# Session 1

The purpose of this session is to teach you the basics of electricity and introduce the concepts of potential difference, current, resistance and power. You will learn how to use Ohm’s law to calculate all of the above parameters in:

• Series circuits
• Parallel circuits
• Series mixed with parallel circuits

By the end of this session the learner will be able to:

• Use DC circuit theory to calculate current, voltage and resistance in DC Networks
• Use an electronic simulator to simulate a DC network containing 5 resistors and a power source
• Compare the results of calculation and simulation in terms of accuracy and validity

BTEC Outcomes Covered

• Unit 6 Pass 1

In the Beginning

All matter in the universe is made up of atoms. The basic structure of an atom is shown next:

The nucleus contains positively charged protons and atomic particles with no charge known as neutrons.

Orbiting the nucleus, rather like the planets in the solar system orbit the sun, we have electrons. Electrons have a negative charge which balances the postive charge of the proton.

The atom in its natural state is electrically balanced. To achieve electrical balance the number of electrons orbiting the nucleus must match the number of protons in the nucleus.

Some materials have many protons and electrons. Copper for example has 29 protons and electrons. The electrons orbit in shells around the nucleus. Each shell can contain a maximum number of electrons. The shells are given letters from the closest to the furthest from the nucleus:

• K Shell = 2 electrons
• L Shell = 8 electrons
• M Shell = 18 electrons
• N Shell = 32 electrons

And so on. Next we have a diagram of a copper atom. Copper like most metals, makes a good conductor.

Note the electrons arranged in shells about the nucleus. Electrons in the outermost shell are known as valence electrons. They are weakly attracted to the protons in the nucleus and easily displaced. When an atom loses an electron we say it has become a positive ion.

Positives ions attract electrons from neighboring atoms. Remember atoms naturally seek electrical balance. When this happens, electrons move or flow between atoms. Electricity is basically the flow of electrons moving between 2 points in a conductor.

Thus we can also define a conductor as being a material that allows electrons to move easily between atoms.

Conversly an insulator is a material that inhibits or completely stops the flow of electrons. Glass, rubber and paper all make good insulators.

Resistance is a measure of how easily a material permits the flow of electrons through it. Electrical resistance is measured in Ohms and given the following symbol Ω (omega).

Potential Difference

There is random movement of electrons happening all the time in conductive materials. Heat causes random electron displacement and is one of the main causes of electrical noise and interference in electrical circuits. The electron movement is rather small and has no specific direction.

In order to get the electrons to flow in significant numbers and in the the same direction we need some type of electrical force. When electrons flow in this way they do work, the amount of work done is proportional to the amount of electrons flowing. The flow of electrons represents a transfer of energy. Electrical power (measured in Watts) is the rate at which energy is transferred.

Power (Watts) = Joules (Energy)/Time(Seconds)

There are two main ways of ‘forcing’ electrons through a material. The first is to create a chemical reaction as in a battery.

The second is by movement of a conductor in a magnetic field. This is known as Faradays law of electromagnetic induction and we will becoming very familiar with it later on in our studies.

The idea is to create a circuit. A circuit has 2 terminals (ends), one terminal produces lots of positive ions (the positive terminal), the other terminal has lots of spare electrons (the negative terminal).

Electrons are attracted to the postive terminal if we provide a path through which they can flow i.e. a circuit. The following diagram illustrates this.

The  difference in the amount of positve ions and spare electrons available at each of the two terminals determines how many electrons can flow through the circuit. Electrical and electronic engineers call this the potential difference between 2 points in a circuit. It is an indication of the potential for electron flow between the 2 points.

Potential difference is measured in volts. Electric current (the flow of electrons) happens when, and only when a potential difference exists between 2 points in a circuit. The amount of electrons flowing in a circuit is measured in amps and is rather confusingly given the letter I in equations.

All of the above information allows us to define the relationship between current, voltage and resistance:

• Voltage: a potential difference of 1 volt is said to exist when a current of 1 ampere flows through a circuit with a resistance of 1Ω
• Current: is a flow of electrons, roughly 6,241,506,000,000,000,000 electrons flowing past a point in a circuit in 1 second = 1 amp

Ohm’s Law states:

• “The current flowing through a circuit is directly proportional to the potential difference (voltage) and indirectly proportion to the electrical resistance (Ohms)”

Ohm’s Law gives the following relationships:

• I = V/R
• V =IR
• R = V/I

Where I = electrical current, V = potential difference and R = electrical resistance.

We can use this very important law to calculate the resistance, flow of current, voltages and power dissipated for any circuit containing a power source and resistors. Lets start with the simplest case.

The Series Circuit

The simplest type of electrical circuit is made up of:

• A power source to supply the potential difference
• A selection of one or more components

The components are connected in a chain. This means there are 3 important things to note:

1. The current flowing in this circuit must be the same regardless of where it is measured. Engineers call values of this type a constant
2. The resistances of all the components must add together to form a total resistance
3. The total voltage supplied by the power source must be used up by all of the components in the circuit (including the power source itself). The amount of voltage used up (we say ‘dropped across’) a component is directly proportional to the resistance of the component.

Lets just briefly visit point 3 before moving on. Imagine you have 2 electrical resistances in series with each other. One of the resistances is twice as big as the other. We label them R and 2R.

Remember in a series circuit, the current is a constant i.e. the same current flows through both resistances. Remember Ohm’s law for voltage.

V = I x R

Let I = 1 Amp and R = 1 Ohm:

First voltage dropped (V1) using R

V1 = 1 x 1 = 1 volt dropped across R

Second voltage dropped (V2) using 2R

V2 = 1 x 2 = 2 volts dropped across 2R

This rule is absolute and should be committed to memory:

“The voltage dropped across a component is directly proportional to the resistance of the component”

Introducing the Resistor

In session 2 you will make and test some simple circuits using just one component, the resistor. Resistors are one of the most common components to be found in any circuit. The main functions of a resistor is to limit or restrict the current flowing through a circuit to a safe value or to set the voltage to a desired level at a particular point in a circuit (a voltage divider)

Below is an illustration of a resistor and an example of how to read the resistor colour code. You should practice using this code until you are proficient in using it.

Series Circuit Example

In the following example we are going to calculate

• The total resistance (Ω)
• The current flowing (I)
• The voltage dropped across each resistor
• The total power dissipated by the circuit (Watts)

Total Resistance

In series you simply add all the resistors together

R1 + R2 + R3 = 1,110Ω

Total Current

From Ohm’s law, current (I) = V/R

12/1,110 = 0.0108 Amps

All 3 Volt Drops

From Ohm’s law, voltage (V) = I x R (remember I is a constant in this circuit)

V1 = 0.0108 x 1000 = 10.8 Volts

Its the same procedure for the other volt drops but notice this.

R2 is 10x smaller than R1, so from previous learning and not using the calculator at all. The volt drop across R2 must be 10x smaller. Therefore:

V2 = 1.08 Volts

Similarly:

V3 = 0.108 Volts

Kirchoff’s Voltage Law (KVL) states:

“The sum of external volt drops in the external circuit (the voltages across all the resistors) must equal the applied voltage (12 volts)”

When you add all 3 volt drops together you will see that they come very close to 12 volts. This is your check, if your total volt drops do not come to within 1% of the applied voltage then you have made an error and will need to check your work.

Power Dissipated

There are 3 calculations for power they are:

• Power = V x I
• Power = I^2 x R (current squared)
• Power = V^2/R (voltage squared)

Using Power = V x I

Power = 12 x 0.0108 = 0.19 Watts

This video goes through the calculation process in detail

This video shows you how to simulate the circuit in Multisim

Parallel Circuit Example

I have included the simulated results in this screen shot so that you can check your working

We are going to find:

• The total resistance (a different process)
• The total circuit current
• The current flowing through each branch of the circuit
• The total power dissipated

For a circuit using exactly the same components as for the series example. You will find the results are very different.

Lets start with the total circuit resistance. I am going to show you the easiest way, its called product over sum. Basically you take the product of 2 resistors (multiply their values) and divide the product by the sum. It only works for 2 resistors at a time.

In our example lets take R2 and R3.

Product (100 x 10)/Sum (100 + 10)

= 9 Ω

Now we take the 9Ω result and use it to product/sum with the remaining resistor in the circuit. You can carry out this technique over and over again for as many resistors as you have in parallel

(9 x 1000)/(9 + 1000)  (Thanks Robert

Because R1 is so large there is very little change in the overall value of the total resistance

Therefore Total Resistance = 9 Ω

The total resistance must be lower than the smallest resistor you have in the circuit. If it is not then you need to check your working.

Now we can work out the total current

I = V/R

I = 12/9

Total Current = 1.33 Amps

This current splits up and flows through the resistors. The smallest resistor will conduct the most current. Lets check the current through R3:

IR3 = 12/10

IR3 = 1.2 Amps

Now it follows that R2 is 10x larger than R3 so 10x less current must flow through it.

IR2 = 1.2/10

IR2 = 0.12 Amps

And of course R1 is 10x larger again so the current through it reduces by a factor of 10

IR1 = 0.012 Amps

Total Power

We have used the power calculations before. For this circuit P = V x I is the easiest to use

Power = V x I

Power = 12 x 1.33

Power = 16 Watts

In conclusion for a parallel circuit:

• The voltage across each branch is the same
• The total resistance is always less than the value of the lowest resistor
• Much more current flows for the same components compared to a series circuit
• Much more power is dissipated because of the extra current flow

This video explains the process in detail

We can see from the illustration that the current, rather like a river, flows from the source. Then splits up and flows through each branch before joining back and returning to the source. We note the following:

• The voltage drop across each branch equals the supply voltage
• The current flowing in each branch depends on the resistance of the branch
• The total resistance of this circuit is always lower than the value of its lowest resistor. In this case the total resistance must be lower than 10Ω

Series and Parallel Circuits

Now we move on to the assessed task. This is the circuit you are going use for your calculation.

Here is the process I want you to follow:

• Choose 5 different value resistors from the table
• Calculate the total resistance of the circuit
• Assuming the circuit has a maximum power dissipation of 1.25 watts choose suitable voltage from the table. Use V = √1.25 x R

• Calculate the total current flowing
• Calculate the voltage drop across R2
• Calculate the current flowing through R3
• Calculate the total power dissipated
• Confirm your results in Multisim

Fill in the relevant sections of this table in your workbook

Save all of the data for this assessed task including all of your calculation, screen dumps showing the simulated values and most important. Make sure you simulation and calculations agree. Within 1% of each other. You will complete your report during session 2 once the practical testing is done.

Here is a video showing the calculation process

Here is a video showing the simulation process

That concludes all of the work required for this session. We sure did cover a lot of ground didn’t we ? 😉

## One thought on “Session 1”

1. I thought I had covered it but it appears that I didn’t. To find the power dissipated calculate the voltage across the bank that the resistor is in, square that voltage and the divide it by the resistance of the resistor. This method will give you the power dissipated, I will add a quick vid showing this later on 😉