Topic 2.1 Current, Voltage, Resistance and Power

Topic 2.1 – Current, voltage, resistance and power.

Learning Objectives:

At the end of this topic you will be able to;

recall and use ;

apply Kirchoff’s laws;

calculate the combined resistance of resistors connected in series and or parallel;

select appropriate values from the E24 series.

perform calculations on voltage divider circuits;

define power as VI and apply the formula to calculate power dissipation.

In module ET1 we discovered much about how logic circuits and some basic voltage amplifiers were used. In this Module we are going to take a much closer look at the way in which a number of components work and how we can design circuits using a variety of components to meet a particular specification.

Fundamental Circuit Concepts.

We will begin by revising some basic circuit ideas that you would have covered during your GCSE Science course.

Types of Circuit.

There are two ways thatcomponents can be connected in circuits, these are referred to as series and parallel.

Components in Series

Components connected in series are only connected to each other at one point.

The circuit below shows 2 identical lamps connected in series

•The same current flows through both components

•The sum of the voltages across each component in the circuit equals the power supply voltage.

Components in Parallel

The components sit side by side and connect to more than one other component.

The circuit below shows 2 identical lamps connected in parallel

•The voltage is the same across all parallel components.

•The sum of the currents in each branch of the circuit equals the current leaving the power supply.

Electric Current.

An electric current is a flow of charge carried by tiny particles known as electrons.

Movement of charge is caused by the force from a voltage supply such as a battery or mains supply acting on electrons which are free to move. When electricity was first discovered scientists thought that current was a flow of positive charge from the positive terminal of the battery to the negative terminal as shown on the left hand diagram below.

The discovery of the electron proved that this original idea was incorrect, and that current is a flow of negatively charged particles from the negative to the positive terminal. This is shown on the diagram on the right.

A decision was made not to change all the books that had been written, since the effect of a negative charge moving in the opposite direction is the same as a positive charge moving the other way. For most people the exact nature of the flow of charge is not important, so the original decision remained and we now call this conventional current.

Electric current is measured in units called Ampères(A),or Amps for short. In our work in electronics we will usually be dealing with much smaller units of current such as mA and μA

The electric current is not used up by the components in a circuit but it transfers energy from the voltage source to the various components making up the circuit.

Electric current is measured using an ammeter.

The ammeter is placed inseries within the circuit, as shown in the circuit diagram on the right.

Voltage.

Voltage is supplied by a battery, cell or power supply and is measured in volts (V).

The voltage of a power supply is the driving force that pushes the current through the components of a circuit. The larger the voltage the bigger the push on the current.

The voltage across a circuit component, e.g. a bulb converting electrical energy into heat and light is sometimes called apotential difference (p.d.).Potential difference is also measured in volts (V).

Potential difference is measured using a voltmeter.

The voltmeter is placed in parallel to the component as shown in the diagram opposite.

From this point on we will simply refer to potential difference simply as voltage.

Resistance.

Resistance is the property of a material which restricts the flow of electricity. Energy is converted into other forms (e.g. light, heat), as the voltage across the component drives the current through it.

The unit of resistance is the Ohm (Ω).In electronics we usually deal with much larger units of resistance as kΩ (1000Ω or 103Ω), andMΩ (1000000Ω or 106Ω)

What causes resistance?

As an electric current flows, charged particles called electrons move through a conductor. These moving electrons collide with the atoms of the conductor, making it more difficult for the current to flow. This causes the resistance.

What factors affect resistance?

Longer wires increase resistance. The electrons collide with atoms more often in a long wire than in a short wire.
Thin wires increase resistance. A thin wire has fewer electrons to carry the current than a thick wire.
The material the wire is made from e.g. nichrome is used for heating elements, because it has a high resistance per metre, so produces heat when current flows through it. Aluminium is used for transmission lines because it has a very low resistance per metre and so less energy is wasted during the transfer of electricity from power stations to your home.

In 1826, based on his work on conduction in metal wires, Georg Ohm formulated a law relating to the current passing through a wire, and the voltageapplied.

He found that for a fixed temperature, the current flowing through a resistor is directly proportional to the voltage across the resistor.

V = I  R

This is known as Ohm’s Law. Ohm also introduced the SI unit of resistance, as the Ohm (Ω).

Using Ohm’s Law, the resistance of any component can be calculated by measuring the current in the circuit, and thevoltage across the component.

The triangle opposite is often used to work out the formula required to solve problems of this type. Simply cover the term you want to find and what is left is the formula to use.

Example:

1.A current of 0.1A flows through a 20Ω resistor. What is the voltage across the resistor?

Formula required :

Calculation :

2.A 10Ω resistor is connected acrossa 5 volt power supply.What is the resulting current?

Formula required :

Calculation :

3. A resistor is placed in a circuit with a voltage of 12 volts and a current flow of 100mA.What is the resistance?

Formula required :

Calculation :

Student Exercise 1 : Ohms Law Practice

1.A 12Vpower supply is connected to a 3kΩ resistor. What is the current flowing in the circuit?

………………………………………………………………………………………………………………………………

2.A 9V battery provides a current of 1.5A to a resistor, What is the value of the resistance?

………………………………………………………………………………………………………………………………

3.A lamp is marked with the following maximum ratings: 6V 0.3A. What is the resistance of the lamp if it is operated at these maximum values ?

………………………………………………………………………………………………………………………………

4.A current of 4mA is driven through an 8 kΩ resistor. What is the voltage of the power supply producing the current.

………………………………………………………………………………………………………………………………

5.A resistor of 15Ω is connected to a battery of unknown voltage. The current in the circuit is measured to be 0.1A. What is the voltage of the battery ?

………………………………………………………………………………………………………………………………

6.A 4.5V battery is connected to a 36Ω resistor. How much current flows in the circuit ?

………………………………………………………………………………………………………………………………

7.A current of 3A leaves a 24Vpower supply. What is the value of the resistor connected to the power supply?

………………………………………………………………………………………………………………………………

8.A current of 0.2A flows through a 45Ω resistor. What is the voltage of the battery producing the current?

………………………………………………………………………………………………………………………………

9.A resistor of 24Ω is connected to a power supply. The current in the circuit is measured to be 0.05A. What voltage does the power supply provide ?

………………………………………………………………………………………………………………………………

10.A 1.5V battery is connected to a 75Ω resistor. How much current flows in the circuit ?

………………………………………………………………………………………………………………………………

Note : More information about the different types of resistors available, their properties and how to determine their value will be found in the Supplementary Pack of Notes which are non-examinablecalled “Practical Circuit Assistance.”

The E24 Series of resistor values

If you look in a component suppliers’ catalogues for resistorse.g. Rapid Electronics, you will find that resistors are available in the range from 1Ω to 10MΩ. Clearly there are 10 million resistor values in between these two values, and it would be completely impractical to expect any manufacturer to produce every single value and make these available for sale.

The manufacturers actually produce a much smaller number of resistors but cover the same range. This subset of resistors are perfectly adequate for 99.9% of all electronic circuits. When a situation requires an exact resistance the most common solution is to use a variable resistor and adjust it to the exact value required.

The resistor values available are as follows:

10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91

and any multiple of 10 these values, e.g. 130Ω, 3900Ω, 180kΩ, 1.2kΩ, 8.2MΩ etc. If you count up the number of values you will find that there are 24 of them, so this is where the name E24 Series comes from.

Kirchoff’s Laws.

Kirchhoff's circuit laws are two equalities that deal with the conservation of charge and energy in electrical circuits, and were first described in 1845 by Gustav Kirchhoff. We will consider some simple interpretations of these here.

Kirchoffs First Law - This can be stated simply as ‘the sum of currents entering a junction, is equal to the sum of currents leaving a junction’ as can be shown by the diagram opposite where I1 = I2 + I3.

Now try these :

Kirchoff’s Second Law – This can be stated simply as ‘the sum of all voltages in a series circuit is equal to the voltage of the supply’ as shown in the circuit opposite, where V1 = V2 + V3

For each of the following circuits complete the readings on the voltmeters.

(a)

(b)

(c)

We will now look at the effect of connecting resistors in series and parallel.

Resistors in series

The diagram shows two resistors of resistance R1and R2connected in series, and a single resistor of resistance Requivalent to them. The current Iin the resistors, and in their equivalent single resistor, is the same. The total voltageVacross the two resistors must be the same as that across the single resistor. If V1and V2are the voltages across each resistor,

But since voltage is given by multiplying the current by the resistance,

Dividing by the current I (which is the same in both cases)

This equation can be extended so that the equivalent resistance R of several resistors connected in series is given by the expression…

Therefore,the combined resistance of resistors in series is just the sum of all the individual resistances.

Resistors in parallel

Now consider two resistors of resistance R1and R2 connected in parallel, as shown.

The current through each will be different, but they will each have the same voltage across them. The equivalent single resistor of resistance Rwill have the same voltage across it, but the current will be the total current through the separate resistors.

We know that

and, using Ohm’s law, , so

Dividing by the voltageV (which is same in both cases),

This equation can be extended so that the equivalent resistance R of several resistors connected in parallel is given by

Therefore,the reciprocal of the combined resistance of resistors in parallel is the sum of the reciprocals of all the individual resistances.

Note that:

for two identical resistors in parallel, the combined resistance is equal to half of the value of each one,
forn identical resistors in parallel, the combined resistance is equal to the value of each onedivided by n
for resistors in parallel, the combined resistance is always less than the value of the smallest individual resistance.
for two resistors in parallel, there is an easier formula:

Or in words

Electronics engineers often use this as a quick way to calculate the equivalent of two resistors in parallel, and you don’t make errors with all those reciprocals.

Be careful though, it only works for 2. If you have more, do the first 2 then put that with the next one etc.

Student Exercise 2:

Calculate the equivalent resistance of the arrangement of resistors shown below.

……………………………………………………………………………………………………………………………………………

2. Calculate the equivalent resistance of the arrangement of resistors. (Hint: First find the resistance of the parallel combination.)

……………………………………………………………………………………………………………………………………………

3.Calculate the effective resistance between the points A and B in the network below.

……………………………………………………………………………………………………………………………………………

4.You have 3 resistors of 60Ω, 60Ωand 30Ω, Show how some or all of these resistors can be connected to make the following resistor values.

(i)150Ω(ii)20Ω(iii)90 Ω(iv)15 Ω

Voltage dividers

Two resistors connected in series with a battery or power pack each have a voltage across them. They may be used to divide the voltage of the supply. This is illustrated below.

The current in each resistor is the same, because they are in series. Thus

Dividing the first equation by the second gives

The ratio of the voltages across the two resistors is the same as the ratio of their resistances.

If the voltage across the combination were 12V and R1were equal to R2, then each resistor would have 6V across it.

If R1were twice the magnitude of R2, then V1 would be 8V and V2would be 4V.

Example:

If the voltage across the combination were 10V and R1werefour as large asR2, What would the voltage be across each resistor?

Voltage across R1 = ………………………………

Voltage across R2 = ………………………………

Also,

and

Dividing these gives,

So,

This is called the voltage divider formula (see next section)

A similar argument can be applied for the voltage across R2. You should be able to show that V2 is given by the following formula:

Example: Calculate the voltage reading on the voltmeter in the following circuit.

......

Potentiometers

A potentiometer is a continuously-variable voltage divider.

In the diagram above a fixed voltage supply of 9V is being used to provide a current through a variable resistor,when used in this way the variable resistor is called a potentiometer, and this can be used to produce a continuously variable voltage.

All three connections to the variable resistor are used. The fixed ends are connected across the battery so that there is the full battery voltage across the ends AB of the resistor. As with the voltage divider, the ratio of the voltages across AC and CB will be the same as the ratio of the resistances of AC and CB.

When the sliding contact C is at the end A, the output voltage VOUTwill be 9V.

When the sliding contact is at end B, then the output voltage will be zero.

So, as the sliding contact is moved from A to B, the output voltage varies continuously from the battery voltagedown to zero.

Student Exercise 3:

1.Find the total resistance in each of the following combinations of resistors.

2.Given the information on this circuit diagram

find:

(a)the voltage of the battery.

(b)the current in the 20 resistor.

(c)the current supplied by the battery.

3.Find the value of VOUT in the following circuits.

4.Plot a sketch graph showing how the output voltage varies as the slider is moved from the bottom to the top of the variable resistor.

5.Using only resistors from the E24 Series, show how you could make the following resistance values using onlytwo resistors in each case.

i)60Ω

ii)500Ω

iii)165Ω

iv)670Ω

v)125Ω

vi)420Ω

6.Determine the values of R1, V1, I1, I2 and R2 in the following circuit.

(a)R1………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(b)V1………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(c)I1………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(d)I2………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(e)R2………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(f) What is the power dissipated in R1 ?

………………………………………………………………………………………………………………

7.Determine the values of V1, I1, R1, I2 and R2 in the following circuit.

(a)V1………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(b)I1………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(c)R1………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(d)I2………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(e)R2………………………………………………………………………………………………………………

………………………………………………………………………………………………………………

(f) What is the effective resistance of R1 and R2 in parallel?

………………………………………………………………………………………………………………

Voltage at a point

It is sometimes more convenient to measure or calculate the voltage at a point in a circuit rather the voltage across acomponent.

Voltages at a point are measured with reference to zero volts or ground.

A voltage at a point can be negative as well as positive

Example

Calculate the voltage at points X,Y, Z, P and Q, in the following circuits, with respect to the zero volt line.

In the left hand circuitIn the right hand circuit

Voltage at X = +7VVoltage at P = +5.4V

Voltage at Y = +6VVoltage at Q = -6.6V

Voltage at Z = +4V

Electrical Power

When electricity flows through a resistor or other components, some energy is dissipated in the resistor or component as heat. The power dissipated in a component can be calculated by using the following formula:

This equation can be combined with Ohm’s law to provide two alternative forms of the equation, should R be known but V or I is not.

Substituting for V we get

Alternatively, substituting for I we get

Example:

Calculate the power dissipated in a 10kΩ resistor, when the voltage across the resistor is 10V.

Student Exercise 4:

1.Calculate the power dissipated in a light bulb, carrying a current of 60mA, when connected to a 6V battery.

…………………………………………………………………………………………………………………………………

2.A 220Ω resistor has a power rating of 0.4W. What is the maximum current that can be allowed through the resistor if the power rating is not to be exceeded?

…………………………………………………………………………………………………………………………………

3.The current flowing through a resistor is 0.125A, and the voltage across the resistor is 12V. Calculate

(i) the value of the resistance,

…………………………………………………………………………………………………………………………………

(ii) the power dissipated in the resistor.

…………………………………………………………………………………………………………………………………

Solutions to Student Exercises.

Exercise 1 : Ohms Law Practice

1.2.

3.4.

5.6.

7.8.

9.10.

Kirchoffs Law Exercises.

(a)

(b)

(c)

Student Exercise 3 :

1.(i)For each branch in the circuit there are two series resistors which can be added together to give a total resistance in each branch of 6Ω.

(ii)Effectively therefore we have two 6Ω resistors in parallel, therefore,

the equivalent resistance of the arrangement of resistors is therefore 3Ω.

2. For the parallel resistors.

Therefore for the whole network –

3.The circuit is effectively a 20Ω resistor and 10Ω resistor in series, connected in parallel with a 100Ωresistor and a 50Ωresistor also connected in series.

So dealing with the series parts first we would have 20 + 10 =30Ωin one branch, and 100 + 50 = 150Ω in the other branch, i.e. a parallel combination of 30Ωand 150Ω. So the effective resistance between the points A and Bwould be –

4.i)60Ω + 60Ω + 30Ω in series = 150Ω

ii)60Ω + 30Ω in parallel = 20Ω

iii)60Ω + 30Ω in series = 90Ω

iv)60Ω + 60Ω + 30Ω all in parallel with each other = 15Ω

Student Exercise 3:

1.a)

f)For parallel circuit

For the whole circuit

2.(a)the voltage of the battery will be the voltage across the series combination of the 4Ω and 3Ω resistor.

(b)the current in the 20 resistor, will be the same as the current in the 8 resistor, and as these are in parallel with the 3 and 4 resistor they will have the same voltage across them, therefore.

(c)the current supplied by the battery will be the sum of the two currents in each branch.