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Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Unit 2: Electric Forces and Fields Electric Forces Other forces can act on objects beside the gravitational force – e.g. Magnetic force – causing a nail to accelerate towards a magnet Electric force – a rubbed comb with sweater attracts small pieces of paper There are two kinds of charges. 1. Positive Charge An object is positively charged if it has more protons than electrons. 2. Negative Charge Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 An object is negatively charged if it has more electrons than protons. Neutral objects An object is neutral when the number of electrons is equal to the number of protons. Protons do not leave their positions and are not free to move. But electrons can be added or removed from atoms. Static electricity (electricity at rest) - Electrostatic charges are collected and held in the object. Law of Electrostatics 1. Like charges repel. 2. Unlike charges attract. 3. Charged objects attract neutral objects http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l1c.html Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Methods of Charging 1. Friction – electrons move from one object to another when it is rubbed. Electrons go to the object that has stronger attraction to electrons (protons do not leave their positions). Ex: Ebonite rubbed with fur becomes negatively charged (Electrons move from the fur to ebonite.) Glass rod rubbed with silk becomes positively charged (Glass loses electrons to silk) Law of Conservation of Charge - In an isolated system the total amount of charge remains constant. Charges only separate or electrons move from one object to another. http://www.regentsprep.org/Regents/physics/phys03/atribo/default.htm http://www.regentsprep.org/Regents/physics/phys03/aeleclab/chargstp.htm http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2a.html 2. Conduction or Direct Contact – An object gets the same charge as the charging object by contact. Protons do not move from atom to atom. http://www.regentsprep.org/Regents/physics/phys03/aeleclab/escope.htm http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2c.html Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Electroscope – device to identify and measure the amount and type of charge. - Pith ball electroscope http://www.fiu.edu/~phydemo/j1-04.htm - Metal leaf electroscope http://www.engr.uky.edu/~gedney/courses/ee468/expmnt/escope.html http://rene.balderacchi.club.fr/illustrations/Electroscope%2002.jpg Charging a metal leaf electroscope by contact http://regentsprep.org/Regents/physics/phys03/aeleclab/induct.htm http://cougar.slvhs.slv.k12.ca.us/~pboomer/labsphys/physlabook/lab36.htm l http://www.toppermost.biz/electroscope.gif Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 3. Induction – There is no contact between the charging object and the charged object. This is a rearrangement of charges. By induction, an object receives the opposite charge. http://www.physicsclassroom.com/mmedia/estatics/isop.html http://www.physicsclassroom.com/mmedia/estatics/itsn.html http://www.physicsclassroom.com/mmedia/estatics/epn.html http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2b.html http://encarta.msn.com/media_701504655_761566543_1_1/Charging_by_Induction.html http://www.shep.net/resources/curricular/physics/P30/Unit2/electroscope.html Grounding charged object by providing a path for the charges to move to/from Earth. http://www.physicsclassroom.com/mmedia/estatics/gep.html http://www.physicsclassroom.com/mmedia/estatics/gen.html http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l2b.html Why does a charged object attract a neutral object? The electrons in the neutral object rearrange themselves so that unlike charges are close to each other. http://www.ux1.eiu.edu/~cfadd/1160/Ch17/Friction.html Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Lightning – is a discharge (transfer of electrons) between 2 clouds or a cloud and the earth. http://www.regentsprep.org/Regents/physics/phys03/alightnin/default.htm p. 539 – key terms (define) Movie: http://www.bchs.calgary.ab.ca/site.php?p=/curriculum/science/physics30/ Charging electroscope and pithball http://departments.weber.edu/physics/amiri/director/dcrfiles/electricity/pith BallS.dcr http://departments.weber.edu/physics/amiri/director/dcrfiles/electricity/elect roscopeS.dcr Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Coulomb’s Law Coulomb’s law states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Fe q1q2 Fe 1/R2 Fe = k q1q 2 R2 Fe = Electric Force in N q1 and q2 are charges in Coulombs (C) R = distance between the charges in m k = Coulomb’s Constant = 8.99 x 109 N.m2/C2 1 C = 6.25 x 1018 elementary charges (e- or e+) Thus, the charge on an electron and proton is: 1 e- = 1.60 x 10-19 C 1 e+ = 1.60 x 10-19 C Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Double the distance, force drops by ¼. Double both charges, force increases by a factor of 4. Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 http://www.usm.maine.edu/~newton/Chy251_253/Lectures/CoulombsLaw/Coulo mb.html Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Example Find the force between a 50 C charge and a -30 C charge separated by a distance of 20 cm. Fe = kq1q2/R2 = (8.99 x 109)(5.0 x 10-5)(3.0x10-5) / (2.0 x 10-1)2 = 337.1 N F = 3.4 x 102 N (attraction) An Elementary Charge is The charge on an electron is The charge on a proton is The charge on an alpha particle is 1.60 x 10-19C 1.60 x 10-19 C 1.60 x 10-19 C 3.20 x 10-19 C Find the force between a proton and another proton separated by a distance of 5.0 x 10-11 m. Ex: Fe = kq1q2/R2 = (8.99 x 109)(1.60 x 10-19 )(1.60x10-19 ) / (5.0 x 10-11 )2 F = 9.2 x 10-8 N (repulsion) Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Ex: Find the electric force between a nucleus (+20e) and an electron separated by a distance of 5.0 x 10-11 m. Fe = kq1q2/R2 = (8.99 x 109 )(20)(1.60 x 10-19 )(1.60 x 10-19 ) / (5.0x10-11 )2 F = 1.8 x 10-6 N (attraction) Newton’s Law of Universal Gravitation The gravitational force between two masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between them Fg 1/R2 Fg m1m2 Fg Gm1m2/R2 Fg = Gravitational Force (N) m1, m2 = Mass 1, 2 (kg) R = Distance between the centers of masses (m) G = Gravitational Constant = 6.67 x 10-11 N.m2/kg2 Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Example Find the gravitational force between a 20 kg object and a 50 kg object if the distance between them is 30 cm. Fg= 7.4 x 10-7 N Differences between Gravitational Force and Electric Force Gravitational Force Electric Force Weak Force Strong Force Only attractive Both attractive repulsive Inversely proportional to the square of Inversely proportional to the square of distance between the masses. distance between the charges. Proportional to the product of the Proportional to the product of the masses charges p. 530 #1, p.531 #1, p. 533 #1, 2 p. 538 #5 - 7 Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 The redistribution of charges on spheres after they touch Example 1. Sphere 1 of charge 10 C is touched to sphere 2 of charge –20 C and then separated. Then sphere 1 is touched to sphere 3 of 25 C and separated again. Find the charge on each sphere. 10 -20 -5 -5 (1) (2) (1) (2) Thus, Sphere 1 Sphere 2 Sphere 3 = = = 10 C -5.0 C 10 C -5 25 10 10 (1) (3) (1) (3) Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 2. A conducting sphere X that has an initial charge of 2.0x10-8 C and an identical conducting sphere Y has an initial charge of –3.0x10-8 C are touched together. (a) After they are separated, what is the charge on sphere X? (b) If the distance between the charges is 2.0 cm after they are separated, what is the electric force between the charges? (a) -5.0x10-9 C (b) 5.6x10-4 N Vector addition of electric forces Example: Three charges are collinear as shown. FBC FAB 3.0C A 4.0C 1.0 m B -5.0C 2.0 m Find the net electric force on B. C Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 2 FAB=kq1q2/R = (8.99x109)(3.0x10-6 )(4.0x10-6 )/12 = 0.10788 N (repulsion) FBC = (8.99x109 )(4.0x10-6 )(5.0x10-6 )/22 = 0.04495 N (attraction) Net force on B = 0.10788+0.04498 FNet = 0.15 N (right) B FAB FBC Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Example Three charges of 30 C, 40C, and 50 C are placed on the vertices of a right triangle as shown. Find the net force on the charge at the right angle. 30 C 450 50 C 1.5 m θ FR F1 F2 40C Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Electric Fields The region around a charge in which there is an electric force experienced by a neutral object or a charged object is the electric field. Electric field is a vector quantity. (Gravitational field produced by a mass is also a vector quantity) Examples for scalar fields are heat and sound fields. The direction of an electric field, produced by a charged object, is the direction of the electric force on a positive test charge. This direction representation is a convention used by all scientists. + + Producing charge + Electric fields are drawn away from positive charges Producing charge - Test charge Electric fields are drawn toward negative charges. Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 E E ++ ++ http://www.mta.ca/faculty/science/physics/suren/FieldLines/FieldLines.html Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 To calculate the magnitude of the electric field intensity: ee Electric force per unit charge E = Electric Field Intensity (N/C) Fe = Electric Force (N -Newton) q = Test Charge (C - Coulombs) Rem: Gravitational field intensity is gravitational force per unit mass, g = Fg/m charge of 2.0 C is placed in an electric field and it experiences a force of 20 N. Find the electric field intensity at that point. Ex1: A E = Fe/q = 20 N / 2.0 x 10-6 C = 1.0 x 107 N/C charge of 5.0 C is placed in an electric field (3.0 x 104 N/C). Find the electric force. Ex2: A E = Fe/q 3.0 x 104 = Fe/ 5.0 Fe = 1.5 x 105 N F = Eq Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Electric Field intensity due to a point charge (the charge producing the field) r Q E = kQ/r2 E = Electric field intensity (N/C) Q = Charge producing the electric field (C - Coulombs) R = Distance from the charge to the point (m) Ex3: Find the electric field intensity at a point 20 cm from a charge of 5.0 C. E = kQ/r2 = 8.99 x 109 (5.0) / (0.2)2 E = 1.1 x 1012 N/C Ex4: Find the electric field intensity 10 cm from an alpha particle. E= kQ/r2=8.99x109x3.20x10-19/0.102 E = 2.9x10-7N/C Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 p. 548 #1, 2 p. 549 #1, 2 p. 550 #1, 2 p. 553 #1 E - - F Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Electric Field Intensity due to two Charged Plates Two parallel plates, one charged positively and the other negatively, creates a uniform electric field. +++++++++++ d E (uniform) --------------- E = V/d E = Electric Field Intensity (N/C or V/m) V = Potential Difference between the plates (V = Volts) d = distance between the plates (m) For charged plates use only these two formulas involving the electric field: E=V d E=F q Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Potential difference or Voltage is the amount of work done to move a unit charge from one point to another against the field. (If the charge is moved perpendicularly through the field no work is done – same as gravity). V=W q V = Potential Difference (V - Volts) W = Work done (J - Joules) q = Amount of charge (C-Coulombs) Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Work is also equal to the change in energy (energy used or released). Thus, V = ΔE q 1 V = 1 J/C Find the electric field intensity between two plates with a potential difference of 50 V and a distance of 5.0 cm between them. Ex: E = V/d = 50 V /5.0 x 10-2 m = 1000 V/m = 1.0 x 103 V/m Note: The electric field intensity between two parallel plates is uniform. The electric field intensity inside a conductor is zero. Charges usually reside on the outer surface of a metal sphere. Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 1 eV = 1.60 x 10-19 J (eV = electron volt) p. 540 #11, 12 p. 548 #1, 2 p. 549 #1, 2 p. 550 #1, 2 p. 553 #1 p. 565 #1, 2 p. 566 #1, 2 Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 p. 434 #5-8 p. 444 #10-13 p. 703 #1-5,8-12 Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Electric field Intensity Practice Example What is the magnitude and direction of the electric field intensity at a point 3.0 m to the right of a positive point charge of 5.4x10-4 C? E=kQ/R2=8.99x109x5.4x10-4/3.02 =5.4x105 N/C, right Example 50C -10C X X 0.45m Y 0.30 m Z What is the magnitude and direction of the electric field at a point Z in the diagram above, due to the charged spheres at points X and Y? E due to X=kQ/R2= 8.99x109 x50x10-6/0.752 =799111.1N/C, right(direction taken by a positive test charge) E due to the Y=kQ/R2= 8.99x109x10x10-6/0.302= 998888.888 N/C, left(direction taken by a positive test charge) Net electric field intensity=998888.88-799111.11 =2.0x105N/C, left Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Example Determine the magnitude and direction of the electric field at point Z in the diagram below, due to the charges at points X and Y. Z 4.0 cm X 3.0 cm 3.0cm Y Charge on X=2.0x10-8 C Charge on Y=2.0x10-8 C XZ=YZ=5.0 cm E due to X=kQ/R2= 8.99x109x2.0x10-8/.052= 71920 N/C E due to Y is the same. 0=53.1301 Ex(x)=71920 cos 53.1301=43152 N/C Ex(y)= 71920 sin 53.1301= 57535.998 N/C Ey(x)= -71920 cos 53.1301= -43152 N/C Ey(y)= 71920 sin 53.1301= 57535.998 N/C The x-components cancel each other. The sum of the y-components= 115071.996 N/C 1.2x105 N/C (up) Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 #14. What is the magnitude and direction of the electric field at Z due to X and Y? X 0.60m Y 0.30 m Z -5 -6 X=2.0x10 C Y=8.0x10 C Ex= 221975.30864 N/C to the right Ey= 799111.11111N/C, right E resultant= 1021086.419 N/C to the right =1.0 x106 N/C Unit 2: Electric Forces and Fields Chapter 2: Electric Field v 1.1 Physics 30 – Electric field intensity practice problems: 1. What is the electric field intensity 0.50m away from a small sphere with a charge of 1.6 x 10-8 C? 2. Calculate the electric field intensity midway between two negative charges of 3.2 x 10-9 C and 6.4 x 10-9 C that are 30 cm apart. 3. What is the magnitude and direction of the electric field at point Z due to charges at points X and Y. -5μC +1.5μC X Y Z 4.5 m 3.0 m 4. Calculate the electric field intensity at point Z. If an electron is placed at point Z, calculate the electron’s acceleration. A= 2.0 μC B= 1.2 μC C= 3.5 μC B 30 cm A 70 cm Z 30 cm C Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Electric Power and Electrical Energy Ex: The wire in a house circuit is rated at 15.0 A that has a resistance of 0.150 a) What is its power rating? b) How much heat does the wire give off in 10.0 mins? a) P = I2R P = (15.0)2(0.150) = 33.8 W b) E = Pt E = (33.8)(600) E = 2.03 x 104 J Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Ex: A heating coil has a resistance of 4.0 and operates on 120 V: a) What is the current in the coil while it is operating? b) What energy is supplied to the coil in 5.0 min? c) If the coil is immersed in an insulated container holding 20.0 kg of water, what will be the increase in temperature of the water in 5.0 minutes? “c” (specific heat) of water = 4190 J/(kg.oC) a) I = V/R = 120 V / 4.0 = 30 A b) E = VIt = (120 V)(30 A)(300 s) = 1.1 x 106 J c) Electrical Energy = Heat Energy E = mcT 1.1 x 106 = 20.0(4190)Δt Δt = 13 °C Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 *E (Heat gained or lost)= mct This is Science 10 formula Ex: An electric heater is rated at 500 W a) How much energy is delivered to the heater in half an hour? b) The heater is being used to heat a room containing 50.0 kg of air. If the specific heat of air is 1100 J/kg°C, what is the change in air temperature after half an hour? Only 50% of the energy heats the air. a) E = Pt E = 500 W(1800 s) = 9.00 x 105 J b) E = mct 9.00 x 105x(50%) = 50.0 kg(1100)Δt = 8.18 °C Cost of Electricity Calculated in terms of kW. h (or MJ). Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Ex: Find the cost of operating a 60 W bulb for 30 days at 8 hours a day if the cost of electricity is 9.0 cents/kW.h E = Pt Cost = kW.h x Price = (60 /1000 kW)(240 h) = 14.4 kW.h Cost= 14.4 kWh x 9.0 cents/kWh = $ 1.30 Ammeters and Voltmeters An ammeter is a device that measures currents. It is always connected in series. Voltmeter is a device, which measures potential difference. It is always connected in parallel. Safety devices in electric circuits Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 1. Grounding Wire – Appliances are connected to the ground, so that any extra electric current goes to the ground. 2. Fuses – Fuse is a metal that melts when a certain amount of current passes through it. If excess current passes through the metal, the fuse melts and the circuit becomes open. 3. Circuit Breakers – A breaker can carry a certain amount of current. If the current exceeds the limit, then the circuit breaker goes off. Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Millikan’s Oil Drop Experiment Millikan wanted to find the elementary charge. He believed that all charges are multiples of this elementary charge. http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapte r02/Text_Images/FG02_04.JPG http://www.britannica.com/nobel/cap/omillik001a4.html http://www68.pair.com/willisb/millikan/experiment.html http://www.juliantrubin.com/bigten/millikanoildrop.html Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Millikan’s Oil Drop Experiment Ex: An oil drop whose mass is found to be 4.95 x 10-15 kg is balanced between two large, horizontal parallel plates 1.0 cm apart, by a potential difference of 510 V, with the upper plate being positive. What is the charge on the drop, both in coulombs and electrons? Is it an excess or deficit of electrons? E = V/d Eq = mg = 510 V/(1.0 x 10-2 m) = 51000 V/m 1. Oil Drop Stationary Fe = Fg (Oil drop is stationary) Eq = mg Eq=mg 51000q=4.95x10-15x9.81 51000q=4.86x10-14 q=9.5x10-19 C=6 e(Excess) 2. Oil Drop Accelerating Upward Fe > Fg Fe – Fg = ma Eq – mg = ma 3. Oil Drop Accelerating Downward Fe < Fg Fg – Fe = ma mg – Eq = ma 4. Oil Drop Moving at a Constant Speed Fe – Fg = ma Fe – Fg = 0 Fe = Fg Eq = mg Ex: An oil drop whose mass is 5.70 x 10-16 kg accelerates upward at a rate of 2.90 m/s2 when placed between two horizontal parallel plates that are 3.50 cm apart. If the potential difference between the plates is 792 V, what is the magnitude of the charge on the oil drop? E = V/d = 792 V / (3.50 x 10-2 m) = 2.26 x 104 N/C Eq – mg = ma (2.26 x 104 J)q – (5.70 x 10-16 kg)(9.81 m/s2) = (5.70 x 10-16 kg)(2.90 m/s2) q = 3.20 x 10-19 C Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Potential Difference Potential Difference (PD) is defined as energy per unit charge – commonly known as Voltage. Formula: V = ∆E q Units: ∆E = change in energy in J q = charge in C J/C or Volt (V) 1 V = 1 J/C The first battery was produced by Volta in 1800. Current Electricity Electrical energy can be transferred over long distances and then can be changed to other forms of energy. Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 When two conducting spheres touch, charges flow from one sphere (high potential) to the other sphere (low potential) until the potentials are equal. Current – Electrons flow continuously from the negative to the positive in a closed circuit. This is known as the electron flow. The flow of positive charges from the positive to the negative was known as the conventional current. http://www.windows.ucar.edu/tour/link=/spaceweather/images/water_pump2_ lg_jpg_image.html Electric Circuits Parts: 1. Sources of Electricity - provides electric potential difference. The change in potential energy of the charge is converted to other energies. Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 a) Electrochemical Cells – Chemical potential energy is converted into electrical energy. i) Cells in Series The total potential difference of the battery is the sum of the potential differences of the cells. Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 ii) Cells in Parallel The potential difference of the battery is the same as the potential difference of one cell. b) Generators – Mechanical energy is converted into electrical energy c) Photoelectric Cells – solar (light) energy is converted into electricity Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 2. Resistors – resist the flow of electrons. Resistors reduce potential energy of charges flowing through it. Unit of resistance is the Ohm (Ω) Factors that determine the amount of resistance in a wire: 1. Thickness – As the thickness increases, resistance decreases. 2. Type of wire – Copper – low resistance. Nichrome – High resistance 3. Length of Wire – Longer the wire, greater the resistance. 4. Temperature – As the temperature increases, resistance increases. Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Resistors in series Resistors in parallel Current is the amount of charge transferred in a second. I=q t I = Current (A-Ampere) q = charge (C) t = time (s) 1 A is the electric current when 1 C of charge moves past a point in a conductor in 1 s. Ex: Find the number of electrons transferred in a circuit if a current of 2.0 A passes through a point in 2 min. I = q/t 2.0 A = q/120 s q = 240 C 240 C / 1.60 x 10-19 C/Electrons = 1.5 x 1021 Electrons Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Ohm’s Law states that the ratio between the potential difference and current is a constant. This constant is called the resistance. V/I =k=R R=V I V = Potential difference (V = Volts) I = Current (A = Amperes) R = Resistance (Ω = Ohms) (Ohm’s law) Electric power = VI measured in watts (W) P = VI P = power in W I = current in A V = voltage in V Proof: P = W t P = Vq t But W = ∆E P =VI and ∆E = Vq Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Ex: A circuit has a resistance of 20 Ω when the PD is 120 V. What is the current through the circuit? R = 20 Ω I=V V = 120 V R I=? = 120/20 I=6A p. 451 #1-4 p. 454 #5-9 p. 465 #1-6 (problems) Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Electrical Power P = VI From Ohm’s law V = IR and I = V R Thus, P = (IR) I P=V V R P = I2R P = V2 R kw·h measures energy. 1 kw·h = 3,600,000 J Not on formula sheet. Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 e.g. What is the heat produced in a 7 Ω resistor in 1.5 min if there is a current of 10 A? R=7Ω E = Pt t = 90 s I = 10 A P = I2R = (10)27 = 700 W E=? E = (700)(90) E = 63,000 J = 63 kJ Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Kirchoff’s Voltage Law Around any complete path through an electric circuit, the sum of the increases in electric potential is equal to the sum of the decreases in electric potential. Kirchoff’s Current law At any junction point in an electric circuit, the total electric current into the junction is equal to the total electric current out. Kirchoff’s Laws Series Parallel Req = R1 +R2+R3+...... V = V1 +V2+V3+....... I = I1 =I2=I3= ... (I is equal throughout) 1/Req = 1/R1 +1/R2+1/R3+...... V = V1 = V2=V3= .....(V is equal throughout) I = I1 +I2+I3+..... Simple circuits Example Find the equivalent resistance when a 4.0 bulb and an8.0 bulb are connected: (a)in series (b)in parallel. Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 (a)R =R1+R2=4.0+8.0=12 (b)1/R=1/R1+1/R2=1/4+1/8 R=2.7 Example Two 20- resistors are connected in parallel with a 50- resistor in series. Find the total resistance. 1/R=1/R1+1/R2=1/20+1/20 R=10 R=10+50=60 Example Find: R= I= I1= I2= V1= V2= V 3= Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 I3= Example 2 Find: V= V1= V2= V3= I= R= Example Find: V1= I1= R= Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 V2= I2= R2= Example Find: R= R3= V1= V3= I1= I 3= V2= Example V= V1= I= I2= I2= Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 R= R2= Example R1= Find: V= V2= I= I2= Example V 1= V3= I1= R= Find: R= I= Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 V1= V3= I2= V2= I1= I3= Example Find: R= V1= V3= V5= I2= I4= Example I= V2= V4= I1= I3= I5= Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Find: R= V1= V3= I2= Example I= V2= I1= I3= Unit 2: Electric Forces and Fields Chapter 4: Current Electricity v 1.1 Find: R= V1= V3= I2= I= V2= I1= I3 =