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Adnan Menderes UniversityFaculty of Arts And Sciences

Physics Department

PHYS 142 Physics LaboratoryElectricity and Magnetism Laboratory Manual

February, 2016AYDIN

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Contents

Introduction..........................................................................................................3

1. Laboratory 1: Electric Field & Electrostatic Potential in the Plate Capaci-tor...................................................................................................................9

2. Laboratory 2: Dielectric Constant of a Material.........................................13

3. Laboratory 3: Ohm’s Law...........................................................................18

4. Laboratory 4: Wheatstone Bridge...............................................................23

5. Laboratory 5: Magnetic Field of Single Coils / Biot-Savart’s Law..............27

6. Laboratory 6: Magnetic Induction................................................................32

7. Laboratory 7: Transformer...........................................................................37

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INTRODUCTION

Purpose of this laboratory course is to teach electricity and magnetism by observa-tions from experiments and prove some principles of electricity and magnetism doingexperiments by using basic measuring devices and measurement techniques. Thisapproach complements the classroom experience of Physics-162 where you learn thematerial from lectures and books designed to teach problem solving skills. Also withthis laboratory students will learn the how experimental uncertainty plays a role inphysical measurements and they will learn how can they minimize this uncertainty.

1 Student Responsibilities

You must be prepared to perform the experiments by reading the lab manual andrelated textbook material before coming to the laboratory. Since each experimentmust be finished during the lab session, familiarity with the underlying theory andprocedure will prove helpful in speeding up your work. If you have questions orproblems with your preparation, you can contact Laboratory Teaching Assistant(LTA), but in a timely manner. You must come to laboratory on time and you haveonly one Makeup Experiment chance (if you have a valid excuse and you must bedocumented), if you miss more than two experiments you will fail from this course, soyou must come and perform all experiments. In laboratory, you should remain alertand use common sense while performing a lab experiment. You are also responsiblefor keeping professional and accurate record of the lab experiments in a laboratorynotebook and you should report any errors in the lab manual to the teaching assistant.At the end of experiment, you must collect and clear the equipments and tables andyou have to prepare an experiment report for each experiment after the laboratory,and bring it at his/her next coming.

2 Grading Policy

Each experiment will be graduated upon 100 points.

1. The student has an mini exam (approximately 15 minutes) for each experimentbefore performing the experiment. [30 points]

2. The student will take experiment performance grades (This part also includesoral exam). [30 points]

3. The student prepare an experiment report. [40 points]

Average of these total grades will be your midterm exam.Final exam include theoretical questions about each performed experiments, and

will be at the end of semester.

3 Laboratory Reports

The laboratory report is a kind of manuscript to communicate your experiment resultsand conclusions to other professionals. In this course you will write the laboratory

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report to inform your LTA, so LTA can learn what you did and learned from theexperiment. But don’t write in your report like this sentence ”In this experiment ilearned the Ohm’s law · · · ”. Your laboratory report should be clear and concise. Weprepare a report format as follow, you can use as a guide. In this lab probably youwill work with more than one lab partners, but your report will be your individualeffort.

3.1 Format of Laboratory Report

Name Surname:

Student Number:

Section and Experiment Date:

Laboratory Teaching Assistant Name: The name of the instructor who youhave performed the experiment with.

Title of Experiment

Purpose In this part explain the purpose of the experiment briefly.

Theory In this part you can benefit from your lecture notes, books, etc. (Dontwrite more than one page for this part)

Apparatus In this part indicate which equipment was used while you perform theexperiment.

Procedure In this part you will write how you perform this experiment in yourown words (not from laboratory manual)

Data and Calculations In this part you must write your data in order ( Youcan express them in a data-table format). You must write formulas that used forcalculations, and your results must be written with significant numbers. You mustwrite units of your all results in all steps of your calculations. You also plot graphson graph paper (milimetric, logarithmic, etc.) in this part if necessary.

Conclusion In this part you must make a comment on your experiment and re-sults (and also on graphs). For example: you can evaluate and compare your resultswith theory, are they in agreement with physics laws, are your graphs as expected withrespect to those laws? If there is error in your results, what can be the physical errorcauses? You must make comment also on those error causes.

Questions In this part you must answer questions about performed experimentwhich are in laboratory manual.

4 Symbols and Units of Some Fundamentals

In this course we will use SI (Systeme International) units for calculations. Some ofthe physical quantities are given in Table 1.

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Table 1: Symbols and units of some physical quantities

Symbol SI unit (Abbreviation)Electrical Field ~E Volt/meter (V/m)

Electrical Potential V Volt (V)Electrical Charge Q, q Coulomb (C)Electrical Current I, i Amper (A)

Power P Watt (W)Current Density ~J Amper/(meter)2 (A/m2)Magnetic Field ~B Tesla (T)

Resistor R Ohm (Ω)Capacitor C Farad (F)Inductor L Henry (H)

Table 2: Some prefixes for powers of ten

Power Prefix Abbreviation Power Prefix Abbreviation10−24 yocto y 101 deka da10−21 zepto z 102 hecto h10−18 atto a 103 kilo k10−15 femto f 106 mega M10−12 pico p 109 giga G10−9 nano n 1012 tera T10−6 micro µ 1015 peta P10−3 milli m 1018 exa E10−2 centi c 1021 zetta Z10−1 desi d 1024 yotto Y

5 Resistor color code

This part is written by the help of the ”Physics Laboratory Manual” which is writtenby David H. Loyd, Angelo State University.

For resistors routinely used in electronic instrumentation, resistance is coded by aseries of colored bands on the resistor. The key to the resistor color coding system isgiven in figure 2 above. The four bands are placed with three equally spaced bandsclose to one end of the resistor followed by a space, and then a fourth band. Thefirst two bands are the first two digits in the value of the resistor, and the third bandgives the exponent of the power of 10 to be multiplied by the first two digits. Thusa resistor with its first three bands labeled Yellow-Violet-Red has a value of 47×102

Ω, or 4700 Ω.

6 Use of Laboratory Instruments

This part is written by the help of the laboratory manual which is written by Dr. J.E. Harriss.

One of the major goals of this lab is to familiarize the student with the properequipment and techniques for making electrical measurements. Some understandingof the lab instruments is necessary to avoid personal or equipment damage. By

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+-

+-

+

-

battery

ac supply

ammeter

galvanometer

potentiometer

heating element

cell

lamp

switch

voltmeter

resistor

transformer

Figure 1: Some electrical circuit symbols.

Figure 2: Resistor color code.

understanding the device’s purpose and following a few simple rules, costly mistakescan be avoided.

Ammeters and Voltmeters: The most common measurements are those ofvoltages and currents. Throughout this manual, the ammeter and voltmeter arerepresented as shown in Figure 3.

Ammeters are used to measure the flow of electrical current in a circuit. Theo-

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Figure 3: Ammeters (A-internal resistance be very small) and Voltmeters (V-voltmeters have ideally infinite impedance).

retically, measuring devices should not affect the circuit being studied. Thus, forammeters, it is important that their internal resistance be very small (ideally nearzero) so they will not constrict the flow of current. However, if the ammeter is con-nected across a voltage difference, it will conduct a large current and damage theammeter. Therefore, ammeters must always be connected in series in a circuit, neverin parallel with a voltage source. High currents may also damage the needle on ananalog ammeter. The high currents cause the needle to move too quickly, hitting thepin at the end of the scale. Always set the ammeter to the highest scale possible,then adjust downward to the appropriate level.

Voltmeters are used to measure the potential difference between two points. Sincethe voltmeter should not affect the circuit, the voltmeters have very high (ideallyinfinite) impedance. Thus, the voltmeter should not draw any current, and not affectthe circuit.

In general, all devices have physical limits. These limits are specified by the devicemanufacturer and referred to as the device rating. The ratings are usually expressedin terms of voltage limits, current limits, or power limits.

7 Preparing Graphs

This part is written by the help of the ”Physics Laboratory Manual” which is writtenby David H. Loyd, Angelo State University.

It is helpful to represent data in the form of a graph when interpreting the overalltrend of the data. Most of the graphs for this laboratory will use rectangular Carte-sian coordinates. Note that it is customary to denote the horizontal axis as x and thevertical axis as y when developing general equations, as was done in the developmentof the equations for a linear least squares fit. However, any two variables can beplotted against each other.

When preparing a graph, first choose a scale for each of the axes. It is not necessaryto choose the same scale for both axes. In fact, rarely will it be convenient to havethe same scale for both axes. Instead, choose the scale for each axis so that the graphwill range over as much of the graph paper as possible, consistent with a convenientscale. Choose scales that have the smallest divisions of the graph paper equal tomultiples of 2, 5, or 10 units. This makes it much easier to interpolate between thedivisions to locate the data points when graphing.

The student is expected to bring to each laboratory a supply of good quality lineargraph paper. A very good grade of centimeter by centimeter graph paper with one

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0 1 2 3 4 5 6

Time (s)

0

5

10

15

20

25

30

Dis

pla

cem

en

t (m

)

0 1 2 3 4 5 6

Time (s)

0

5

10

15

20

25

30

Dis

pla

cem

en

t (m

)

IDEAL ONE

(a) (b)

(c)

FALSE

EXCEPTABLE

Graph (a)

d(m) t(s)

7.57 1.00

11.97 2.00

16.58 3.00

21.00 4.00

25.49 5.00

Graph (b), (c)

d(m) t(s)

0.0 0.00

5.1 1.00

11.0 2.00

14.5 3.00

20.7 4.00

25.6 5.00

29.4 0.00

Figure 4: Graph of displacement versus time; a) plotted by employing list squares fitmethod (Graph (a) data is used), b) False one (Graph (b) data is used), c) Exceptableone (Graph (b) data is used).

division per millimeter is the best choice. Do not, for example, ever use 1/4 inchby 1/4 inch sketch paper or other such coarse scaled paper as graph paper. In somecases special graph paper like semilog or log-log graph paper may be required. Figure4 (a) is a graph of the data for displacement versus time from Figure 4 Graph (a)for which the least squares fit was previously made (You can learn least squaresfit method from related books). Figure 4(b) is the false one, do not connect datawith each other. Figure 4(c) is the exceptable one, this graph is plotted by sense ofproportion according to data and theoretical knowledge. Note that scales for eachaxis have been chosen, to spread the graph over a reasonable portion of the page.Also note that because the data have been assumed linear, a straight line has beendrawn through the data points. The straight line is the one obtained from the leastsquares fit to the data. For most experiments, the variables will take on only positivevalues. For that case the scales should range from zero to greater than the largestvalue for any data point. For example, in Figure 4(a) the displacement is chosen torange from 0 to 30 meters because the largest displacement is 25.49, and the timescale has been chosen to range from 0 to 6 seconds because the largest time is 5.00seconds. Also note that the scales should not be suppressed as a means to stretchout the graph. For example, if a set of data contains ordinates that range from 60to 90, do not choose a scale that shows only that range. Instead a scale from 0 to100 should be chosen, and there is nothing that can be done in that case to make thegraph range over more than about 30% of the graph paper. Scales should always bechosen to increase to the right of the origin and to increase above the origin.

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Experiment 1Electric Field & Electrostatic Potential

in the Plate Capacitor

1 Purpose

To analyze the relationship between the electric field strength and the voltage (elec-tric potential) where the plate spacing is held constant, to analyze the relationshipbetween the electric field strength and the plate spacing where the voltage is heldconstant and to measure the potential as a function of the distance in the manner ofthe plate capacitor phenomenon.

1.1 Keywords

Partial derivative, Derivative operator, Charge, Charge density, Coulomb force, elec-tric field, electrostatic potential energy, electrostatic potential.

2 Theory

According to the Maxwell’s equations;

~∇× ~E ≡ −1c

∂ ~B

∂t(1)

where ~∇ ≡ [Derviative operator], ~E ≡ [Electric field vector], ~B ≡ [Magnetic field vector]and c ≡ [Speed of light]. For the electro-static case (in which the charges are steady)

~∇× ~E ≡ ~0 (2)

However, in terms of ~F ≡ [Coulomb Force] & q ≡ [A test charge], ~E ≡ ~F/q ≡[Electric field vector] ≡ [Coulomb force per unit charge]. Thus for the electrostaticscase;

~∇× ~E ≡ ~0 ⇒ ~∇×~F

q≡ ~0 ⇒ ~∇× ~F ≡ ~0 (3)

which means that the Coulomb force is a conservative force at this point. Hence,the Coulomb force can be expressed as a derivative of a scalar (i.e. it can be derivedfrom a scalar) s.t.

~F ≡ −~∇U (4)

where the scalar U has to be in the unit of energy according to the dimensionalanalysis. (Here, this scalar ”U” with the dimension of energy is nothing but theelectrostatic potential energy difference). Then, dividing the both sides of the aboveequation by q;

~F

q≡ −1

q~∇U ⇒

~F

q≡ −~∇U

q⇒ ~E ≡ −~∇U

q(5)

where ~E ≡ ~F/q ≡ [Electric field vector] again. Hence;

~E ≡ −~∇U

q⇒ ~E ≡ −~∇φ (6)

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where φ ≡ U/q ≡ [Electrostatic potential difference] ≡ [Electrostatic potential energy difference per unit charge] ≡[voltage] Thus, for two different reference points with coordinates a and b;

∫ b

a

~Ed~x ≡∫ b

a(−~∇φ)d~x (7)

In this experiment, if it is assumed that one of the capacitor plates lies in the yz-planein 3-dim. cartesian space and the other one is parallel to it, it can also be assumedthat electric field is in the x-direction and uniform.

⇒∫ b

a

~Ed~x ≡∫ b

aEdx ≡

∫ b

a(−~∇φ)d~x

⇒ E

∫ b

adx ≡ −φ(b)− φ(a) ⇒ E(b− a) ≡ φ(a)− φ(b) ≡ φ

⇒ Ed ≡ φ ⇒ E ≡ φ/d (8)

where ”b − a ≡ d” is nothing but the distance between the parallel-plates and the”φ(a) − φ(b) ≡ φ” is the potential difference between those plates and ”E ≡ | ~E|” isthe magnitude of the electric field vector.

3 Apparatus

1. Two plates

2. Electric Field Magnitude Meter

3. Resistor

4. DC Power Supply (DC Voltage Source)

5. Two Digital Multimeters

6. Connecting Cables

7. Various Supporting Units

4 Procedure

4.1 Relation Between the Magnitude of the Electric Field Eexp.(kV/m)& the Electric Potential φ(V )

1. Construct the experimental setup shown as in Fig.1.

2. Fix the separation of plates as 10× 10−2m.

3. Turn on the DC power supply and set the input voltage and current values 12Vand 0.5A respectively.

4. Check the rotators of electric field magnitude meter and set the output of itto ”0” value adjusting the power supply and the knob-2 ”← 0 →” behind theelectric field magnitude meter when there is no capacitor potential applied.

5. Set the capacitor voltage values to 50V, 80V, 100V, 120V, 150V respectively, de-termine the corresponding experimental value of the electric field magnitudeEexp. and record it on the Table-1 for each case. (Check the leds expressing thescaling of the electric field magnitude meter regularly during the measurements)

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Figure 1: Experimental Setup

Table 1

Etheo.(kV/m) Eexp.(kV/m) φ(V )5080100120150

6. Turn off the DC power supply.

7. Calculating the theoretical values of the electric field magnitude Etheo.(kV/m)complete the Table-1 and determine the error percentages comparing Etheo.(kV/m)values with Eexp.(kV/m) values for each case.

8. Plot an Eexp.(kV/m) v.s. φ(V ) graph using the Table-1 and reach the distance”d” between the plates by means of the graph. Determine the error percentageassuming that the measured value of the distance is correct.

4.2 Relation Between the Magnitude of the Electric Field Eexp.(kV/m)& the Distance d (m)

9. Set the plate distance to 2 × 10−2m moving the plate which does not containthe electric field magnitude meter.

10. Adjust the electric field magnitude meter output to ”0” value using the knob-2”← 0 →” again.

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11. Turn on the DC power supply and set the capacitor voltage (the voltage differ-ence between the two plates of the capacitor) to 200V .

12. Determine the Eexp.(kV/m) and record it to the table-2.

13. Continue the same process for the plate distance values 4×10−2m, 6×10−2m, 8×10−2m, 10× 10−2m respectively.

Table 2

Etheo.(kV/m) Eexp.(kV/m) d (×10−2m)246810

14. Turn off all the electronic devices and cut the power connections.

15. Calculating the theoretical values of the electric field magnitude Etheo.(kV/m)for each case complete the table-2.

16. Determine the error percentage of Eexp.(kV/m) only for the d = 8 × 10−2mvalue.

17. Sketch an Eexp.(kV/m) v.s. d (m) graph and analyze the graph.

5 Questions

1. What do you expect for the electric field in this experiment if the capacitorplates are completely in an ionized liquid (Liquid which is electrically conduct-ing)?

2. Reach the magnitude of the Coulomb force which is felt by a test charge ”q”located at exactly the middle of the capacitor plates if the magnitude of theelectric field is measured as 2 kV/m in this experiment.

Prepared by : Onur GENC.

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Experiment 2Dielectric Constant of a Material

1 Purpose

To examine the relation between the charge ”Q” & the voltage ”φ” in the manner ofplate capacitor, to determine the electric permittivity constant of the vacuum ”ε0”,to examine the relation between the charge Q on the plate capacitor and the distance”d” between the plates when the potential difference between the plates (voltage) ”φ”is constant, to examine the relation between the charge Q on the plate capacitor andthe potential difference between the plates (voltage) when the distance d between theplates is constant.

1.1 Keywords

Charge, electric permittivity constant, Vacuum, Gauss Surface, directional area,

2 Theory

According to the Maxwell’s equations;

~∇ ~E ≡ ρ

ε0︸︷︷︸Differential form of the Gauss’ Law

ρ ≡ [charge density (total charge over volume)]

(1)

⇒∫

~∇ ~EdV ≡∫

ρ

ε0dV ; dV ≡ [infinitesimal volume element] (2)

However, according to the Divergence Theorem;

⇒∫

~∇ ~EdV ≡∮

~Ed~S ; d~S ≡ [infinitesimal surface element] (3)

⇒∫

~∇ ~EdV ≡∫

ρ

ε0dV ≡

∮~Ed~S (4)

⇒ Q

ε0≡

∮~Ed~S

︸︷︷︸Integral form of the Gauss’ Law

Q ≡ [Total charge enclosed by the surface of magnitude S]

(5)As in the previous experiment, if it is assumed that one of the capacitor plates lies inthe yz-plane in 3-dim. cartesian space and the other one is parallel to it, it can alsobe assumed that electric field is in the x-direction and uniform in this experimenttoo. Additionally, if it is also assumed that the area is right-handed and the Gaussiansurface chosen around the capacitor plate from which the electric field is originated iscylinder of the same diameter with the capacitor plate in the manner of the symmetryof the problem;

⇒∮

~Ed~S ≡∫

~Ed ~A (6)

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where d ~A is the infinitesimal area element, defining the surface of the mathematicalGauss cylinder which is parallel to the electric field. Hence;

∮~Ed~S ≡

∫~Ed ~A ≡

∫EdA ≡ E

∫dA ≡ EA ≡

∮~Ed~S

⇒∮

~Ed~S ≡ Q

ε0≡ EA

where A is the magnitude of the area of the plate from which the electric field is orig-inated. On the other hand, the relation among the Capacitance ”C” of the capacitor,voltage φ on the capacitor and the total charge Q accumulated on the capacitor platefor a capacitor for which there is vacuum between the plates is expressed as Q ≡ Cφand Ed ≡ φ from the previous experiment, thus;

Q

ε0≡ EA ⇒ Q

ε0≡ φ

dA ⇒ ε0 ≡ Qd

φA(7)

andQ

ε0≡ EA ⇒ φ

Cε0≡ φ

dA ⇒ C ≡ ε0A

1d

(8)

However, the above equations are valid for the vacuum between the capacitor platescase. If there exists a dielectric material which is not vacuum between the capacitorplates, the voltage φ on the capacitor is reduced by the relative permittivity constantεr ≡ ε/ε0, where ε ≡ [electric permittivity constant of the dielectric material], due tothe electron polarization of the dielectric material s.t.

φD ≡ φ

εr(9)

where ”φD” is the potential difference between the capacitor plates when there is adielectric material between them. Thus the capacitance ”CD” of the capacitor in thepresence of the dielectric material in terms of the capacitance C when there is nodielectric material comparing with respect to the same amount of charge accumulated,is;

CD ≡ Q

φD≡ εr

Q

φ≡ εrC ≡ CD (10)

Then for the case φ ≡ φD using the Q ≡ Cφ the charge relation is;

QD

Q≡ CDφD

Cφ≡ CDφ

Cφ≡ εr ≡ QD

Q(11)

where ”QD” is the charge accumulated on the capacitor plate in the existence of thedielectric material and Q is the charge accumulated on the capacitor plate in theexistence of the vacuum.

3 Apparatus

1. Plate Capacitor

2. Plastic Plate

3. Universal Measuring Amplifier

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4. High Voltage Supplier (0 kV - 10 kV )

5. Digital Multimeter


7. Capacitor (0.22 µF )

4 Procedure

4.1 Relation Between the Accumulated Charge ”Q” & the CapacitorPotential ”φc”

1. Construct the experimental setup shown as in the figure below.


2. Fix the separation of plates as 0.95× 10−2m.

3. Turn the high voltage supply on and adjust the potential values to φc to 0.5kV ,1kV , 1.5kV , 2.0kV , 2.5kV , 3.0kV , 3.5kV , 4.0kV respectively, determine thecorresponding voltage value φ using the multimeter and record it on the table-1for each case. (Do not forget to discharge the capacitor of 0.22µF after eachmeasurement)

Table 1

φc(kV) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0φ(V )Q(C)

ε0(C/V m)

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4. Turn off the high voltage supply.

5. Calculate the Q values benefiting from the relation Q ≡ Cφ and record ontable-1 for each case. (Note: C = 0.22µF )

6. Plot a Q v.s. φC graph using the table-1 and reach the ”ε0” by means ofthe graph. Determine the error percentage assuming that the theoretical value”ε0theo.

= 8.8542× 10−12C/V m” of the ε0 is correct. (Note:A = 0.0531m2)

4.2 Relation Between the Accumulated Charge ”Q” & the PlateDistance ”d (m)”

7. Turn on the high voltage supply and set it to 1.5 kV .

8. Adjust the capacitor plate distance ”d” to 0.2 ×10−2m, 0.3 ×10−2m, 0.4×10−2m, 0.5 ×10−2m, 0.6 ×10−2m, 0.7 ×10−2m respectively, determine thecorresponding voltage value φ using the multimeter and record it on the table-2for each case. (Do not forget to discharge the capacitor of 0.22µF after eachmeasurement)

Table 2

φ (kV )d (×10−2m)

1/d (1/(10−2m))Q (C)

ε0 (C/V m)

9. Turn off the high voltage supply.

10. Calculate the Q values benefiting from the relation Q ≡ Cφ and record ontable-1 for each case. (Note: C = 0.22µF )

11. Plot a Q v.s. 1/d graph using the table-2 and reach the ”ε0” by means ofthe graph again. Determine the error percentage assuming that the theoreticalvalue ”ε0theo.

= 8.8542× 10−12C/V m” of the ε0 is correct. (Note:φc = 1.5kV &A = 0.0531m2)

4.3 The Measurement of the Permittivity Constant of the Di-electric Material (Plastic Plate)

12. Place the plastic plate between the capacitor plates.

13. Fix the separation of plates as 0.95× 10−2m again.

14. Turn the high voltage supply on and adjust the potential values of φc to 0.5kV ,1kV , 1.5kV , 2.0kV , 2.5kV , 3.0kV , 3.5kV , 4.0kV respectively, determine thecorresponding voltage value φD using the multimeter and record it on the table-3 for each case. (Do not forget to discharge the capacitor of 0.22µF after eachmeasurement)

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Table 3

φc (kV ) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0φ (V )

QD (C)Q (C)QD/Q

(QD/Q)average

15. Turn all the electronic devices off and cut all the connections.

16. Calculate the QD values using the relation QD ≡ Cφ and record on table-3 foreach case. (Note: C = 0.22µF )

17. Complete the Q (C) row benefiting from the table-1.

18. Calculate the QD/Q ratio for each case, determine an average (QD/Q)average

value and reach the electric permittivity constant εD of the plastic plate.

5 Questions

1. In this experiment in some steps the vacuum equations are used. In this labora-tory, is the existence of the vacuum possible? If it is, how? If it is not, explainthe situation in terms of physical arguments.

2. If the voltage φ in the experiment is measured as 0.5kV what would it be if the”plate thickness” increases to two times of its initial value?

Prepared by : Onur GENC.

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Experiment 3Ohm’s Law

1 Purpose

1. To test the validity of Ohm’s law, to construct a circuit using resistors, wires anda breadboard from a circuit diagram and to construct series and parallel circuits.

1.1 Keywords

Ohm’s law, serial connection, parallel connection, current, electric potential.

2 Theory

One of the fundamental laws describing how electrical circuits behave is Ohm’s law.According to Ohm’s law, there is a linear relationship between the voltage dropsacross a circuit element and the current flowing through it. Therefore the resistanceR is viewed as a constant independent of the voltage and the current. In equationform, Ohm’s law is:

V = IR (1)

Here V is the voltage applied across the circuit in volts (V ), I is the current flowingthrough the circuit in units of Amperes (A) and R is the resistance of the circuit withunits of ohm (Ω).

Equation (1) implies that, for a resistor with constant resistance, the current flowingthrough it is proportional to the voltage across it. If the voltage is held constant,then the current is inversely proportional to the resistance. If the voltage polarity isreversed, the same current flows but in the opposite direction. If Ohms law is valid,it can be used to define resistance as:

R =V

I(2)

where R is a constant, independent of V and I.The current (I) is a measure of how many electrons are flowing past a given point

during a set amount of time. The current flows because of the electric potential (V )applied to a circuit. If one point of the circuit has a high electric potential, it meansthat it has a net positive charge and another point of the circuit with a low potentialwill have a net negative charge. Electrons in a wire flow from low electric potentialwith its net negative charge to high electric potential with its net positive chargebecause unlike charges attract and like charges repel.

As these electrons flow through the wire, they are scattered by atoms in the wire.The resistance of the circuit is just that; it is a measure of how difficult it is forthe electrons to flow in the presence of such scattering. Materials that have a lowresistance are called conductors and materials that have a very high resistance arecalled insulators. Two or more resistors can be connected in series, connected oneafter another, or in parallel, typically shown connected so that they are parallel toone another.

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When two resistors R1 and R2 are connected in series, the equivalent resistance Rs

is given by Rs = R1 + R2. Thus the circuit in Figure 3.1 behaves as if it containeda single resistor with resistance Rs that is, it draws current from a given appliedvoltage like such a resistor. When those same resistors are connected in parallelinstead, we use a different formula for finding the equivalent resistance.

Figure 1: Schematics of circuits illustrating resistors connected in series and in par-allel

Table 1

Series ParallelVs = V1 + V2 Vp = V1 = V2

Is = I1 = I2 Ip = I1 + I2

Rs = R1 + R21

Rp= 1

R1+ 1

R2

3 Apparatus

1. DC power supply

2. Resistor connection board

3. Digital multimeter

4. Various resistors

5. Connecting cords

4 Procedure

4.1 Part A

1. Adjust the power supply to 0 V.

2. Construct the circuit shown in figure 3.2.

3. Set the power supply to five different potential values and read off the voltageand current values from digital multimeters. Record your measurements inTable 3.2.

4. Plot V versus I graph with your data and find resistance value from slope ofthe graph.

19

Figure 2:

5. Calculate the error percentage of the resistor.

6. Repeat the experiment for two different resistors.

Table 2

R1 R2 R3

V I V I V I

4.2 Part B

1. Construct the circuit shown in Figure 3.3.

2. Use two different resistors while constructing the circuit.


4. Plot V versus I graph with your data and find the total resistance value fromslope of the graph.

5. Calculate the error percentage of the total resistor.


Table 3

R1 and R2 R2 and R3 R1 and R3

V I V I V I

20

Figure 3:

4.3 Part C

1. Construct the circuit shown in figure 3.4.

2. Use two different resistors while constructing the circuit.


4. Plot V versus I graph with your data and find the total resistance value fromslope of the graph.

5. Calculate the error percentage of the total resistor.


Figure 4:

Table 4

R1 and R2 R2 and R3 R1 and R3

V I V I V I

21

5 Questions

1. What is the main difference between an ohmic and non-ohmic conductor? Giveexamples for ohmic and non-ohmic conductors.

2. The current flowing in a circuit containing three resistors connected in parallelis I = 1.0A (Figure 3.5). Find the current and voltage values of all the resistors(R1 = R2 = R3 = 15Ω).

Figure 5:

Prepared by ; S. Gokce CALISKAN.

22

Experiment 4Wheatstone Bridge

1 Purpose

To determine some unknown, connected in series and connected in parallel resistors’resistance values by means of the Wheatstone Bridge, in the Kirchhoff’s Principles(relaying on the conservation of the charge and the energy) concept.

1.1 Keywords

Charge, electrical energy, voltage, Kirchhoff’s principles, Wheatstone bridge, resis-tivity.

2 Theory

For a Wheatstone Bridge circuit in the figure below;

Figure 1: Wheatstone bridge circuit

when the circuit is at equilibrium (i.e. there is no current measurement in thegalvanometer), it can be asserted according to physics principles that the potentialdifference (the voltage) between the points ”A” & ”B” equals to zero. Thus for theequilibrium case;

I1R1 ≡ I2R2 & I3R3 ≡ I4R4 (1)

However, according to Kirchhoff rule of incoming and outgoing current magnitudessum up to zero at a point on a circuit, I1 ≡ I3 & I2 ≡ I4 for the equilibrium situation,hence from the above equations;

I1R1

I1R3≡ I2R2

I2R4⇒ R1

R3≡ R2

R4(2)

for the equilibrium case in Wheatstone Bridge. On the other hand, for a wire theresistance of the wire is expressed in terms of the resistivity of the wire ”ρ”, the

23

length of the wire ”L” and the cross-sectional area of the wire ”A” s.t.

R ≡ ρL

A(3)

3 Apparatus

1. Power Supply 5 V

2. Resistor Connection Board

3. Various Resistors

4. Wire Resistor Board with Various Wires of Diameter and Material

5. Wire Length Changeable Bridge


7. Digital Multimeter (As Galvanometer)

4 Procedure

Construct the experimental setup shown as in the figure below.


4.1 Resistance Value of a Resistor

1. Connect a resistor of 10 Ω as an unknown resistor R3 and 5 Ω as the knownresistor R1 to the circuit.

2. After checking all connections turn on the power supply.

3. Start to change the wire lengths ”l1” & ”l2” on the wire length changeablebridge until determining the equilibrium point (The point at which the currentvalue is read as zero on the multimeter) and record the length values on table-1.

24

Table 1

l1 (×10−2 m) l2 (×10−2 m) R1(Ω) R3(Ω)

4. Turn off the power supply and remove the 10Ω resistor (as R3 in this part) fromthe circuit.

5. Calculate the unknown resistance value ”R3” and calculate the error percentageassuming that its theoretical value 10Ω is correct.

4.2 Resistance of Wires

6. Connect one of the wires of the wire resistor board to the circuit as unknownresistor ”R3” and record the wire type and the diameter ”d” on that wire ontable-1.

7. After checking all connections again turn on the power supply.

8. Start to change the wire lengths ”l1” & ”l2” on the wire length changeablebridge until determining the equilibrium point (The point at which the currentvalue is read as zero on the multimeter) and record the length values on table-1.

9. Continue the process for all the remaining wires on the wire resistor board.

Table 2

Wire Type d(×10−3m) l1(×10−2m) l2(×10−2m) R3(Ω) r(×10−3m) 1/r2( 110−6m2 )

10. Turn off the power supply.

11. Calculate the resistance ”R3” values for each wire type, calculate the radius”r” and ”1/r2” values for constantan wire and complete the table-1.

12. Calculate the resistivity ”ρ” of the brass wire and reach the error percentageassuming that the theoretical value of it ρ = 7.00× 10−8 Ωm is correct.

13. Plot a R3 v.s. 1/r2 for the constantan wire and analyze the graph.

25

5 Questions

1. If the diameter of a brass wire with a length L=1m increases to two times ofits initial value what can you say about the resistance? (resistivity is constant)

2. A Wheatstone Bridge such as in Figure 1 has a V = 10V . With R1 = 5.00Ωand the unknown resistor R3 at room temperature, the bridge is balanced withL1 = 34.0cm and L2 = 66.0cm. What is the value of R3?

Prepared by ; S. Gokce CALISKAN.

26

Experiment 5Magnetic Field of Single Coils / Biot-Savart’s

Law

1 Purpose

1. To measure the magnetic flux density in the middle of various wire loops withthe Hall probe and to investigate its dependence on the radius and number ofturns.

2. To determine the magnetic field constant µ0.

3. To measure the magnetic flux density along the axis of long coils and compareit with theoretical values.

2 Theory

The Biot-Savart law is an equation describing the magnetic field (B) generated byan electric current (I). It gives the magnetic field due to an infinitesimal length ofcurrent; the total field can then be found by integrating over the total length of allcurrents:

d−→B =

µ0I

4π

d~l × r

r2or d

−→B =

µ0I

4π

d~l × ~r

r3(1)

Figure 1: Drawing for the calculation of magnetic field along the axis of a wire loop

Here r is a unit vector that points from the position of the charge to the pointat which the field is evaluated. r is the distance between the charge and the pointat which the field is evaluated. µ0 = 1.2566 × 10−6H/m is the permeability of freespace. The direction of ~B is perpendicular to the direction of both I and r by righthand rule.

The vector d~l is perpendicular to ~r so that :

d ~B =µ0I

4π

d~l

r2(2)

27

and from the drawing, it can be written as :

d ~B =µ0I

4π

d~l

R2 + z2(3)

d ~B can be resolved into a radial d ~Br and an axial d ~Bz components. The d ~Bz

components have the same direction for all conductor elements d~l and the quantitiesare added; the d ~Br components cancel one another out, in pairs. Therefore

Br(z) = 0 (4)

and

Bz(z) = B(z) =∮

dBcosθ =µ0I

4π

∮cosθdl

R2 + z2(5)

Because of cosθ = RR2+z2

B(z) =µ0IR

4π(R2 + z2)3/2

∮dl =

µ0IR2

2(R2 + z2)3/2(6)

If there is a small number of small number of identical loops close together, themagnetic field is obtained by multiplying by the number of turns (n).

i) At the center of loop (z=0) we obtain:

B(0) =µ0nI

2R(7)

ii) To calculate the magnetic field of a uniformly wound coil of length l and Nturns, we multiply the magnetic field of one loop by the density of turns (N/l) andintegrate over the coil length.

Figure 2:

28

dB =µ0IR2

2(R2 + z2)3/2

N

ldz (8)

x = R tanφ and dx = R sec2 φdφ. After a little calculation we obtain

dB =µ0IN

2l

∫ φ2

φ1

cosφdφ (9)

Then

B(z) =µ0IN

2l(sinφ2 − sinφ1) =

µ0IN

2l

(a√

R2 + a2− b√

R2 + b2

)(10)

here a = z + l/2 and b = z − l/2.

At the center of the coil (z = 0):

B(0) =µ0IN

2l

(R2

l2+

14

)−1/2

(11)

3 Apparatus

1. Teslameter

2. Hall probe

3. Induction coils

4. Universal constanter

5. Multimeter

6. Conduction loops

4 Procedure

ATTENTION: Coil current is 0.5 A, so never touch the circuit unless the constanter is totally off!

Part A : Magnetic Field of Coil

1. Set up the experiment as shown in Figure 3.

2. Turn on the teslameter and adjust the zero point.

3. Operate the power supply at 18 V and 0.5 A.

4. Measure the magnetic field at different positions within the coils and record theresults in the tables.

5. Measure the length of each coil and write to the related part on tables. (Φ isthe diameter of coil)

6. Calculate the magnetic fields in the middle of different coils using equation 11and compare with the experimental values.

29

Table 1: For 300 turnsand Φ = 33mm ,l=......mm

z distance (cm) B (mT)

0 (center)



0 (center)



0 (center)

7. Calculate the percentage error for magnetic field in the middle of the coils.

8. Plot the curve of magnetic field along the axis of coil for N=150.

Figure 3: Experimental set-up

Part B : Magnetic Field of LoopATTENTION: Loop current is 5 A, so never touch the circuit unless the constanter is totally off!

1. Set up the circuit in Figure 3.

2. Turn on the teslameter and adjust the zero point.

3. Operate the power supply at 18V and 5A.

4. Measure the magnetic field at the center of the single conductor loops withdifferent diameters and record the results in the Table 4.

5. Plot the magnetic field at the center of single turn, as a function of the radius.

6. Measure the magnetic field at the center of the loops using the loops with a 6cm radius and different numbers of turn. Record the data in the Table 5.

30

Table 4: For single loops

Radius (cm) B (mT)

Table 5: For radius r = 6cm.

Number of turns B (mT)123

7. Plot the magnetic field at the center of the loops with n turns as a function ofthe number of turns.

8. Calculate the magnetic field constant µ0 using one of the graphs plotted forloops and find the percentage error.

5 Questions

1. In the figure below, point P2 is at perpendicular distance R=25.1 cm from oneend of straight wire of length L=13.6 cm carrying current i=0.693 A. (Notethat the wire is not long.) What is the magnitude of the magnetic field at P2 ?

2. A solenoid 1.30 m long and 2.60 cm in diameter carries a current of 18.0 A.The magnetic field inside the solenoid is 23.0 mT. Find the length of the wireforming the solenoid.

3. The figure below shows an arrangement known as a Helmholtz coil. It consistof two circular coaxial coils, each of 200 turns and radius R=25.0 cm, separatedby a distance s=R. The two coils carry equal currents i=12.2 mA in the samedirection. Find the magnitude of the net magnetic field at P, midway betweenthe coils.

6 References

1. Phywe Physics Laboratory Manual

31

Experiment 6Magnetic Induction

1 Purpose

Measuring the induction voltage as a function of :

• the current in the field coil at a constant frequency,

• the frequency of the magnetic field at a constant current,

• the number of turns of the induction coil at constant frequency and current,

• the cross-sectional area of the induction coil at constant frequency and current.

2 Theory

The basic physics phenomenon studied in this lab is the Faraday’s Law, which is thediscovery of Faraday and Henry that a changing magnetic flux induces an electromo-tive force (emf) and thus a current in a conductor as follows :

Uind = −dΦdt

= − d

dt

∫~Bd ~A (1)

Any change in the magnetic environment of a coil of wire will cause a voltage(emf) to be induced in the coil. The negative sign, which is also denoted as Lenzlaw, describes the tendency of the system to oppose the change in magnetic flux.

We know that Φ = B∫

~Bd ~A so

Uind = −dΦdt

= − d

dt

∫~Bd ~A (2)

For n parallel conductor loops the equation can be written as :

Uind = −ndΦdt

(3)

The induced emf in a coil is equal to the negative of the rate of change of magneticflux times the number of turns in the coil.

On the other hand Ampere equation which states that in a conductor a currentgenerates a magnetic field of which the closed field lines around the currents :

∮~B · d~s = µI (4)

There µ is the magnetic conductivity (a material’s constant). We consider the mag-netic flux density in a long coil is constant, so that ~B

∮d~s = Bl = µI. (There l is

the length of the coil which must be significant higher than the diameter. In air µcan be approximated by the magnetic constant µ0 = 1.26 · 10−6 V ·s

A·m .) Also it can bewritten as Φ = B

∫d ~A = ~B ~A.

32

For a long coil with N turns, the absolute value of ~B can be approximated by thefollowing equation :

B = NµI

l(5)

If an alternating current I(t) = I0 · sinwt with a frequency f = ω/2π flows throughthe field coil, then from equation 5, the field density in the field coil is a function oftime and alternates in phase with current.

B(t) =Nµ

l· I0 sin(2πft) (6)

From equation 3,

Uind = −nAdB(t)

dt= −nA2πf

Nµ

lI0 cos(2πft) (7)

where n is the induction coil’s number of turns and A its cross-sectional area. Theinduced voltage alternates with same frequency as the current but is phase-shiftedby π/2.

3 Apparatus

1. Field coil (750mm, 485turns/m)

2. Induction coils

3. Function generator

4. Multimeter

5. Connecting cords

4 Procedure

ATTENTION: The effect of frequency should be studied between 1 kHz and 12 kHz,since below 0.5 kHz the coil practically represents a short circuit and above 12 kHzthe accuracy of the measuring instruments is not guaranteed.

Part A : Measuring the induction voltage as a function of current in thefield coil and calculation of magnetic constant µ

1. Set up the experiment as shown in Figure 1.

2. Tune the current in the field coil by turning up the amplitude of the sinus signalof the digital frequency generator.

3. Start with an amplitude of 0.5 V and increase to a maximum of 10 V in stepsof 0.5 V.

4. Record current and induced voltage datas in Table 1 for f=10 kHz and secondarycoil with n=300 turns and d = 41mm.

5. In order to vary the magnetic field, the current in the field coil has to be altered.

33

6. To calculate µ0, we consider relation 7. The time dependence can be disre-garded, if we always measure current and voltage in intervals of one periodT = 1/f

7. Plot a graph Uind versus I0 and obtain magnetic constant by the help of equation8. (The magnetic field constant is included in the slope s of the induced voltage’slinear dependence of the current. Uind = s(t) · I0)

Uind = −nA2πfNµ

lI0 (8)

Figure 1: Experimental set-up for magnetic induction

Part B : Measuring the induction voltage as a function of frequency ofmagnetic field at constant current.

1. Choose the current in the field coil between 20 mA and 40 mA.

2. The effect of frequency should be studied between 1 kHz and 10 kHz, sincebelow 0.5 kHz the coil practically represents a short circuit and above 12 kHzthe accuracy cannot be guaranteed.

3. Increase the frequency in steps of 0.5 kHz and record datas in Table 2.

4. In order to maintain a constant current in the field coil for various frequencies,you have to adjust amplitude very accurately for each frequency.

5. Plot a graph Uind versus f, calculate the slope of the graph (Uind = s · f) andcompare it with theoretical value of s (theoretical value of s will be calculatedfrom equation 8.)

Part C : Measuring the induction voltage as a function of the numberof turns of the induction coil at constant frequency and current.

34

Table 1: f=10 kHz and secondarycoil with n=300 turns and d =41mm

Amplitude (V) Current (A) Uind (V)0.51.01.52.02.53.03.54.04.55.05.56.06.57.07.58.08.59.010.0

Table 2: I0 = 25 mA and sec-ondary coil with n=300 turns andd = 41mm

Frequency (kHz) Current (A) Uind (V)0.51.01.52.02.53.03.54.04.55.05.56.06.57.07.58.08.59.010.0

Table 3: For fixed I0 and d =41mm

n Uind (V) I0 (A)300200100

Table 4: For fixed I0 and n = 300

d (mm) Uind (V) I0 (A)

1. Choose signal-amplitude for f=10 kHz (fixed I0) and maintain these settingsthroughout the measurements.

2. Note down the induced voltage and diameter (d) for given number of turns onTable 3.

3. Plot a graph Uind versus n.

Part D : Measuring the induction voltage as a function of the cross-sectional area of the induction coil at constant frequency and current.

1. Choose signal-amplitude for f=10 kHz (fixed I0) and maintain these settingsthroughout the measurements.

2. Note down the induced voltage for given diameter (d) on Table 4.

3. Plot a graph Uind versus A.

35

4. The cross-sectional area is the circular area enclosed by coil:

A = π

(d

2

)2

(9)

5 Questions

1. A flat loop of wire consisting of a single turn of cross-sectional area 8 cm2 isperpendicular to a magnetic field that increases uniformly in magnitude from0.5 T to 2.5 T in 1 s. What is the resulting induced current if the loop has aresistance of 2 Ω.

2. A 30 turn circular coil of radius 4 cm and resistance 1 Ω is placed in a magneticfield directed perpendicular to the plane of the coil. The magnitude of themagnetic field varies in time according to the expression B = 0.0100t+0.0400t2,where B is in teslas and t is in seconds. Calculate the induced emf in the coilat t=5 s.

6 References

1. Phywe Physics Laboratory Manual

36

Experiment 7Transformer

1 Keywords

Faraday’s law, Magnetic flux, AC voltage, Transformer.

2 Purpose

The secondary voltage on the open circuited transformer is determined as a function

1. of the number of turns in the primary coil,

2. of the number of turns in the secondary coil,

3. of the primary voltage.

The short-circuit current on the secondary side is determined as a function

1. of the number of turns in the primary coil,


3. of the primary current.

With the transformer loaded, the primary current is determined as a function

1. of the secondary current,


3. of the number of turns in the primary coil.

3 Theory

This theoretical part is written by the help of the book “Physics Principles with Ap-plications” (Douglas C. Giancoli).

Principle: An alternating voltage is applied to one of two coils (primarycoil) which are located on a common iron core. The voltage induced in thesecond coil (secondary coil) and the current flowing in it are investigatedas functions of the number of turns in the coils and of the current flowingin the primary coil.

A transformer is a device for increasing or decreasing an ac voltage. A transformerconsists of two coils of wire, known as the primary and secondary coils. The two coilscan be interwoven; or they can be linked by a soft iron core (laminated to preventeddy current loses). The idea in either case is that the magnetic flux produced by acurrent in the primary should pass through the secondary coil. When an ac voltageis applied to the primary, the changing magnetic field it produces will include an acvoltage of the same frequency in the secondary. However, the voltage will be different

37

according to the number of loops in each coil. From Faradays law the induced voltagein the secondary is

Us = −Nsdφ

dt(1)

where Ns is the number of turns in the secondary coil, and dφdt is the rate at which

the magnetic flux changes. The input primary voltage, Up, is also related to the rateat which the flux changes;

Up = −Npdφ

dt(2)

where Np is the number of turns in the primary coil. We have assumed thereare no losses so that all the flux produced in the primary reaches the secondary.We divide these two equations to find

Up

Us=

Ns

Np(3)

This transformer equation tells how the secondary (output) voltage is related tothe primary (input) voltage. Although voltage can be increased (or decreased) with atransformer, we dont get something for nothing. Energy conservation tells us that thepower output can be no greater than the power input. A well-designed transformercan be greater than 99 percent efficient, so little energy is lost to heat. The powerinput thus essentially equals the power output; since power P = UI, we have

UpIp = UsIs or (Is/Ip) = (Np/Ns) (4)

4 Apparatus

1. Multitab transformer

2. Coils (140 turns, 6 tappings)

3. Iron cores

4. Rheostat

5. Digital multimeters

6. Clamping device

7. Connecting cords

5 Procedure

1. Set up the experimental system as shown in Figure 1.

2. The multi-range meters should be connected as shown in Figure 2, while thevoltmeter can be used through a double-pole two-way switch for the primaryand secondary circuit.

The iron yoke should be opened only when the supply is switchedoff, as otherwise excessive currents would flow.

38

Figure 1: Experimental set-up for investigating the laws governing the transformer.

Figure 2: Connection of the multi-range meters.

3. When loading the rheostat, the maximum permissible load of 6.2 A for 8 min-utes must not be exceeded.

4. At constant supply voltage, the primary current is adjusted using the rheostat inthe primary circuit, with the secondary short-circuited. When the transformeris loaded, the rheostat is used as the load resistor in the secondary circuit.

5.1 Part 1:

1. Measure the secondary voltage for same turns (Np = Ns = 140) and record thedata in Table 1.

39

2. Draw graph of secondary voltage on the unloaded transformer as a function ofthe primary voltage.

3. Make a comment for your graph.

5.2 Part 2:

1. Fixed the primary voltage at 15 V (Up=15V) and turns N at 140 (Np=140).

2. Measure the secondary voltage for given Ns turns in Table 2.

3. Fixed Ns=140 and alter Np like table and measure secondary voltage.

4. Plot a graph for U versus N . Make a comment for your graph.

5.3 Part 3:

1. Similar with Part 1 for same turns measue the secondary current for values intable (Np = Ns=140)..

2. Plot a graph for Ip versus Is. Make a comment for your graph.

5.4 Part 4:

1. Similar with Part 2 fixed the current (Ip=2A) for primary.

2. For alternating Ns and fixed Np=140, measure the secondary current.

3. For alternating Np and fixed Ns=140, measure the secondary current.

4. Plot a graph for Is versus N . Make a comment for your graph.

Table 1: Determination of secondary voltage for same turns

Up (V) Us (V)2.04.06.08.010.012.014.0

Table 2: Determination of secondary voltage for various turns

Ns(Np = 140) Us (V) Np(Ns = 140) Us (V)14 -42 4284 84112 112140 140

40

Table 3: Determination of secondary current for same turns

Ip (V) Is (V)1.01.52.02.53.03.54.0

Table 4: Determination of secondary current for various turns

Ns(Np = 140) Is (A) Np(Ns = 140) Is (A)14 -42 4284 84112 112140 140

6 Questions

1. What is a transformer? How does a transformer work? (Hint: what is the mainphysical principle? think it!)

2. A transformer for a transistor radio reduces 120 V ac to 9 V ac (Such a devicealso contains diodes to change the 9 V ac to dc). The secondary contains 30turns and the radio draws 400 mA. Calculate (a) the number of turns in theprimary, (b) the current in the primary, and (c) the power transformed.

3. Transformer operates only on ac. However, if a dc voltage is applied to theprimary there will be an induced current in the secondary. How is it possible?Explain it with physical laws.

7 References

1. Douglas C. Giancoli, Physics Principles with Applications, Prentice-Hall, INC.Englewood Cliffs, New Jersey 07632.

2. PHYWE, Physics University Experiments, 2013.

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