eng1021 electronic principles

8/13/2019 ENG1021 Electronic Principles

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ENG1021 Electronic Principles

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Learning Package 1

What is electricity?


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1 Introduction

Welcome to the Electronic Principles module of the course. The course is split upinto Learning Packages.

At the start of each Learning Package you will be presented with a Do you knowall this already? section which directs you to some questions on the subject thatappear within the Learning Package. If you are 100 % confident with the subjectthen you may continue with the next Learning Package, but you should be surethat you are able to solve all of the problems that I have selected before movingon.

Within each Learning Package you will be given the basic concepts which will bereinforced by problem solving. I will give detailed explanations to the solutions toall of the problems in which I will highlight common pitfalls and suggest

alternative solutions and observations. You should always read the solutions andsuggestions. However, it is important that you attempt the problems before youstudy the solutions. This is the only way that you will build up confidence intackling similar problems in electronics.

Often at the end of each Learning Package you will also be given informationabout where to find opportunities for further reading.

This introduction package differs from the others in that there are no setproblems. You will familiarise yourself with the concepts of charge, current,voltage and resistance. Do not struggle to understand (or remember) everydetail, most of the content will be expanded upon in later chapters, and you canalways go back and read them later (when they may make more sense).

2What is electricity?

All substance is made of atoms, and all atoms are made of smaller particles mainly protons, electrons and neutrons. Each proton has a positive charge, andeach electron has the equivalent but opposite negative charge (neutrons do nothave a charge). In an atom, the number of protons equals the number ofelectrons, so that the atom is electronically neutral.

In many substances, the atoms can be persuaded to part with one or moreelectrons temporarily by the application of energy. The free electrons can thenflow through the material. The flow of electrons is what is called current, and thisis the basis for all of electrical engineering and electronics. The amount of currentis measured in amperes or amps for short.

When an electrical current flows through a substance such as a metal wire, itproduces two effects heat and magnetism. It is these effects that are used to


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create all of the electrical and electronic devices that are used today. Forexample, when heat is produced this can be exploited to make electric fires,electric cookers, and if the heat is so intense that the wire glows, it can produceelectric light. The magnetism that is produced can be exploited to make electro-magnets or electric motors.

The energy that has to be provided to move the electrons is usually described asthe voltage and is measured in volts. Batteries have a voltage, and when aconductor is placed between the terminals, the electrons flow through theconductor. Similarly, the electric socket that you plug your electrical devices in tohas a voltage which in the UK is quoted in volts. There is a close relationshipbetween the voltage and the subsequent amount of current that flows, and thiswill be discussed in more detail in the next Learning Package.

3 The circuit

When describing electricity, the term circuit is often used. You may have comeacross the term open-circuit or short-circuit before. A simple circuit is shownbelow. It consists of a battery, which has a positive (+) and a negative (-)terminal. A wire is connected from the positive terminal to a component, which inthis case is a light bulb. The other terminal, the negative, is also connected to theother side of the bulb. So now we have a complete circuit there are no breaksanywhere.

Notice that Ive used standard symbols to represent a battery and a bulb, and Ivefollowed the standard convention when showing wiring of using straight lines withright angle bends. This kind of diagram is often referred to as a circuit diagramor a schematic diagram.

So, within the battery is an electronic charge. This would have been createdduring manufacture, and the way I think of it is that the battery has twocompartments. In the one, which would be labelled negative, is a compartmentthat has millions of free electrons. In the other, labelled positive, there is a lack ofelectrons. When the negative terminal is connected to the positive terminal,electrons flow out of the negative compartment, through the wire, and into thepositive compartment. It will carry on doing this at a fairly constant rate until thenumber of electrons in each compartment is equal, when the battery would bedescribed as flat.

The important point here is that the two terminals are joined together. If therewas a break in the wire anywhere, then the electrons would not be able to flowand there would be no current. That would be an open-circuit. So for current toflow there has to be a closed-circuit, which means no breaks. You can placedevices in that circuit, such as light bulbs, as long as the current can flow throughthem. If you just have a wire between the two terminals and nothing else, thatwould be described as a short-circuit and is usually something that you areadvised not to do because it can create very high currents.


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Potential Difference (voltage) between two points in an electric field is the workdone per unit charge moved from point to point in the field.

To assist understanding, an analogy with water is often made in texts. You maylike to consider the charge mentioned above as water in a tank which is sitting ona table. Energy (effort) was used lifting the water tank onto the table (or pumpingwater from a lower level into the tank), and that energy would be released if thetank were to be emptied. If the table had been higher, more energy would havebeen required to lift the same quantity of water into it. If we were to connect apipe to the base of the tank and bring the pipe to the ground then we would havewater pressure at the end of the pipe due to the height of the water. The higherthe tank, the higher that pressure will be. The height of the tank (or pressure ofthe water) in this case is analogous to the voltage.

In electricity, charge may move from a more negative (or lower) point (orpotential) to a more positive (or higher) point (or potential), in our water analogywe have moved water from a lower point to a higher point (or potential).

We can continue to use this analogy to give a feel for electrical current. Electricalcurrent is the rate of flow of charge. Charge will only flow between two points ifthere is a voltage or potential difference between the points. If we connect aflexible pipe to the bottom of our water tank, and move the end of the hosevertically to positions between the ground and the level of the water in the tank,water will flow out of it, providing that there is a difference in height between theend of the hose and the water level in the tank. The flow of water is analogouswith the flow of charge (or current), and there is no flow unless there is adifference in height for the water or a voltage exists (potential) for the charge.

This brings us to an important property of voltage. Like height, it is a relativemeasurement, and only meaningful if measured across or between two points.Often one of these points is taken as being earth or ground which is alwaysregarded as having zero volts.

If the pipe is made narrower, then the rate at which water flows out of the pipewill be less. A narrower pipe has more resistance and impedes flow rate of waterfor a given tank height or pressure. Electrical resistance impedes flow rate ofelectrons (current) for a given voltage.

So there is a potential for the energy to be released, but while the water sits in

the tank on the table it is insulated from the floor. Current is the movement orflow of electrical charge. If charge remains static no current flows. Charge willremain static unless there is a difference in potential or voltage.

4 Measuring voltage and current

In any electrical circuit the voltage or the current can be measured. Voltage ismeasured with a voltmeter and, as mentioned in the previous section, has to be


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measured between two points. Current is measured with an ammeter where theammeter has to be placed in an electrical circuit so that the current can flowthrough it.

5 Where next?

The next Learning Package is called Ohms law which explores to relationshipsbetween voltage, current, resistance and power in more detail. However, beforeleaving the topic of electricity, you may like to have a look at the on-line book thathas been supplied with this module. It is entitled Fundamentals of ElectricalEngineering and Electronics by Tony L Kuphaldt with help from other people.Have a look at the first section entitled DC, within that all of the material thatcomes under the heading of Basic Concepts of Electricity, such as:

Static electricity

Conductors, insulators and electron flow Electric circuits

Voltage and current

Resistance

Voltage and current in a practical circuit

Conventional versus electron flow

These will help you to understand the concepts discussed in this LearningPackage.


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Learning Package 2

Ohms law and power


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Ohms law and power

1 Do you know all this already?

If you are in doubt about Ohms law and power, please attempt the selfassessment questions that appear later in this Learning Package. If you cananswer all of the questions correctly you may omit this Learning Package. If not,please read on.

2 Introduction

This section links the electrical quantities current, voltage, resistance and power.It also introduces multiple units, which are prefixes attached to units that areintended to allow us to write very large or very small numbers more concisely.

3 Ohms law

In the previous Learning Package you were introduced to the concepts of voltageand current and you were told that there was a relationship between thesequantities. This relationship is known as Ohms law, and states that the voltagedivided by the current is a constant (if the temperature remains constant). Theconstant is given the name resistance, and the symbol R. Usually Ohms law isexpressed as:

V = I R (1)

where V is the voltage in volts, I is the current in amperes and R is the resistancein ohms.

Example:

A battery has a voltage of 9 V. It is connected to a device which has a resistanceof 100 . What is the current in the circuit?

First, note the units. The voltage is 9 volts, which is written as 9 V. Theresistance is 100 ohms, which is written as 100 , where is the upper caseGreek character omega.

Rearranging Ohms law:

I = V/R = 9/100 = 0.09 A

The solution is 0.09 amperes, or 0.09 amps, which is written as 0.09 A.


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4 Power

You were also told in the previous Learning Package that when an electriccurrent flows, it generates heat. Heat is a form of energy, and the rate at whichheat is generated is the power, measured in watts. You may have come across

this when buying devices that use electrical heating devices such as fire or kettlewhere the manufacturer would be described the device as having 2 kilowattpower, for example.

The equation for electrical power is:

P = I V (2)

where P is the power in watts, I is the current in amperes and V is the voltage involts.

Example:

What is the power in the circuit described earlier, where the battery is 9 V and thedevice has a resistance of 100 ?

We calculated the current as 0.09 A, so:

P = I V = 0.09 9 = 0.81 W.

The power generated is 0.81 watts, which is written as 0.81 W.

4.1 Other forms of the power equation

We can also calculate the power in a circuit using just the voltage and resistance,or just the current and resistance by combining the power equation with Ohmslaw.

P = I V

V = I R

So, substituting for V is the power equation, we get:

P = I (I R) = I2 R (3)

Alternatively, substitute for I in the power equation:

P = (V/R) V = V2/R (4)

Example:

Using the same example again, we have V = 9 v, I = 0.09 A, and R = 100 .


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P = I2 R = 0.09 x 0.09 x 100 = 0.81 W

P = (V/R) V = V2/R = 9 x 9 / 100 = 0.81 W

5 Resistors

Resistors are the most common components used in electrical and electroniccircuits. Essentially they are components with a fixed resistance value, althoughusually that value is only approximate. For example, a resistor that states that ithas a value of 100 may actually have a value of 99 or 101 . The range ofvalues is usually referred to as the tolerance, and can be 20%, 10% 15% or2%. If a resistor is stated as having a value of 100 and a tolerance of 20%,then its value can be as little as 100 (20% of 100) = 100 (0.2 x 100) = 100 -20= 80 , or as high as 100 + (20% of 100) = 100 + (0.2 x 100) = 100 + 20 = 120 .

Resistors have their values written on them as four coloured bands. The colourscorrespond to the values shown in the following table.

Colour 1stdigit 2nddigit Multiplier Tolerance

Silver 10-2

=0.01 10%

Gold 10-1=0.1 5%

Black 0 0 100=1 20%

Brown 1 1 101

=10 1%

Red 2 2 102=100 2%

Orange 3 3 103=1000

Yellow 4 4 104=10000

Green 5 5 105=100000

Blue 6 6 106=1000000

Violet 7 7 107=10000000

Grey 8 8 108=100000000

White 9 9 109=1000000000

The first two bands form a number which is then multiplied by the value of thethird band as a power of 10. The fourth band gives the value of the tolerance.


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Note that if the fourth band is not present, then the tolerance is taken to be20%.

For example, a resistor with coloured bands of Brown, Green, Red, Silver wouldhave a resistance of 15 x 102or 1500 and a tolerance of 10%. 10 % of 1500 is150, so the actual value of the resistor is in the range 1350 to 1650 .

Most components, like resistors, are available at various values. Instead of aninfinite choice, the values that are available are limited. Not only that, the rangeof value available depends on the tolerance. There is a wider range of values forlower tolerance resistors.

The following Table shows the range of values available for resistors of differenttolerance. The values that are used are known as the preferred values.

Tolerance Preferred Values

5% 10 11 12 13 15 16 18 20 22 24 27 30

10% 10 12 15 18 22 27

20% 10 15 22

Tolerance Preferred Values

5% 33 36 39 43 47 51 56 62 68 75 82 91

10% 33 39 47 56 68 82

20% 33 47 68

These preferred values are then multiplied by factors of 10. So you could have15 , 150 , 1500 , 15000 etc. So, for example, it is possible to have aresistor with a value of 6800 with a tolerance of 5, 10 or 20%, but you can only

have a resistor with a value of 6200 with a tolerance of 5%.

Another point to note is the standard symbol for the resistor. British andEuropean circuit diagrams use the following symbol for resistors (Figure 1).


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Figure 1 The British Standard resistor symbol

However, it is common to find the American standard symbol (which used to bethe British standard also). This is shown in Figure 2. Throughout this module thetwo symbols for resistors will be used.

Figure 2 The American Standard resistor symbol

In some circuits you may want to use a variable resistor one where the valuecan be changed. A device which is simply a variable resistor is called a rheostat.It has two connections, and then either a slider or knob that you turn to vary theresistance. A similar device is a potentiometer. This also has a slider or a knobthat turns, but it has three connections, as shown in Figure 3.

Figure 3 Rheostat and potentiometer


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6 Units

You need to understand the conventional way that multiple units are used. Theywill be used extensively throughout this course without further explanation, soyou need to make sure that you understand them now.

In a previous table where the colour code of resistors was shown, a multiplierwas referred to which used powers of 10. So, for example, 10 was seen to be 10to the power of 1, written as 101. Similarly 100 was described as 102or 10 to thepower of 2, and a thousand, 1000, was described as 103, and so on. These areways of writing powers of 10 so that you dont have to write a string of zeroesafter every value. You may note that the power is the same as the number ofzeroes. Alternatively, the power is the same as the number of times you multiply10 by itself. So 100 is 10 x 10 and 1000 is 10 x 10 x 10. The following tableshows the powers of 10.

Value power 10 multiplied by itself metric value symbol

1 100

10 101 10

100 102 10x10

1000 103 10x10x10 kilo k

10000 104

10x10x10x10

100000 105 10x10x10x10x10

1000000 106 10x10x10x10x10x10 mega M

1000,000,000 109 giga G

1000,000,000,000 1012 tera T

You can see that 100= 1. This is true for any number, X, so that X

0= 1. Ive also

introduced the metric terms, kilo for a thousand, mega for a million, giga for athousand million and tera for a million million. The terms may be familiar fromkilogram, a thousand grams, megawatt, a million watts.

So, for example, a resistor with a value of 120,000 could also be described ashaving a value of 120 kilo ohms or 120 k. Similarly, a resistor with a value of1,200 could be said to have a value of 1.2 k.


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What about values that are less then 1? We can also use powers of 10, but nowthe powers are negative. For example, 0.1 is equivalent to 1/10

thwhich in powers

of 10 is written as 10-1. Similarly, 0.01 is equivalent to 1/100thor 1/10thx 1/10thand is written as 10-2. The following table completes the story.

Value power 10 multiplied by itself metric value symbol

1 100

0.1 10-1 1/10

0.01 10-2

1/10 x 1/10 centi c

0.001 10-3 1/10 x 1/10 x 1/10 milli m

0.0001 10-4

1/10 x 1/10 x 1/10 x 1/10

0.00001 10-5

0.000001 10-6 micro

0.000000001 10-9

nano n

0.000000000001 10-12

pico p

Some of these terms may be familiar from centimetre, a hundredth of a meter, ormillisecond, a thousandth of a second.

Note that the choice of multiple unit prefix is up to you and an answer is notincorrect because you have not chosen the correct prefix. For example 470000 may be written as 470 k or 0.47 M. It may be difficult to decide which of thelatter two forms to use and either would be suitable.

In general, the values that you calculate or measure will be over a wide range, sothey might be thousands of millions or millionths. It is difficult to read numbers ifthey are very long, especially if they have either lots of training or leading zeroes.Therefore a standard called Scientific Notation is often used. In ScientificNotation, all numbers are written as one digit, decimal point, the remaining digits

then multiplied by a power of ten. For example, 1.234567 x 106would be how thenumber 1,234,567 would be written in Scientific Notation. Similarly, 1.234567 x10

-6would the way that 0.000001234567 would be written.

7 Additional reading

As in the previous Learning Package, you are encouraged to read some of thesections from the on-line book Fundamentals of Electrical Engineering and


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Electronics. In this Learning Package the relevant section is headed Ohmslaw, and the parts that you should read are:

How voltage, current and resistance relate

An analogy for Ohms law

Power in electric circuits

Calculating electric power

Resistors

The remaining parts of that section can be omitted as they are not relevant,although we may return to some of them later.

In addition, you may want to read the section entitled Scientific Notation and

Metric Prefixes. The relevant parts are:

Scientific notation

Arithmetic with scientific notation

Metric notation

Metric prefix conversion

Again, the remainder of the parts can be ignored for now.

Finally, tucked away in another section is some information about resistor colourcodes in the section entitled Reference, and in a section called Experimentsand sub-section called DC Circuits there are a couple of experiments calledPotentiometer as a voltage divider and Potentiometer as a rheostat which youmay find useful.

8 Self assessment exercises

Now attempt the following Problems. These problems will assess yourunderstanding of Ohms law, power, the electrical quantities and you should beable to exercise your use in multiple units. The solutions are given below with fullexplanations.

Problem 1A 15 V dc source is connected across a 1 k resistance. (a) Draw aschematic diagram. (b) Calculate the current I through the resistance. (c) Howmuch current flows through the voltage source? (d) If the voltage is doubled, howmuch is the current I in the circuit?


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Problem 2A current of 2 A flows through a 12 resistance connected across abattery. (a) How much is the battery voltage? (b) How much power is dissipatedby the resistance? (c) How much power is supplied by the battery?

Problem 3 Calculate the resistance, R, in ohms, for each of the followingexamples: (a) 1 mA drawn from a 12 V source; (b) 4 mA drawn from a 15 Vsource; (c) 150 mW dissipated with 36 V applied; (d) 16.2 W dissipated with acurrent of 30 mA.

Problem 4How much will it cost to operate a 1500 W heater for 36 hours if thecost of electricity is 7 cents/kWh?


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Solutions

Problem 1a

See Figure 4.

Figure 4 The circuit diagram

Problem 1b

We use Ohms law to find the current I

V= IR

Firstly we have to rearrange Ohms law by dividing both sides of the equation byR:

VI

R= (5)

Substituting our values for V = 15 V dc and R = 1k = 1 10 3we get

3

15

1 10x= 15 103amperes = 15mA

Problem 1c

Since current in all parts of the circuit is the same then the same current mustalso flow through the voltage source, that is, the current through the voltagesource is also 15mA.

Problem 1d


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If the voltage of the source is doubled, then multiplying both sides of Equation 5by 2 gives:

22

VI amperes

R

=

and shows that the current will also be doubled. The current Iwill therefore be

2 15 mA = 30 mA

An alternative solution is to double the voltage and recalculate Iusing Equation5.

The old voltage was V = 15 volts so the doubled voltage is V = 30 volts.Substituting into Equation 5 as before gives

3

30

1 10 = 30 103

amperes = 30 mA

Problem 2a

This first part of this question requires Ohms law again. We use the form:

V = IR

We are told that 2 amps of current flows in a circuit with a resistance of 12.Substituting these values for I and R respectively into Ohms law gives:

V = 2 12 = 24 volts

Problem 2b

Now that we have found the voltage V we may calculate the power using

P = V I.

Substituting for I and V (I = 2 amps and V = 24 volts)

P = 24 2 = 48 watts

Problem 2c

The power dissipated in the resistor must have been supplied by the battery toheat up the resistor; therefore the power supplied by the battery is also 48 watts.

Problem 3

Parts a) and b) of this problem use a rearrangement of Ohms law. We require anequation for resistance R alone so we divide both sides of equation 1 by I to give:


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V IR

I I=

Cancelling the Is and swapping left and right hand sides of the equation gives:

VRI=

(6)

Problem 3a

The current I is 1 mA = 1103 amperes while the voltage V is 12 volts.Substituting these values into Equation 6 gives:

3

3

1212 10

1 10R = =

Using the k for kilo prefix the answer is:

Resistance R = 12 k.

Problem 3b

Similarly substituting into Equation 6 for I equal to 4 mA = 4 103amperes andthe voltage V equal to 15 volts gives:

3 3

3

15 1510 3.75 10

4 10 4R = = =

Using the k for kilo prefix the answer is:

Resistance R = 3.75 k .

Problem 3c

We can attempt the solution of this problem in two ways. I prefer the first waybecause we only have to remember the two equations for voltage and power(Equations 1 and 2).

First way (preferred)

We have values for the power P and the voltage V and require the resistance R.Ohms law will give us a value for R, but we need I. I can be found from Equation2, by dividing both sides of the equation by V and rearranging.

PI

V=

Substituting for P and V into the equation gives:


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33150 10 4.1667 10

36I amperes

= =

Substituting I and V into Equation 6 gives:

3

336 8.64 10 8.64

4.1667 10R k= = =

Second way

Using the Equation 4 directly:

2VR

P=

and substituting for V = 36 volts and P = 150 mW = 150 10 3gives:

23

3

36 129610

150 10 150R = =

R = 8.64 103= 8.64 k

Problem 3d

Again we can attempt the solution of this problem in two ways and the first wayonly requires knowledge of Equations 1 and 2.

First way (preferred)

Now we have values for the power P and the current I and require the resistanceR. Again we can use Ohms law to find R, but we need a value V now. V can befound from Equation 2, by dividing both sides by I and rearranging.

PV

I=

Substituting for P and I into the equation gives:

3

3

16.2

0.54 1030 10V volts= =

Substituting for I and V into Ohms law gives:

36

3

0.54 100.018 10 18

30 10R k

= = =

Second way


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Using the Equation 3 directly:

2

PR

I=

and substituting for P = 16.2 watts and I = 30 mA gives:

6 6

3 2

16.2 16.210 0.018 10 18

(30 10 ) 900R k

= = = =

Problem 4a

This exercise relates power and energy. Electrical energy is the power multipliedby the time that the power is used for.

Energy = power time (7)

Hopefully this is instinctively true from this exercise. When we pay our electricitybills we pay for energy used. It would seem reasonable that the longer theelectric heater is on (time) then the more energy we would use and the more wewould pay.

Before we can calculate the payment we must first determine how much energyhas been used. There are two common units for measuring energy, joules andkilowatt hours. They are both in units of power multiplied by time. Joules arewatts multiplied by seconds and kilowatt hours (kWh) are kilowatts multiplied byhours. We could calculate the energy used in either joules or kilowatt hours, butsince this exercise deals with power in kilowatts (1500W =1.5 kW) and time in

hours and the payment details are in terms of kilowatt hours then we willcalculate the energy for this question in kilowatt hours. The heaters powerconsumption is 1500 W or 1.5 kW. It is used for 36 hours. Let us substitute thesevalues into the energy equation, Equation 7:

Energy = 1.5 36 = 54 kWh

Now that we know how much energy is used we calculate the payment bymultiplying the charge per kilowatt hour by amount of energy used (in kilowatthours) which is 7 cents/kWh. So the answer is:

Cost = 54 7 = 378 cents = 3.78 dollars

11 Further reading

At this point you may like to have a look at the relevant sections of the on-linebook Fundamentals of Electrical Engineering and Electronics.


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Under the section entitled DC, you may like to have a look at the part calledBasic Concepts of Electricity, which reinforces some of the theory given at thebeginning of this Learning Package. Following that you could have a look at allthe parts under the section heading Ohms Law.

12 Where next?You are encouraged to study the Learning Package entitled Series and Parallelcircuits next.


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Learning Package 3

Series and parallel circuits


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Series and parallel circuits


If in doubt, please attempt the self-assessment questions in this LearningPackage. If you can answer all of the questions correctly you may omit thissection. If not, please read on.

2 Introduction

This section introduces series and parallel circuits. The meaning of the termsseries and parallel are introduced. Then methods for calculating the totalresistance of series and parallel combinations of individual resistors areexplained. Polarity of voltages and currents are demonstrated. Methods ofadding voltages and currents are then shown. We will start with series circuitsand then highlight differences and similarities with parallel circuits.

3 Series circuits

3.1 Potential difference

In the circuits that weve seen so far there has only been one resistor. In acircuits with one resistor, the whole of the voltage of the battery as applied acrossthe resistor. In other words, if we put a voltmeter across the resistor which wasconnected to a 9V battery, it would give a reading of 9V. The voltage across the

resistor is called the voltage drop or the potential difference across the resistor(often abbreviated to pd). If the voltage drop across the resistor is V, the currentthrough the resistor is I and the resistance itself is R, then Ohms law applies.

V = I x R (1)

As mentioned before, a voltage has to be measured between two points. It istherefore convenient to think of one point as the reference point, and measure allvoltages relative to that reference point. In the case of a battery, the negativeterminal would most likely be chosen as the reference point. This means that thepositive terminal is 9 V higher than the negative terminal.

The voltage of the battery is often called the electromotive force and given thesymbol E, and is often shortened to the emf. Therefore the battery is supplyingan electromotive force which is pushing the electrons around the circuit. As theelectrons travel around the circuit they lose energy, so that the energy at oneside of the resistor will be higher than at the other, and this difference is whatweve called the potential difference.


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So, the energy at the positive terminal is E volts. When we get to the end of theresistor that is connected to the positive terminal the energy is still E (ignoring thevery small drop along the wire). Now pushing the electrons through the resistortakes more effort, and uses up the energy, so that by the time we look at theenergy at the other end of the resistor it will have dropped by an amount V volts,

and this end of the resistor is connected to the negative terminal of the battery.All of the energy that was available, namely E, has to be used up by the time youget around the circuit. So if the energy lost in the resistor is V, we have to be ableto say that V = E. So we can say, that:

E = V = I x R (2)

So, in the case of a circuit with a single resistor, the voltage difference acrossthat resistor equals the electromotive force of the battery. Now lets look atcircuits with more resistors.

3.2 Series resistors and current in series circuit

A circuit in which two or more resistors are connected together in series meansthat one end of resistor is connected to one end of another, as shown in Figure 1.

Figure 1 Two resistors in series

We have already said in an earlier Learning Package that the current flows allthrough the circuit, and in the circuit in Figure 1 this means that the same currentflows through both resistors. From the previous discussion we know that there isa potential difference across R1, which I shall call V1, and a potential differenceacross R2, which I shall call V2. In Figure 1 the polarity of the voltages has beenshown. We know that the battery has a positive and a negative terminal, and thatthe electron current flows from the negative terminal, around the circuit to the

positive terminal. In other words the current flows from the negative to thepositive. Similarly, the potential difference across a resistor is shown as having apositive side and a negative where the positive side will have a higher voltagethan the negative side, and the current will flow from the negative to the positiveagain.


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Also, we know that the electromotive force, E, of the battery must equal the totalpotential difference around the circuit (this will be discussed in more detail in alater Learning Package). So, we can say that:

E = V1 + V2 (3)

For example, if the two resistors are equal, the battery is 9 V, and we have avoltmeter with the negative lead attached to the negative terminal of the battery,then going around the circuit we would find that:

At the positive terminal of the battery we would measure 9 V;

At the left-hand (+) terminal of R1 we would measure 9 V;

At the right-hand (-) terminal of R1 we would measure 4.5 V;

At the left-hand (+) terminal of R2 we would measure 4.5 V;

At the right-hand (-) terminal of R2 we would measure 0 V;

In other words, with equal value resistors, the voltage difference across eachresistor would be equal to half the electromotive force of the battery.

Returning to a series circuit with two arbitrary resistors, we know that thepotential difference across each resistor when added together equal the emf ofthe battery. Using Ohms law we can replace the potential difference across eachresistor by the current times the resistor value.

V1 = I1R1 (4)

V2 = I2R2 (5)

E = V1 + V2 = I1R1 + I2R2

However, we already know that the current through each part of the circuit mustbe the same. So I1 = I2 which we will call I, and the equation becomes:

E = IR1 + IR2 = I(R1 + R2) = IReq

We can replace the two resistor values with one equivalent resistor, Req. Thevalue of this equivalent resistor is:

Req= R1 + R2 (6)

So the total resistance of the circuit is the sum of the individual resistors. This istrue for any number of series resistors. For any number of resistors this would bewritten as:

RT = R1 + R2 + R3 + (7)


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You should now appreciate what a series circuit is, and calculate the totalresistance (or equivalent resistance) of a series string of resistors. You have alsolearnt why the current in all parts of a series circuit is the same. You should nowbe able to attempt the following Problem. Do this now without reference to theworked solutions if possible.

Problem 1 (Resistor in series)

A circuit has 20 V applied across a 10 resistance R1. How much is the currentin the circuit? How much resistance R2must be added in series with R1to reducethe current by one half? Show the schematic diagram of the circuit with R1andR2.

Solution

The first part of this question is a simple Ohms law exercise. The voltage V is 20volts, the resistance R1 is 10 and we are asked to calculate the current I.

From the simple re-arrangement of Ohms law which we used in the last LearningPackage:

202

10

VI amperes

R= = =

We are asked to add some resistance R2 to reduce the current by one half. Onehalf of the above current is 1 ampere, so we must firstly calculate what total (orequivalent) resistance RT is required to cause a 1 ampere current to flow in acircuit with a 20 volt voltage source. Re-arranging Ohms law again:

2020

1T

VR

I= = =

We are not asked for the total resistance RT, we are asked for the additionalresistance R2. Therefore we must use the equation for resistors in series:

RT = R1 + R2 +

In this case we only have two resistors so the equation is simply:

RT = R1 + R2

Subtracting R1 from both sides of this equation gives:

RT R1 = R2

and then substituting in our values for RT , and R1 gives:

R2 = 20 10 = 10


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You may have arrived at your value for RT using by noting that from Ohms law,to half the current we must double the resistance (assuming that the voltagestays the same). So

RT = 2 R1 = 20

Either way is correct.

The schematic diagram for the circuit is shown in Figure 2.

Figure 2

3.3 Power in series circuit

The power dissipated in any resistor is given by the power equation (which wasintroduced in the previous Learning Package), but the total power in the series

circuit is the sum of the powers dissipated in the individual resistors. You shouldnow be able to attempt Problems 2 and 3. Do this now without reference to theworked solutions if possible.

Problem 2 (potential difference and power)

Draw the schematic diagram of 20, 30 and 40 resistances in series. (a) Howmuch is the total resistance of the entire series string? (b) How much currentflows in each resistance, with a voltage of 18 V applied across the series string?(c) Find the voltage drop across each resistance. (d) Find the power dissipated ineach resistance.

Solution

The schematic (or circuit) diagram is shown in Figure 3.

Problem 2a


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Figure 3

The total resistance is given by the formula for resistors in series, Equation 7above. In this problem we have three resistors which I will call R1, R2 and R3.

I have labelled the resistors so that I can identify the voltage (IR) drops later inthe solution. The total resistance is:

RT = R1 + R2 + R3 = 20 + 30 + 40 = 90

Problem 2b

We should know that the same current flows in all parts of a series circuit andtherefore the same current flows in each resistor in this circuit. We calculate theseries current using RT and Ohms law in the following form.

18

0.290T

V

I amperesR= = =

Problem 2c

This part of the problem requires us to calculate the IR drops across theresistors.

Let the voltage drop across R1 be V1, the voltage drop across R2 be V2 and thevoltage drop across R3 be V3. The IR (or voltage) drops are all given by Ohmslaw. Therefore using our value for I and the three resistances we have:

V1 = I R1 = 0.2 20 = 4 volts

V2 = I R2 = 0.2 30 = 6 volts

V3 = I R3 = 0.2 40 = 8 volts

We can check our answers because we know that the total of the IR dropsshould add up to the source voltage of 18 volts. Lets do that:


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V1 + V2 + V3 = 4 + 6 + 8 = 18 volts

which is the source (or applied) voltage.

Problem 2d

The final part of this question asks us to calculate the power dissipated in eachresistor. We can do this using:

P = IV

P = I2R

or

2VP

R=

I will obtain the solution using P = IV. Let the power dissipated in R1 be P1, thepower dissipated in R2 be P2 and the power dissipated in R3 be P3. Using ourvalues for I and the voltage drops across each resistor gives

P1 = I V1 = 0.2 4 = 0.8 watts

P2 = I V2 = 0.2 6 = 1.2 watts

P3 = I V3 = 0.2 8 = 1.6 watts

You may want to verify that the same solutions are found if the alternativeequations for power are used. Note that if we add up the individual powers,

PT = P1 + P2 + P3 = 0.8 + 1.2 + 1.6 = 3.6 watts

we get the same result as calculating the power dissipated in the circuit using theapplied voltage and the current,

P = I VT = 0.2 18 = 3.6 watts

Problem 3

Draw a schematic diagram showing two resistances R1and R2in series across a100 V source. (a) If the IR voltage drop across R1 is 60V, how much is the IRvoltage drop across R2? (b) Label the polarity of the voltage drops across R1andR2. (c) If the current is 1A through R1, how much is the current through R2? (d)How much is the resistance of R1 and R2? How much is the total resistanceacross the voltage source? (e) If the voltage source is disconnected, how muchis the voltage across R1and across R2?


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Solution

The circuit diagram is shown in Figure 4.

Figure 4

Problem 3a

We do not need I to determine the IR drop across R2, because we know that thesum of the IR drops is equal to the applied voltage. If V1 is the voltage dropacross R1 and V2 is the voltage drop across R2 the applied voltage

VT = V1 + V2. (30)

VT = 100 volts and V1 = 60 volts Therefore

V2 = VT V1 = 100 60 = 40 volts (31)

Problem 3b

When I (and you) produced the circuit diagram at the start of this question wedecided on the polarity of the voltage source. I decided that the top of the batteryis positive. This determines the polarity of the voltage drops. If we assumeelectron current flow (current flows from negative to positive) then the currentflows anti-clockwise around the circuit as shown by the arrow in Figure 5. Forcurrent to flow in the anti-clockwise direction, and remembering that the samecurrent flows in all parts of a series circuit, then the left of R1 must be positivewith respect to the right of R1 (which in turn will be negative). By a similarargument, the left of R2 will be positive with respect to the right of R2. These

polarities are marked on Figure 5.


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Figure 5

Problem 3c

They cannot catch us with this one, can they? The current flowing in all parts of aseries circuit is the same, and since R2 is in the same series circuit as R1 thecurrent must also be the same. That is the current through R2 is 1A.

Problem 3d

We know that the IR drop across R1 (V1) is 60 V and that the IR drop across R1(V1) is 40 V Using Ohms law again:

IR1 = 40 V, since I = 1 A, R1 = 40

and

IR2 = 60 V, since I = 1 A, R2 = 60.

The total resistance across the voltage source

RT = R1 + R2 = 40 + 60 = 100 .

Problem 3e

If the voltage source is disconnected then no current can flow, since there is nolonger a potential difference across the resistors. Then I = 0 and the IR drops (orvoltage across) IR1 and IR2 must both be zero.

3.4 Series and opposing voltages, tips for analysis

Youve seen circuits with more than one resistor in series, so what happens ifthere are more batteries in series? Put simply, if the batteries are facing thesame way, then the voltages are added up. By the same way I mean that thepositive terminal of the first battery is connected to the negative terminal of thesecond and so on. In other words they are all trying to push the current around


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the circuit in the same direction. If the batteries are connected such that someare trying to push the current in the opposite direction, then their voltages aresubtracted.

Now please attempt the following Problem 4. Try the problem before looking atthe solution.

Problem 4 (Series adding/opposing voltages)

Figure 6 shows the circuit for keeping a 12.6 V car battery charged from a 14.8 Vdc generator. Calculate the current I and show the direction of electron flow forcurrent between point A and B.

Figure 6

Solution

The problem introduces a battery charger as a practical example of opposingvoltages. We will work through the problem and then consider the effect ofconnecting the charger to the battery the wrong way round.

Consider the circuit associated with Problem 4 as shown in Figure 6. V1represents voltage of the battery, while V2 represents the voltage of the charger.These two voltage sources have their positive ends connected together; thevoltage therefore oppose. The overall voltage is therefore:

V2 V1 = 14.8 12.6 = 2.2 volts

We could redraw the circuit replacing both sources with a single 2.2 volt source.We have to consider what polarity this replacement 2.2 volt source will have. Thisis shown in Figure 7. The polarity will be the same as the battery charger voltage

V2 since V2 is the higher voltage. The current is then given by Ohms law:

2.23.367

0.6

VI amperes

R= = =

+

-

-

+V1=12.6 V V2=14.8

0.6A B


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Figure 7 The opposing voltages are equivalent to a 2.2 volt source

Now consider what happens if we connect the battery charger the wrong wayround. Figure 8 shows the new circuit. The circuit looks almost identical to Figure6, but notice that the polarity of the battery has changed. Voltage source V 1 nowhas its negative terminal connected to the positive terminal of voltage source V2.This means that their voltages will add and the overall voltage will be:

V2 + V1 = 14.8 + 12.6 = 27.4 volts

Figure 8 Now the battery is connected the wrong way round

We could redraw the circuit as we did above, but this time replacing both voltagesources with a single 27.4 volt source. This is shown in Figure 9. The polarities ofthe original voltage sources would have individually caused an electron current to

flow in the same direction, that is, anticlockwise around the circuit. The polarity ofthe replacement voltage source is therefore the same as both of the originalsources. The current is again given by Ohms law:

27.445.67

0.6

VI amperes

R= = =


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Figure 9 The series aiding voltages are equivalent to a 27.4 volt source

This is a very high current and if no other protection was provided would certainlymelt the wires of typical battery chargers.

The example serves to show a practical application of series aiding and seriesopposing voltages.

3.5 Open and short circuits in a series path

In many practical circuits it is possible to introduce a short-circuit or an open-circuit. A short circuit has zero resistance, and can cause high currents to appear

in a circuit. It may be caused by component failure or by contamination. Opencircuits, on the other hand, have infinite resistance and prevent current fromflowing. This is usually caused by a connection breaking between a componentand the board that it is on. Now attempt Problem 5.

Problem 5

Three resistors of 100, 200 and 300 are in series across a 24 V source. (a) Ifthe 200 resistor shorts, how much voltage is across the 300 resistor? (b) Ifthe 300 resistor opens, how much voltage is across the 100 and 200 resistors?

Solution

The problem concerns a series circuit containing 3 resistors 100, 200 and300. I have drawn the circuit in Figure 10.

Problem 5a


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The first part of the problem asks us to find the voltage across the 300 resistorif the 200 resistor goes short circuit. The situation is depicted in Figure 11. Theshort circuit is considered to have no (zero) resistance, and all of the currentflows in the short circuit, with no current flowing in the 200 resistor. Effectively,it is as if the 200 resistor has been removed from the circuit and replaced by a

piece of wire. The circuit therefore reduces to a 100 resistor and a 300resistor in series. The total series resistance is therefore:

RT = 100 + 0 (the short circuit) + 300 = 400

Figure 10 The initial series circuit for Problem 5.

Figure 11 The circuit for Problem 5 if the 200 resistor goes short circuit

The current in the circuit is given by Ohms law

2460

400T

VI mA

R= = =

The voltage across the 300 mA resistor is also given by Ohms law

V300 resistor = IR = 60 103300 = 18000 103= 18 V

Problem 5b


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This part of the Problem considers the situation if the 300 resistor breaks orgoes open circuit. The situation is depicted in Figure 12. The open circuit isassumed to have infinite resistance. It is as if the 300 resistor has beenremoved from the circuit and its connecting wires have been left unconnected.No current can flow in the circuit and since the same current (zero) flows in all

parts of a series circuit, no current flows in the 100 and the 200 resistors.Since no current flows in these resistors there is no voltage (IR) drop across theresistors. That is the voltage across the 100 and the 200 resistors is zero inboth cases.

Figure 12 The circuit for Problem 5 if the 300 resistor goes open circuit.

Another way of seeing that the current is zero is by considering the open circuitresistor to have a resistance of (infinity) and calculate the total resistance. Wehave

RT = 100 + 200 + =

The current is then the supply voltage divided by infinity which equals zero.

240

T

VI

R= = =

4 Parallel circuits

4.1 Introduction to parallel circuits and branch currents


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Figure 13

Figure 13 shows a simple parallel circuit where we have two resistors, this timejoined at both ends to each other. The branches of the circuit with the resistor inare parallel to one another, hence the name.

In this instance we can see that the positive terminal of the battery is connectedto the left-hand side of both resistors, and that the negative terminal of thebattery is connected to the right-hand side of both resistors. If we were to place avoltmeter across either of the resistors it would measure the same value as theelectromotive force of the battery, since they are all effectively the same twopoints. So, in this case the potential difference across each resistor must equalthe electromotive force:

E = V1 = V2

V1 = I1R1 and V2 = I2R2

So, E = I1R1 and E = I2R2

I1 = E/R1 and I2 = E/R2

This gives us the value of the current in each branch. Although we will discussthis more formally later in the module, we can intuitively see that if the currentflowing out of the negative terminal of the battery reaches the point where thetwo branches are connected, the current has to split so that some of it goesaround one branch and some around the other. If we think of the water analogy,then it is clear that the amount of current going into each branch must add up to

the total current coming out of the battery, since the current hasnt got anywhereelse to go, and it doesnt accumulate in puddles at junctions. So,

I = I1 + I2 = E/R1 + E/R2 = E(1/R1 + 1/R2)

If we introduce an equivalent single resistor again, then we could say that:

E = IReq

I = E/Req

Then:

E/Req= E(1/R1 + 1/R2)

1/Req= 1/R1 + 1/R2 (8)

In a circuit with two resistors in parallel, the equivalent resistor is found using thisequation. Namely, that the reciprocal of the equivalent resistance equals the sum


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of the reciprocals of the two resistances. This can be generalised for any parallelcircuit. For example, with many resistors in parallel, the equation would be:

1 2 3

1 1 1 1..........

eqR R R R= + + + (9)

In the special case of a parallel circuit with two resistors in parallel, the equationcan be rearranged.

1/Req= 1/R1 + 1/R2 = (R2 + R1)/R1R2

Req= R1R2/(R1 + R2) (10)

You now know what a parallel circuit is. You will have seen that there are ruleswhich apply to parallel circuits which differ slightly to series circuits. Generallythey differ by the transposition of the terms current and voltage. For examplein a series circuit, the current flowing in all parts of the circuit is the same, while ina parallel circuit the applied voltage is the same across all parallel branches.

Also, in a series circuit the applied voltage Eis the sum of the individual voltagedrops, while in a parallel circuit the main current I is the sum of the individualbranch currents. Bear these differences and similarities in mind when trying toremember the rules.

Now attempt Problem 6. Do this without reference to the worked solutions ifpossible.

Problem 6 (Resistors in parallel)

A 6 R1and a 12 R2are connected in parallel across a 12 V battery. (a) Drawthe schematic diagram. (b) How much is the voltage across R1and R2? (c) Howmuch is the current in R1 and R2? (d) How much is the main-line current? (e)Calculate REQ.

Solution

Problem 6a

The diagram is shown in Figure 14.


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Figure 14

Problem 6b

Since all parts of the circuit are in parallel and we know that the voltage is thesame across all parallel branches, then the voltage across R1 and R2 is the sameas the battery voltage which is 12 V.

Problem 6c

Let us call the current through 6 resistor R1, I1 and the current through 12 resistor R2, I2. The current though R1

11

12

26

TV

I amperesR= = =

and

2

2

121

12

TVI amperesR

= = =

Problem 6d

The total or mainline current IT is the sum of the two branch currents.

IT = I1 + I2 = 2 + 1 = 3 amperes

Problem 6e

We can calculate the equivalent resistance by considering what single resistancewould cause the same main-line current of 3A to flow. This is given by Ohms law:


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124

3

Teq

T

VR

I= = =

A handy check to see if you have calculated a parallel resistance properly is thatthe calculated resistance must always be less than the any of the values of the

resistors in the branches. In this case the calculated resistance is 4 and thetwo parallel resistors are 6 and 12 . If our calculated value had been greaterthan 6 we must have made a mistake in our calculations.

4.2 More parallel circuits and branch currents and power

You will remember the rather uncomfortable formulae for parallel resistors:

1 2 3

1 1 1 1..........

eqR R R R= + + +

1 2

1 2

eq

R RRR R

=+

if there are only two resistors

Dont forget that if the resistors all have the same value then the equivalentresistance Req of the parallel combination is simply the value of the resistancedivided by the number of resistors in the parallel combination. For example, ifthere are four resistors with the same value, R, in parallel, the equivalentresistance is:

1 2 3 4

1 1 1 1 1 1 1 1 1 4

4

eq

eq

R R R R R R R R R RR

R

= + + + = + + + =

=

Now attempt Problem 7 and Problem 8.

Problem 7 (Resistors in parallel equation and power)

For the circuit in Problem 6, how much is the total power supplied by the battery?

Solution

We are asked to calculate the total power supplied by the battery in the aboveproblem. The total power is the sum of the power dissipated in each branch ofthe parallel circuit. The current through the first branch I1 is 1 A. The supplyvoltage is 12 V and this is the same across all parallel branches. Therefore thepower in the first branch is,

P1 = I1 V = 1 12 = 12 W


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The current in the second branch I2 is 2 A. Again the same voltage 12 V existsacross this branch, therefore the current in the second branch

P2 = I2 V = 2 12 = 24 W

The total power

P = P1 + P2 = 12 + 24 = 36 W

We could have used the main-line current to give the total power supplied by thebattery. Since the battery supplies 3 A and its voltage is 12 V, then it mustsupply:

P = I V = 3 12 = 36 W

of power.

The same result can be arrived at using the equivalent resistance Req and either:2 2

12 14436

4 4eq

VP W

R= = = =

or

P = I2Req = I2 4 = 32 4 = 9 4 = 36 W

All of the above solutions are correct and there is no single correct solution. Dotry different solutions so that you become practised in using them.

Problem 8

Find the REQof the following groups of branch resistances: (a) 10 and 25 ; (b)five 10 k resistances; (c) two 500 resistances; (d) 100 , 200 and 300 ;(e) two 5 k and two 2 k resistances; (f) four 40 k and two 20 k resistances.

Solution

Problem 8a

Here we have two resistors, with unequal values. We can use Equation 10 withR1 = 10 and R2 = 25 . The equivalent resistance:

10 25 2507.14

10 25 35eqR

= = =

+

Problem 8b


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Our luck is in as the resistor values are equal (10 k each), so we simply dividethe value of the resistances (10 k) by the number of resistances (5). Theequivalent resistance

102

5

eq

kR k

= =

Problem 8c

Now we only have two resistances of the same value, so the equivalentresistance of the group is half of the value of one resistance.

500250

2eqR = =

Problem 8d

Here we have to use Equation 9.

Let the 100 resistor be R1, the 200 resistor be R2 and the 300 resistor beR3. Substitute the values into Equation 9 to find the equivalent resistance:

1 1 1 1

100 200 300eqR= + +

10.01 0.005 0.00333

eqR= + +

10.01833

eqR=

1

0.01833eqR =

Req = 54.56

Problem 8e

We could do this using Equation 9, but we can take advantage of the fact that wehave two sets of resistors of the same value. So we can calculate the equivalentvalue for each set and then consider that we have a two resistor parallel group

containing the equivalent resistors.

Firstly take our two 5 k resistors and consider their parallel combination. Theyhave the same value, so the value of the parallel combination is 5k/2 = 2.5 k.

Similarly we can treat the next two 2 k resistors as a parallel combination togive 2k/2= 1 k.


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Now take each parallel combination and use Equation 10 to give a finalequivalent resistance:

2.5 1

2.5 1eq

k kR

k k

=

+

2.5 1

3.5eq

k kR

k

=

2.5714

3.5eq

kR = =

Problem 8f

Again we can take advantage of two groups of resistors of the same value.Taking the first group, we have four 40 k resistors, this groups equivalent

resistance

1

4010

4eq

kR k

= =

In the second group we have two 20 k resistors and so the equivalentresistance of this group is

2

2010

2eq

kR k

= =

Now we have a pair of equal value equivalent resistors (both 10 k), so theoverall equivalent resistance is:

105

2eq

kR k

= =

4.3 Short circuits and open circuits in parallel circuits

If the main line of a parallel circuit opens there is no current flow from that point.However, if a branch opens there is no current flow in that branch, but otherbranches are unaffected, although the total current will drop as the equivalent

resistance changes. A short circuit in any branch is an extreme fault conditionand will cause the whole circuit to draw excessive current.

Now attempt Problem 9.

Problem 9

In Figure 15, what is REQif R2: (a) opens; (b) shorts?


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Figure 15 The circuit diagram for Problem 9

Solution

Problem 9a

If R2 opens then we can consider it either as being removed from the circuit(leaving only R1 and R3) or as a resistor with infinite resistance. This is shown inFigure 16.

Figure 16 Circuit for Problem 9, but R2 is open circuit

If we consider it as being totally removed then we can use the simplified parallelresistance formula:

1 3

1 3

60 20 120015

60 20 80eq

R RR

R R

= = = =

+ +

Alternatively we can consider the open circuit R2 as having a voltage of infinity.The equivalent resistance of the three resistor parallel combination (REQ) is givenby

1 2 3

1 1 1 1

eqR R R R= + +


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1 1 1 1

60 20eqR= + +

Note that:

1 0=

so the open circuit R2 has no effect in the equation

1 1 10

60 20eqR= + +

1 60 20

60 20eqR

+=

60 20

60 20eqR

=

+

This is the same equation as the equation above.

Therefore REQ = 15

I have considered the problem this second way, just to demonstrate that we canconsider the open circuit as an infinite resistance and arrive at the same result.Normally I would assume that an open circuit resistor in a parallel circuit haseffectively been removed.

Problem 9b

Here we consider the case where R2 is short circuited. The equivalent circuit isshown in Figure 17. We assume the short circuit has zero resistance. We canconsider this short circuit as a wire that bypasses the other parallel resistors sothat no current can flow in them. The short circuited R2 is equivalent to replacingR2 with a zero ohm resistor. Remember that the addition of parallel resistancecan only lower the equivalent resistance REQ. The other parallel resistors cannotlower zero ohms (resistors cannot have negative resistance), therefore theequivalent resistance REQ must be zero.


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Figure 17 The circuit of Problem 18 if R2 is short circuited

Alternatively we can substitute the value zero into the three resistor parallelequivalent equation to find REQ (the hard way). (REQ) is given by:

1 2 3

1 1 1 1

eqR R R R= + +

1 1 1 1

60 0 20eqR= + +

Note that:

1

0=

so the short circuit R2 is dominant in the equation:

1 1 1

60 20eqR= + +

1

eqR=

1

eq

R =

Therefore REQ = 0

Again I have used the second method to show that the mathematics will producethe same answer as the more intuitive approach above. In practice any parallelcircuit with a short circuit in any branch will have zero resistance, as the parallel


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resistors cannot reduce the zero resistance of the short circuit any lower thanzero.

5 Potential and current dividers

Weve seen that in a circuit with two resistors in series, R1 and R2, theelectromotive force is divided between the resistors as V1 and V2 and that E =V1 + V2. This is a commonly found circuit in electronics and is called a potentialdivider. The input voltage is the voltage source, E, and the output voltage is thepotential difference across R2, namely V2. To find its value, first apply Ohms lawto find the current in the circuit.

E = I(R1 + R2)

I = E/(R1 + R2)

Next use Ohms law to find the value of the potential difference across R2,namely V2:

V2 = IR2 = E/(R1 + R2) x R2 = ER2/(R1 + R2) (11)

We also saw earlier that when a circuit contains two resistors in parallel, theequivalent resistance is found using:

1/R = 1/R1 + 1/R2

A quantity that isnt often used but is useful in parallel circuit is conductivity,which is the reciprocal of resistance and has the symbol G. Taking the parallel

circuit and replacing resistances with conductances, the equation for theequivalent conductance is:

G = G1 + G2 (12)

A circuit with two parallel branches is sometimes referred to as a current divider.The current generated by the battery is divided down each parallel branch. Thevalue through R2 can be found as follows:

We saw earlier that the equivalent resistance is R = R1R2/(R1 + R2), so thecurrent from the battery is:

E = IR1R2/(R1 + R2)

I = E(R1 + R2)/R1R2 = E/R2 + E/R1

This divides into two branches. Since the potential difference across eachparallel branch is E, the current in each branch is:

I1 = E/R1


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I2 = E/R2

The total current being the sum of these two currents.

Thus I2 in terms of I is:

I2 = E/R2 = IR1R2/R2(R1 + R2) = IR1/(R1 + R2)

If we use conductances instead of resistances, the equation becomes:

I2 = IG2/(G1 + G2)

6 Further reading

At this point you may like to have a look at the relevant sections of the on-linebook Fundamentals of Electrical Engineering and Electronics.

Under the section entitled DC, you may like to have a look at the part calledSeries and parallel circuits.

7 Where next?

You are encouraged to study the Learning Package entitled Kirchoffs lawsnext.


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Learning Package 4

Kirchoffs laws


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Kirchoffs laws


If in doubt, please attempt the self assessments questions in this LearningPackage. If you are not sure about any of the terminology, please read on.

2 Introduction

Although simple in concept, the application of Kirchoffs laws can be quiteconfusing. The main source of the confusion is usually concerned with thedirection and polarity of currents and voltages. The situation is not helped by thevarious methods of solution, which are easy to mix up. Although most books onelectronics described many different methods, here I only want you to learn howto use one of the methods, namely branch currents.

3 Kirchoffs laws

In the previous Learning Package I said that the voltage drops around a circuitmust equal the electromotive force. I used this to derive the equation for theequivalent resistance in a series circuit. Similarly, in deriving the equivalentresistance of a parallel circuit I used the idea that current at a junction wouldbehave in such a way that the current flowing out of a junction must equal thecurrent flowing into a junction.

These two ideas were introduced by Kirchoff and are known as Kirchoffs laws.Formally stated they are:

Kirchoffs voltage law - The algebraic sum of all voltages in a loop must equalzero;

Kirchoffs current law The algebraic sum of all currents entering and exiting anode must equal zero

Lets take the current law first, as I believe this is the simpler of the two.


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Figure 1

At a node, or a junction, where wires are joined together, the sum of the currentflowing into the node must equal the sum of he current leaving the node. If thiswasnt the case there would be a surplus which would form a pool of charge atthe node, and this doesnt happen. If Figure 1 we have two currents flowing intothe node, I1 and I2, and two currents flowing out, I3 and I4. The convention isthat current flowing into a node should have the opposite sign to current flowingout of a node. It doesnt matter which, but I will assign current flowing in to anode as positive and current flowing out as negative. Then:

I1 + I2 I3 I4 = 0

Or put another way:

I1 + I2 = I3 + I4

This equation is the mathematical equivalent to Kirchoffs current law. The sum ofall the currents entering and leaving a node is zero.

Now lets try Kirchoffs voltage law. First of all I need to explain what is meant bya loop. When we were looking series circuits, there was only one loop, and thatwas the complete circuit. Usually circuit are more complex, such as that shown inFigure 2.

Figure 2

In Figure 2 we have a circuit with three resistors. Its not as simple as a seriescircuit or a parallel circuit as there is a mixture of the two. There are a number ofways that this circuit could be tackled to analyse what is going on. However, I justwant to use it to illustrate what is meant by a loop in Kirchoffs voltage law.


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If we start at the point marked A, we could go A B E F A and that wouldbe a loop. Lets call it Loop 1.

We could also go from A again and this time travel around a different loop: A B C D E F A. Lets call this one Loop 2.

Finally, a third loop starts at B and goes B C D E B. Lets call this oneLoop 3.

So, there are a possible three different loops in this circuit. Kirchoffs voltage lawapplies to each one.

In Figure 2 Ive already drawn in the polarity of the potential differences in thecircuit, and the currents in each branch. Although I am confident that Ive drawnthese correctly, it wouldnt matter if I was wrong. If it turned out that one of thecurrents actually goes the other way, then when I do the analysis, that current willcome out negative. Similarly, if I calculate the potential difference and it comes

out negative, that would indicate that I had drawn the polarity of the potentialdifference the wrong way round.

Now, lets apply Kirchoffs voltage law. When analysing a loop, you travel aroundthe loop and note down all the potential differences and electromotive forces.

Again, it doesnt matter if you travel clockwise or anticlockwise. Lets say we goclockwise, and that we note the potential difference as positive if we travel from - to +, and as negative if we travel from + to -.

Let the potential difference across R1 be V1, across R2 be V2 and across R3 beV3.

Loop1

Starting at A, travel clockwise to B. The potential difference across R3 goes from+ to - so this would be recorded as negative, -V3.

Next we go from B to E and travel through R1, where the potential differencegoes from + to -, so we record a negative voltage again, -V1.

Finally we go from E to F to A and pass through the battery, where the voltagegoes from - to + so we record the voltage as positive, E.

Kirchoffs voltage law states that the sum of the voltages around a loop is zero.So:

-V3 V1 + E = 0

Rearrange to get:

E = V1 + V3


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Loop 2

Starting form A and going clockwise to B we find the potential across R3 again,which we record as V3.

From B to C to D we find R2 with a potential difference that goes from + to -,so we record a negative potential difference, -V2.

Finally we go from D to E to F to A and pass through the battery, where thevoltage goes from - to + so we record the voltage as positive, E.

Using Kirchoffs voltage law we get:

-V3 V2 + E = 0

Rearranging we get:

E = V2 + V3

Loop 3

Starting from B and going clockwise to C then D we find R2 and record V2.

From D to E to B we find R1, and this time we find the potential difference goingfrom - to + so we record a positive potential difference, V1.

Using Kirchoffs voltage law we get:

-V2 + V1 = 0

Rearranging gives:

V1 = V2

We end up with three equations. This last one should be no surprise as all it issaying is that the voltage across two parallel branches are equal, which we knowalready.

If we were analysing this circuit we would now go on to substitute using Ohmslaw so that, for example, V1 = I1R1 and so on. Ill come back to this. Before that I

just want to see if I can help to explain Kirchoffs voltage law using an analogy.

In the previous example weve seen Kirchoffs voltage law applied, but it doesntquite explain why the voltages should sum to zero. I like to think of it using ananalogy, where potential difference is equivalent to a change in height. Think ofthe battery as an elevator, and the resistors as steps. If I redraw the circuit ofFigure 2, you should be able to see what I mean.


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Figure 3

In Figure 3, since E, F and D are joined together, they must be at the sameheight (potential). The elevator (battery) takes you up from F to A. Thisrepresents the highest point. Alternatively, you can get to A by going up the stairsfrom E to B and then form B to A, or you could go up the stairs from D to C andthen C to B to A. Either way, you end up at the same height. So if you travelaround a loop, you climb to a particular height and then have to come back downby the same amount to end up where you started. Thats why the sum total of the

height that youve travelled in going around a loop is zero. Even though you haveto climb up and down to get around the loop, if you end up where you startedthen you cant have gained or lost any height.

3 The method of branch currents

If we return to the circuit that was used earlier, as seen in Figure 2, then in ouranalysis of the loops we ended up with 3 equations, repeated here:

E = V1 + V3

E = V2 + V3

V1 = V2

We cant solve these because we dont know the values of the potentialdifferences. First, we can replace the potential differences by the current timesthe resistance using Ohms law.


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E = I1R1 + I3R3 (1)

E = I2R2 + I3R3 (2)

I1R1 = I2R2

The next stage in the solution is to use Kirchoffs current law, which tells us that:

I3 = I1 + I2

So now we can get rid of I2 in Equation 2 for example by substituting:

I2 = I3 I1

E = (I3 I1)R2 + I3R3

E = I3R2 I1R2 + I3R3 = I3(R2 +R3) I1R2 (3)

We now have equation 1 and Equation 3 which both have two unknowns, namelyI1 and I3, and so these can be solved.

Rather than solve a theoretical circuit, lets put some numbers in. Let:

E = 10 V

R1 = 100

R2 = 200

R3 = 300

What are the values of the currents in the circuit?

Starting from the beginning again, we have (using E = 10 V):

10 = V1 + V3

10 = V2 + V3

V1 = V2

We cant solve this because we dont know the values of the potentialdifferences. Replace the potential differences by the current times the resistanceusing Ohms law.

10 = I1x100 + I3x300 (4)

10 = I2x200 + I3x300


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I1x100 = I2x200

The next stage in the solution is to use Kirchoffs current law, which tells us that:

I3 = I1 + I2 (5)

So now we can get rid of I2 in Equation 2 for example by substituting:

I2 = I3 I1

10 = (I3 I1)x200 + I3x300

10 = I3x200 I1x200 + I3x300 = I3x(200 +300) I1x200

10 = I3x500 -I1x200 (6)

Multiply Equation 4 by 2:

20 = I1x200 + I3x600 (7)

Add Equations 6 and 7

10 + 20 = I3x500 -I1x200 + I1x200 + I3x600

30 = I3x1100

I3 = 30/1100 = 0.028 A = 28 mA

Substitute in Equation 6:

10 = 0.028x500 -I1x200

10 = 14 -I1x200

I1x200 = 14 10 = 4

I1 = 4/200 = 0.02 A = 20 mA

Finally, using Equation 5:

0.028 = 0.02 + I2

I2 = 0.028 0.02 = 0.08 A = 8 mA

We have analysed the circuit using the method of branch currents. The steps thatyou need to take are:

Step 1 Identify the currents in each branch of the circuit. If in doubt guess thedirection;


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Step 2 Label each potential difference with its polarity, using the concept thatelectron current will flow from - to +.

Step 3 Apply Kirchoffs voltage law to at least two loops in the circuit, and derivethe equation for each loop

Step 4 Apply Kirchoffs current law to a node in the circuit to get the relationshipbetween currents;

Step 5 Manipulate the equations so that you end up with two equations with twounknowns;

Step 6 Solve the equations.

4 Additional circuits

The circuit that weve just analysed had one battery in it. I now want to look at a

slightly more complex circuit which has an additional battery in one of the otherbranches. This is shown in Figure 4.

Figure 4

Even before Ive put any values on the components Ive made a start. Ive put inthree branch currents, I1, I2 and I3. Ive guessed the direction of the current flow.If they turn out to be wrong, my current values will be negative. Ive then put +and - across all the potential differences in accordance with the rule thatelectron current flows from - to +. Now to analyse the circuit.


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For no particular reason I am going to choose Loop 1 as A - B E F A, andLoop 2 as A B C D E F A.

Loop 1

Starting from A and going clockwise, the first potential difference is across R2and Im going from + to - so its V2. next theres the battery. Again Im goingfrom + to - so the potential difference is negative, and is recorded as E2.Next is E to F to A, in which I pass through the other battery from - to + so thepotential difference is recorded as positive, E1, and then through R1 from + to - so its negative, -V1. Putting all this together I get:

-V2 E2 + E1 V1 = 0 (8)

Loop 2

From A, going clockwise I eventually pass through R3 going from + to -, so I

record V3. Then its back to F and through to A as in Loop 1. So I get:

-V3 + E1 V1 = 0 (9)

Finally, using Kirchoffs current law I get at Node B:

I3 + I2 = I1 (10)

Using Ohms law I can substitute in Equations 8 and 9:

-I2R2 E2 + E1 I1R1 = 0 (11)

-I3R3 + E1 I1R1 = 0 (12)

Then get rid of I3 in Equation 12 using Equation 10:

-(I1 I2)R3 + E1 I1R1 = 0

-I1R3 + I2R3 + E1 I1R1 = 0

I2R3 + E1 I1(R1 + R3) = 0 (13)

Lets put some values in:

E1 = 10V, E2 = 5V

R1 = 100 , R2 = 200 , R3 = 300 .

Re-writing Equations 11:

-I2x200 5 + 10 I1x100 = 0


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I1x100 + I2x200 = 5 (14)

Re-writing Equation 13:

I2x300 + 10 I1(100 + 300) = 0

-I1x400 + I2x300 = -10 (15)

Multiply Equation 14 by 4:

I1x400 + I2x800 = 20 (16)

Add Equations 15 and 16:

I2x1100 = 10

I2 = 10/1100 = 0.009 A = 9 mA

Substitute in Equation 13:

-I1x400 + 0.009x300 = -10

-I1x400 + 2.7 = -10

-I1x400 = -12.7

I1 = 12.7/400 = 0.03175 A = 31.75 mA

Finally, substitute in Equation 10:

I3 + 0.009 = 0.03175

I3 = 0.02275 A = 22.75 mA

The three branch currents have been successfully calculated. By luck they allturn out positive, which means that I guessed correctly when I chose the directionof the currents. Had I got any of the directions wrong, the current would haveturned out with a negative value.

Please now attempt the following Problems.

Problem 1


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Figure 5 Circuit for Problem 1

In Figure 5, use the method of branch currents to solve for I 1, I2and I3. Assumethat V1 = 30 volts and V2 = 90 volts. Once the currents have been calculated,determine the values of VR1, VR2and VR3.

Solution

This problem is an example of branch current analysis. Figure 5 shows thecurrents I1, I2and I3 marked on it. Polarities are also marked on the resistors. Ihave used the convention of electron current flow, which flows from the negativeterminal of the voltage sources to the positive terminal.

Note: Conventional current flow (from positive to negative) could have been usedand the end results would be the same, but I would advise you to stick to oneconvention to avoid confusion.

Since we will use Kirchoffs law to solve the problem, we need to sum thevoltages around two loops. There are three possible loops - one which includesboth voltage sources and two which include a voltage source and R3. Any twoloops will do. Also the direction in which we sum the voltages (that is, go roundthe loop) is optional. It is vital, however, to make sure that the polarities of thevoltages are correct. I think of voltages here as we normally do, that is the

voltage is increasing (going from negative to positive) as we go round the loopthen it is a positive voltage. If the voltage is decreasing (going from positive tonegative) as we go round the loop then it is a negative voltage.

So as we do not introduce too many negative voltages (and increase the risk ofmathematical error) we will choose loops and directions which allow us to addvoltages across the resistors in a positive sense if we can. This will leave the


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voltages across the voltage sources as negative, but when if we take them to theother side of the equals sign (transpose them) they will become positive.

If we choose to go around the loop which includes V1 and R3 in an anticlockwisedirection, starting at the negative end of R3, and using Kirchoffs voltage law weget:

VR3 + VR1 - V1 = 0

Transposing gives:

VR3 + VR1 = V1

By Ohms law:

I3R3 + I1R1 = V1

Substituting the available values gives:

I318 + I1120 = 30 (16)

Similarly if we choose to go around the loop which includes V2 and R3 in aclockwise direction, starting at the negative end of R3, and using Kirchoffsvoltage law we get:

VR3 + VR2 - V2 = 0

Transposing gives

VR3 + VR2 = V2

By Ohms law

I3R3 + I1R2 = V2


I318 + I2180 = 90 (17)

Using Kirchoffs current law at the junction of R1, R2 and R3 gives (consideringcurrents into the branch point as positive and those leaving the branch point as

negative):

I3 - I1 - I2 = 0

Rearranging:

I3 = I1 + I2 (18)


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138 0.16 + 18 I2 = 30

2

30 138 0.16

18I

=

I2 = 0.44 A

I3 = I1 + I2

I3 = 0.16 + 0.44

I3 = 0.6A

We can now determine the voltages across R1, R2and R3 using Ohms law:

VR1 = I1R1

VR1 = 0.16 120

VR1 = 19.2 V

VR2 = I2R2

VR2 = 0.44 180

VR2 = 79.2 V

VR3 = I3R3

VR3 = 0.6 18

VR3 = 10.8 V

Problem 2

In Figure 6, use the method of branch currents to solve for I1, I2, I3, VR1, VR2, andVR3.


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I13 - I32 = 4 (21)

Similarly if we choose to go around the loop which includes V2 and R3 in aclockwise direction, starting at the negative end of R3, and using Kirchoffsvoltage law we get:

VR3 + VR2 - V2 = 0

Transposing gives

VR3 + VR2 = V2

By Ohms law

I3R3 + I1R2 = V2


I310 + I210 = 40

This can be simplified by dividing both sides of the equation by 10:

I3+ I2= 4 (22)

Using Kirchoffs current law at the junction of R1, R2 and R3 gives (consideringcurrents into the branch point as positive and those leaving the branch point asnegative):

I3 + I1 - I2 = 0

Rearranging:

I2 = I1 + I3 (23)

If we substitute the right hand side of Equation 23 into Equations 22 we have:

I3+ (I1 + I3) = 4

Collecting I1 and I3 terms together:

I1 + 2I3 = 4 (24)

To remove I3add Equation 24 to Equation 21:

4I1= 8

I1= 2 A

To find I3 we substitute I1 into Equation 24 giving:


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2+ 2I3 = 4

2I3 = 2

I3 = 1A

Finally, to find I2we substitute I1and I3into Equation 23:

I2 = 2+ 1 = 3A

We can now determine the voltages across R1, R2and R3 using Ohms law:

VR1 = I1R1

VR1 = 2 15 = 30 V

VR2 = I2R2

VR2 = 3 10 = 30 V

VR3 = I3R3

VR3 = 1 10 = 10 V

5 Further

eng1021 electronic principles

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