Chapter 4

Direct Current and Circuits AnalysisChapter 4

Flow of electronsThe direction of conventional current flow is that of positive charge. In a wire, it is actually negatively charged electrons that move, so they flow in direction opposite to the conventional current. Positive conventional current always flows from a high potential to a low potentialExercise 2The current in an electron beam in a cathoderay tube is measured to be 70 A. How many electrons hit the screen in 5.0 s? (e = 1.6 1019 C)a. b. c. d. 2.2 1011 electrons8.8 1013 electrons2.2 1015 electrons8.8 1018 electronsAnswer is CBila user click button C kotak Correct! akan keluar. Selain itu, kotak Try again! akan keluar.7Ohms Law++++----Metal conductorl AIVExperimentally, it is found that the current in a wire is proportional to the potential difference between its ends:

The ratio of voltage to current is called the resistance:

8Exercise 1 A wire carries a steady current of 0.1 A over a period of 20 s. What total charge passes through the wire in this time interval?Solution Formula: Current, I =

Rearrange: dQ = I(dt) Substitute:dQ = (0.1 A)(20 s)Answer:= 2 CUser perlu click pada button Solution utk kluarkan kotak jawapan6A------Electric current is the rate of flow of charge through a conductor. It is measured in Ampere, A or coulomb per second.Click on symbol of current

II Current, I = Unit in Ampere, A or Coulomb per second, C s-1

Bila user cllick button I box disebelahnya akan keluar. Jika boleh charge negative(bulatan merah) kelihatan bergerak dari kiri ke kanan dalam silinder hijau.

5Exercise 4 Formula:

Rearrange:

Substitute: = 5 W

Answer: = 7.5 W

20 30 10 V15 V Calculate the power dissipated by resistors 20 and 30 solution

Click on button solution utk mengeluarkan kotak jawapan13The resistance of a conductor is directly proportional to its length, l and inversely proportional to its cross-sectional area, A of the conductor.Resistivity resistance++++----Metal conductorl AIVAnimation spt yg tlh disediakanButton resistance boleh ditekan utk ke slide 14.9Exercise 3 A chrome wire has a radius of 0.50 mm and a resistivity of 1.5 106 m. What is the resistance per unit length of this wire?

A 0.20-m-long metal rod has a radius of 1.0 cm and a resistance of 3.2 105 . What is the resistivity of the metal?

SolutionSolutionFormula: Resistance

Rearrange:Resistance per unit length,

= 1.5 x 10-6 mArea, A = r2 = (0.5 x 10-3 m)2 = 7.85 x 10-7 m2

Substitute:

Answer: =1.91 m-1 Formula:Resistivity,

Resistance, R = 3.2 x 10-5 Length, l = 0. 2 mArea, A = r2 = (1.0 x 10-2 m)2 = 3.14 x 10-4

Substitute: Answer: = 5.02 x 10-8

Buttn Solution perlu di click utk mengeluarkan kotak jawapan.11Power The energy lost by one coulomb of charge as it passes through the resistor is V. Therefore the energy, E lost by Q coulomb of charge is

Power is the rate of change of energy:

RIQVE = QV

powerExercise Click pd Button power utk keluarkan kotak disebelah dan button exercise utk ke slide 1712EMF and terminal voltageElectric circuit needs battery or generator to produce current these are called sources of emf.Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf:

This resistance behaves as though it were in series with the emf.Resistor in series

A series connection has a single path from the battery, through each circuit element in turn, then back to the battery.Resistor can be combined in series or parallel. In this circuit 3 resistors are connected in series. It is noted that:The current through each resistor is the same. the voltage depends on the resistance.The sum of the voltage drops across the resistors equals the battery voltage.Equivalent resistanceButton Equivalent resistance hyperlink ke slide 2015Resistor in parallel

A parallel connection splits the current:the voltage across each resistor is the sameThe total current is the sum of the currents across each resistorEquivalent resistanceA parallel circuit is one in which two or more components are connected to two common points in the circuitButton Equivalent resistance hyperlink ke slide 2117Exercise 5

Rre.m.fVoltmeterAmmeterFigure above show a source of battery with e.m.f of 15 V and internal resistance, r of 0.5 . If a resistor, R of 4 is added to form a complete circuit. What are the reading of voltmeter and ammeter?Exercise 6

For each of the circuit shown, calculate the Equivalent resistance of the circuit. Values of the current I1 and I2. Voltage across points AB. answerClick pd kotak answer utk mendapatn jawapan20

Some circuits cannot be broken down into series and parallel connections. Therefore, Kirchhoff's rules can be used to solve some of the problems in circuit analysis. Kirchoffs rulesKirchhoffs first law

What do you notice about the traffic at the busy junction? AnswerThe numbers of car entering is equal to the numbers of car leaving the junctionWhen current flow in the circuit the rate of electrons flowing into and out of a junction are the same.User perlu click pada Answer.23BatteryR1R2switch6A2 A4 AKirchhoff s law is based on the principle of conservation of charge. Whatever charge goes into a junction must emerge from it.

Statement of first Kirchoffs lawAt a junction,Sum of current entering = sum of current leaving

12345Introduction Activity 1Activity 3SummaryActivity 224Try it!!R2R3R1R42.5A3.5 Ai1i2i3i4i5i6Key in your answer in the box belowcurrentansweri1 i2i3i4i5i6User perlu masukkan jawapan yg betul pada ruangan answer, utk menyemak jawapan button correct? perlu di click. Jika jawapan betul tanda tick akan muncul, selain tu tanda cross x akan muncul,Contoh utk i1 dan i2 ditunjukkan.Jawapan:i1 = 2.5 Ai2 = 1.0 Ai3 = 1.0 Ai4 = 3.5 Ai5 = 3.5 Ai6 = 2.5 A

Tick / berwarna hijau manakala cross x berwarna merah

25Kirchhoffs Second LawThe roller coaster gains potential energy when it is pulled up to the top of the track. As the roller coaster move down the energy converted to kinetic energy. Similarly in an electric circuit, the source of emf such as battery raises the electrical potential energy of the charge. As the electric charges flow in the circuit the electrical potential energy converted to the other form of energy. Energy is conserved.

Kirchhoffs second law based on conservation of energyKirchhoffs Second LawR2R3R1iiiE1E2ABCDFor a closed loop ABCDA, the current or emf is considered:Positive (+) if it is in same direction with the loop and, Negative (-) if it is in opposite direction with the loop.When applying Kirchhoffs second law, direction of the loop is arbitraryCurrent, iE1E2+++---correct?correct?correct?Click on or +-Bulatan kuning perlu kelihatan berputar untuk menunjukkan arah loop.

Jawapan untuk table Current, i = +E1 =+E2 = -

Untuk semakan button correct perlu di click, tick / (hijau) utk betul dan cross x (merah) utk salah.27E1E2E3R1R2R3i1i2i3ABCEFPossible loopE1E2E1E2E3E3i1i2i1i2i3i3AFEBABEDCBAFDCA+-+-+-+-+-+-+-+-+-Click on each of the possible loop in the circuit.For each loop, state the sign + or for each emf and current. User perlu click pada button loop yg dikehendaki. Bila salah satu loop di pilih hny loop yg berkenaan sahaja akan muncul pada litar. Cth, utk AFEBA, loop kecil atas akn muncul, utk BEDCB loop kecil bwh akn muncl manakala AFDCA loop bsr akn muncl.

Kemudian user perlu menentukan simbol + atau yang mewakili stp E atau i, jika jawapan betul backgroud kotak cell bertukar menjadi warna hijau, jika salah popup TRY AGAIN akan keluar.

Jawapan AFEBAE1 =+E2 =+i1 = -i2 = -

BEDCBE2 =-E3 =-i2 = +i3 = +

AFDCAE1 = +E3 = -i1 = -i3 = +28R2R3R1iiiE2ABCDKirchhoffs Second LawIn a close loop (ABCDA) :E = (iR)

E = E1+E2 (iR) = iR1 + iR2 + iR3User perlu click pada kotak Kirchhoffs Second Law utk mengeluarkan kotak hijau berserta keterangannya.29Exercise 1E2 = 45VE1 =80 V30 i1i2i3ABCDEF40 1 1 Determine the value of current i1, i2 and i3answeri1 = -2.49 Ai2 = 5.40 Ai3 = 2.92 Aworking stepsUser perlu click pada answer utk mengeluarkan jawapan atau click kotak working steps.30E2 = 45VE1 =80 V30 i1i2i3ABCDEF40 1 1 Apply first law at any junction (example junction B):i1 + i2 = i3 Apply second law for any possible close loops:Loop AFEBA45 V = i1 (30 ) + i3 (1 ) + i3 (40 ) 2nd law states that E = iRLoop BCDEB45 V + 80 V = i2 (1 ) + i3 (1 ) + i3 (40 ) Solve the equations by using simultaneous equation or matrix method .User perlu click pada button ungu untuk mdptkan maklumat dibwhnya..Button hijau hny keluar apabila button ungu pada baris ke3 ditekanApabila Button hijau pada baris ke5 ditekan, loop atas dlm circuit akan muncul disertai dengan equation 2(baris ke6)Apabila Button hijau pada baris ke7 ditekan, loop bawah dlm circuit akan muncul disertai dengan equation 3(baris ke8)solve the equation muncul selepas baris ke 8 muncul.

31Exercise 2E 12 V5 A2 AABEF20 10 2RDetermine the value of resistor R and emf EanswerR = 0.67 E = 70 VCDUser perlu click pada answer untuk mendapatkan jawapan.32Summary current is the rate of flow of charge through a conductor:

The direction of current is taken to be the direction of flow of the positive charge, this is opposite to the direction of the flow of electrons.Ohms law: Resistance of a wire directly proportional to its length and inversely proportional to its cross-sectional area: where is resistivity of the material

Power is the rate of change of energy:

Terminal voltage = emf lost voltage

where r is internal resistanceResistor in series: the voltage depends on the resistanceThe current through each resistor is the same.Resistor in parallel:The current through each resistor is the same. the voltage depends on the resistanceEquivalent resistanceSeries:

Parallel:

SummaryKirchhoffs first law states that, at a junctionsum of current entering = sum of current leavingKirchhoffs first law is based on the principle of conservation of charges.Kirchhoffs second law states that for a close loopsum of emfs = sum of the products of the current and resistance Kirchhoffs second law is based on the principle of conservation of energy.