simple dc circuits - university of plymouth€¦ · 13 v (d) 15 4 v end quiz. solutions to...

Basic Engineering

Simple DC Circuits

F Hamer, R Horan & M Lavelle

The aim of this package is to provide a short selfassessment programme for students who wantto understand simple circuits and, in particular,how to add resistors in series and parallel.

Copyright c 2005 Email: chamer, rhoran, [email protected] Revision Date: February 7, 2005 Version 1.0

http://www.plymouth.ac.uk/mathsmailto:protect egingroup catcode ` active def { }catcode `%active let %%let %%catcode `#active def mailto:[email protected]:protect egingroup catcode ` active def { }catcode `%active let %%let %%catcode `#active def

Table of Contents

1. Introduction2. Resistors in Series3. Resistors in Parallel4. Resistor Combinations5. Final Quiz

Solutions to ExercisesSolutions to Quizzes

The full range of these packages and some instructions,should they be required, can be obtained from our webpage Mathematics Support Materials.

http://www.plymouth.ac.uk/mathaid

Section 1: Introduction 3

1. IntroductionConsider the diagram below:

I IR V

it shows a device or part of an electrical circuit with a potential dif-ference V (S.I. units volt, symbol V ) across it and an electric currentI (with S.I. unit ampere, symbol A) flowing through it.The resistance R of the device or circuit is defined by

R =V

I

The S.I. units of resistance are: volts ampere1.

Because resistance is an important and frequently used concept, itsunit, volts ampere1, is called an ohm (symbol ).


Quiz If a current of 0.045 A is measured to pass through a long wirewhen a potential difference of 1.5 V is applied to it, what is the resis-tance of the wire?

(a) 33.3 (b) 3.33 (c) 0.0675 (d) 0.03

For many conducting materials R is constant for a wide range ofapplied potential differences. This is known as Ohms law:

V = IR

The law states that if the potential difference is, say, doubled thentwice as much current will flow through the resistance.

N.B. if in a material, R depends on the current or on the appliedpotential difference, we say that the material does not obey Ohmslaw. In this package it is assumed that all resistors obey Ohms law.


A fundamental principle in science is that electric charge is conserved.In electric circuits this is expressed by Kirchoffs law for currents:the total current that flows into a junction must also flow out of it.In the junction below I1 flows in and I2 and I3 flow out, so Kirchoffslaw says:

I1 = I2 + I3 I1

I2

I3

Quiz Use Kirchoffs law to choose the correct answer for this diagram:

I3

I4

I1

I2

(a) I1 + I2 = I3 + I4 (b) I1 + I4 = I2 + I3(c) I2 = I1 + I3 + I4 (d) I2 + I4 = I1 + I3

Below we will use these rules to describe combinations of resistors.

Section 2: Resistors in Series 6

2. Resistors in SeriesTwo resistors connected in series, as in the diagram below,

R1 R2 RT=

are equivalent to a single resistor with resistance RT given by

RT = R1 + R2

The values of any number of resistors in series can also be added.Example 1 If two resistors of R1 = 3 and R2 = 5 are added inseries, their total resistance is given by:

RT = 3 + 5 = 8

If a current of 4 A flows through this system, then, from Ohms law(V = IR), there is a potential difference of 43 = 12 V across R1 anda potential difference of 4 5 = 20 across R2. The total potentialdifference is thus 12 + 20 = 32 V . This is equivalent to the samecurrent (4 A) flowing through an 8 equivalent resistor.

Section 2: Resistors in Series 7

Exercise 1. These exercises refer to the diagram below:

R1 R2 RT=

(a) If R1 = 6 and R2 = 3, what is the equivalent resistance RT?

(b) If R1 = 0.08 and R2 = 0.17 , what is the value of RT?

(c) If R1 = 0.5 and R2 = 0.2 , what is the value of RT?

(d) If the equivalent resistance is measured to be RT = 27 and it isknown that R2 = 15 , what must the value of R1 be?

(e) If R1 = 2 and R2 = 3, and a current of 3 A is measured toflow through them, what is the potential difference across each ofthe individual resistors and what is the total potential difference?

Section 3: Resistors in Parallel 8

3. Resistors in ParallelTwo resistors connected in parallel, as in the diagram below,

R1

R2

RT=

are equivalent to a single resistor with resistance RT given by

1RT

=1

R1+

1R2

A proof of this result is given on the next page.In the above diagram there are two different ways for current to flowthrough the circuit, and so the equivalent resistance RT is less thaneither of R1 and R2. This is a general property of resistors in parallel.


Proof:It is helpful to drawthe currents in thediagram:

R1

R2

RT=

I1

I2

ITIT

It is important to realise that the potential difference across the re-sistors in parallel is the same, call it V . From Ohms law:

V = I1R1 = I2R2 I1 =V

R1and I2 =

V

R2

This is equivalent to a total resistor RT with a current IT flowingthrough it:

V = ITRT IT =V

RTFrom Kirchoffs law, IT = I1 + I2, so from the equations above:

V

RT=

V

R1+

V

R2

Cancelling the common factor of V yields the desired result.


Example 2 If resistors of 4 , and 8 are added in parallel, theirequivalent (or total) resistance is given by:

1RT

=14

+18

=28

+18

=38

see the package on Fractions

RT =83

.

Exercise 2.These exercises refer tothe diagram. In eachcase, calculate the value ofwhichever of R1, R2 or RTis not given.

R1

R2

RT=

(a) R1 = 6 , R2 = 3 (b) R1 = 6 , R2 = 4 (c) RT = 0.04 , R2 = 0.2 (d) R1 = 104 , RT = 2.5 103


Quiz If two parallel connected resistors have a total resistance of 18 and one is a 72 resistor, what is the value of the other?

(a) 90 (b) 54 (c)725

(d) 24

Any number of resistors in parallel can also bedescribed in this way. Thus three resistors inparallel have total resistance:

1RT

=1

R1+

1R2

+1

R3

R1

R2

R3

Quiz If three resistors, R1 = 4, R2 = 6 and R3 = 12 , are addedin parallel, what is their equivalent resistance?

(a) 22 (b) 2 (c) 36 (d)63144

Quiz If four identical resistors, each of resistance R = 8, are addedin parallel, what is their equivalent resistance?

(a) 2 (b) 4 (c) 32 (d) 0.5

Section 4: Resistor Combinations 12

4. Resistor CombinationsThe above rules can also be used to calculate the equivalent resistanceof combinations of series and parallel resistors. To see how this is doneconsider the following example.Example 2 To calculate the equivalent resistance of the followingnetwork 2 4

6

7

first combine those resistors in the parallel arrangement which are inseries (here this is just on the top row):

6

6

7


Then add the resistors in parallel:1

RT=

16

+16

=13

RT = 3.

Now we have reduced the network to two resistors in series:3 7 10

=

So the initial network is equivalent to a resistor of 10 .

In summary, the general procedure is:

first add up the series connected resistors that are part of aparallel arrangement;

then calculate the equivalent resistance for all these equivalentparallel resistors (the network should then have the form of aseries of resistors in series);

finally add all these series resistors to obtain the total equivalentresistance.


Quiz What is the equivalent resistance of the network below?

5 3

16

(a) 24 (b)163

(c)316

(d) 2

Quiz If R1 = 5 , R2 = 3 , R3 = 8 and R4 = 4 , what is theequivalent resistance of the network below?

R1 R2

R3 R4

(a) 20 (b)211144

(c)275

(d)245


Quiz If R1 = 100 , R2 = 300 and R3 = 40 , what is the equiva-lent resistance of the network below?

R1

R2

R3

(a) 240 (b) 115 (c)300410

(d) 40.75

Quiz If R1 = 5 103 , R2 = 30 , R3 = 60 and R4 = 400 , whatis the equivalent resistance of the network below?

R1

R2

R3

R4

(a) 5490 (b) 990 (c) 920 (d) 5420

Section 5: Final Quiz 16

5. Final Quiz6

3

3I

4A

2A

Begin Quiz All questions refer to the above circuit diagram.

1. What is the current I?(a)

65

A (b) 6 A (c) 5 A (d) 1 A

2. What is the equivalent resistance of the two parallel resistors?

(a) 2 (b) 1 (c)56

(d)65

3. What is the total potential difference?

(a) 30 V (b) 22 V (c)3013

V (d)154

V

End Quiz

Solutions to Exercises 17

Solutions to ExercisesExercise 1(a)

If two resistors of R1 = 6 and R2 = 3 are added in series, asshown in the picture below

R1 R2 RT=

the equivalent resistance is given by:

RT = R1 + R2 = (6 + 3) = 9 .

Click on the green square to return


Exercise 1(b)

If two resistors of R1 = 0.08 and R2 = 0.17 are added in series,as shown in the picture below

R1 R2 RT=

their total resistance RT is :

RT = R1 + R2 = (0.08 + 0.17) = 0.25 .



Exercise 1(c)

If two resistors of R1 = 0.5 and R2 = 0.2 are added in series, asshown in the picture below

R1 R2 RT=

their total resistance RT is :

RT = R1 + R2 = (0.5 + 0.2) = 0.7 .



Exercise 1(d)If an unknown resistor, R1 , and a resistor R2 = 15 , are added inseries, as shown in the picture below

R1 R2 RT=

and if the equivalent resistance is RT = 27 then RT = R1 + R2 .Therefore the resistance R1 is found from the equation

R1 = RT R2 = (27 15) = 12 .



Exercise 1(e)If a current of 3 A flows through the system shown in the picturebelow

R1 R2 RT=

then, from Ohms law (V = IR), there is a potential difference of

V1 = IR1 = 3A 2 = 6 Vacross R1 and a potential difference of

V2 = IR2 = 3A 3 = 9 Vacross R2. The total potential difference is thus

VT = V1 + V2 = (6 + 9) V = 15 V .

N.B. This is equivalent to the same current (3 A) flowing through anequivalent resistor RT = R1 + R2 = (2 + 3) = 5.Click on the green square to return


Exercise 2(a)If two resistors R1 = 6 and R2 = 3 are added in parallel, thetotal resistance RT is determined through the equation:

1RT

=16

+13

=16

+26

=36

=12

,

therefore RT = 2 .Click on the green square to return


Exercise 2(b)If two resistors R1 = 6 and R2 = 4 are added in parallel, the totalresistance RT is determined through the equation:

1RT

=16

+14

=212

+312

=512

,

therefore RT =125

= 2.4 .



Exercise 2(c)If two resistors, one of unknown resistance R1, and another of R2 =0.2 are added in parallel, and the total resistance is RT = 0.04 then the resistance of R1 is determined through the equation:

1R1

=1

RT 1

R2

=1

0.04+

10.2

=1004 10

2= 25 5 = 20 ,

therefore R1 =120

= 0.05 .



Exercise 2(d)If two resistors, one of R1 = 104 , and another of unknown resistanceR2 are added in parallel, and the total resistance is RT = 2.5 103 then the resistance of R1 is determined through the equation:

1R2

=1

RT 1

R1

=1

2.5 103 1

104=

102.5

104 1 104

= (4 1) 104 = 3 104 ,

therefore R2 =1

3 104 =

13 104 = 3.3 103 .


Solutions to Quizzes 26

Solutions to QuizzesSolution to Quiz:If a potential difference of 1.5 V is applied to a wire and an electriccurrent of 0.045 A is measured to pass through it, the resistance R ofthe wire is given by

R =V

I=

1.5 V0.045 A

=1

0.03 = 33.3 .

In this calculation the resistance is calculated in units of ohms

= volts ampere1 .End Quiz


Solution to Quiz:

I3

I4

I1

I2

According to Kirchoffs lawfor currents: the total currentthat flows into a junction mustflow out of it. In the junctionshown in the picture I2 flows inand I1, I3 and I4 flow out, there-fore:

I2 = I1 + I3 + I4 .

End Quiz


Solution to Quiz:If two parallel connected resistors have a total resistance of RT = 18 and one resistor is R1 = 72, then the resistance of R2 is found fromthe equation:

1R2

=1

RT 1

R1

=118 1

72=

472 1

72

=372

=124

,

therefore R2 = 24 .End Quiz


Solution to Quiz:If three resistors, R1 = 4 , R2 = 6 and R3 = 12 , are addedin parallel, then the total resistance RT is determined through theequation:

1RT

=1

R1+

1R2

+1

R3

=14

+16

+112

=312

+212

+112

=612

=12

,

therefore RT = 2 .End Quiz


Solution to Quiz:If four identical resistors, each of resistance R = 8 , are added inparallel, then the total resistance RT is determined via the equation:

1RT

=1R

+1R

+1R

+1R

=4R

=48

=12

,

therefore RT = 2 .

N.B. If n identical resistors are combined in parallel, then the totalresistance is

RT =R

n.

End Quiz


Solution to Quiz:We have to calculate the equivalent resistance of the network below

5 3

16

=

(5 + 3)

16

RT=

Combining in parallel the resistance R1 = (5 + 3) = 8 obtainedin the first step with the resistance R2 = 16 , one can calculate thetotal resistance via

1RT

=18

+116

=216

+116

=316

.

Thus RT =163

. End Quiz


Solution to Quiz:The calculation of the equivalent resistance of the network is shownin the picture below

R1 R2

R3 R4

=

R5

R6

RT=

First combine the resistors in series. From R1 = 5 and R2 = 3 , wehave R5 = (5+3) = 8 and similarly from R3 = 8 and R4 = 4 ,R6 = (8 + 4) = 12 . Now the equivalent resistors R5 and R6 arein parallel, so RT is found from the equation

1RT

=1

R5+

1R6

=18

+112

=3 + 224

=524

,

so RT =245

. End Quiz


Solution to Quiz:

R1

R2

R3

To find the equivalent resistanceof this network, we first calculatethe equivalent resistance R12 ofR1 = 100 and R2 = 300 inparallel

1R12

=1

R1+

1R2

=1

100+

1300

=3 + 1300

=4

300=

175

,

so R12 = 75 . Now R12 and R3 = 40 are in series, therefore thetotal resistance is

RT = R12 + R3 = (75 + 40) = 115 .

End Quiz


Solution to Quiz:

R1

R2

R3

R4

To find the total resistance ofthe network shown in the pic-ture, first calculate the equiva-lent resistance R23 of R2 = 30and R3 = 60 in parallel

1R23

=1

R2+

1R3

=130

+160

=260

+160

=360

=120

,

so R23 = 20. Now this resulting resistor R23 and R1 = 5 103 and R4 = 400 are all three in series, so the total resistance is

RT = R1 + R23 + R4 = (5000 + 20 + 400) = 5420 .

End Quiz

Table of Contents1 Introduction2 Resistors in Series3 Resistors in Parallel4 Resistor Combinations5 Final Quiz Solutions to Exercises Solutions to Quizzes

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