Water Analogies

The challenge of learning the concepts of electricity is that electrons are hard to see and it’s hard for people to tell what is going on in a circuit.  Analogies to water have been made to help people understand different concepts encountered in electrical engineering.  Current, as you might have guessed, is compared to the flow of water while voltage is the difference in water pressure between two points.  More water analogies will be made throughout the course to help you understand new concepts.

This water analogy provides another example of how there can be voltage, but no current.  Think of the build up of pressure behind a dam.  The dam pushes back on the water, allowing no water to flow.  This situation is similar to a battery.  An ideal 12 V battery always has a potential difference between its terminals of 12 V, but no current flows until the battery is connected to a circuit.

#### Final Remark

Current is a through variable and voltage is an across variable.  Current flows through circuits, voltage does not.  Rather, voltage is the potential for current to flow.  When referring to voltage, never say “the voltage through the resistor.”  Instead, say “the voltage across the resistor” or “the voltage at a node.”  This article was written and edited by Ryan Eatinger, Kansas State University (reatinge@ksu.edu), thanks for the donation.

That covers a lot of introductory concepts!  If you have any questions feel free to comment.