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HolisticTuition Home Form 5 Physics Next > The study of matter 1 End CashPlants Chapter 2: Electricity HolisticTuition Physics: Chapter 2 Home Objectives: (what you will learn) < Back Next > 2 End CashPlants 1) electric fields & charge flow 2) electric current & potential difference 3) series & parallel circuits 4) electromotive force & internal resistance 5) electrical energy & power HolisticTuition Electric Fields Home Electric field: region where a charged body experiences a force It is shown by a field pattern that are lines of forces. line of force = path of a test charge in the field < Back Next > direction = motion of a free positive charge electric field pattern + 3 – End CashPlants Positive point charge Negative point charge HolisticTuition Home Electric Fields Electric lines of force Between a positive and a negative point charge < Back Next > 4 Between two positive point charges End CashPlants HolisticTuition Home Electric Fields Electric field between two parallel metal plates that are oppositely charged. < Back Next > 5 End CashPlants Electric field between two opposite charges. HolisticTuition Electric Fields Home Experiments to show existence of electric fields. – + < Back Next > – + + F F – Ball coated with conductor hangs vertically in the centre because it is neutral. Ball oscillating between 2 plates, after it touches one side causing a force, F to repel the ball due to like charges. Positive ions Negative ions 6 + End CashPlants – Candle flame spreading sideways between 2 plates due to attraction between oppositely charged ions and metal plates. HolisticTuition Electric Fields Home Electric fields cause charges to move. Net movement of charges = electric current < Back Next > In the late 1700s scientists chose the direction of electric current to be the direction in which positive charges move in an electric field. They did not know that electrons and protons were the negative and positive charge particles, and that the electron moved much more easily. In a copper wire, the outer electrons of the copper atom move relative to the nucleus of the atom. + 7 End CashPlants Current, I electrons So, the charge carriers (electrons) move in the opposite direction to the current. - HolisticTuition Home Electric Charge Basic unit of electric charge = Coulomb (C) Charge of a proton or electron = ± 1.60 10-19 C A Coulomb of charge is a lot, at 6.25 x 1018 electrons – most objects have charges in the µC (10-6 C) range. < Back Next > Electric charge, Q = It units Q in Coulomb, I in Ampere, t in second C=As Electric current = Rate of flow of electric charge 8 End CashPlants I= Q t , t = time HolisticTuition Home Potential Difference Potential difference (V) between 2 points in an electric field = work done (W) in moving 1 coulomb of charge (Q) between the 2 points. W Work done V= = Q Charge < Back Next > Potential difference between 2 points A Moving 1 coulomb of charge Unit of potential difference: 9 End CashPlants Volt (V) = J C = J C-1 B HolisticTuition Electric Current Home Ohm’s Law The current (I) in a conductor is directly proportional to the potential difference (V) across the conductor if the temperature is constant. V I < Back Next > I = constant Ohmic conductor A conductor that obeys Ohm’s Law. 0 V Switch I A 10 End CashPlants Rheostat Conductor V Circuit used to find the relationship between current I and potential difference V for a conductor. HolisticTuition Electric Current Home Non-ohmic conductor A conductor that does not obey Ohm’s Law. Examples I I I < Back Next > 0 V Dilute sulphuric acid 0 V Filament lamp V 0 Junction diode A circuit element is non-ohmic if the graph of current versus voltage is nonlinear. 11 End CashPlants A filament lamp is a non-ohmic conductor since its resistivity, like most materials, varies with temperature. As the filament gets hot, the resistance increases quickly. HolisticTuition Home Resistance The resistance, R of a conductor is defined as the ratio of the potential difference V across the conductor to the current I in the conductor. V Resistance, R = I < Back Next > The unit of resistance is the ohm (Ω). I conductor I V 12 End CashPlants Potential difference, V = IR HolisticTuition Resistance Home Factors that affect the resistance of a conductor: a. length of wire, l b. cross-sectional area, A c. type of material with resistivity, p d. temperature, T Based on a constant temperature: < Back Next > pl Resistance, R = A R 13 End CashPlants R T/oC 0 Metal 0 T/oC Semi-conductor HolisticTuition Home Series Circuit When resistors are connected in series: a. Same current I is in all the resistors b. Potential difference, < Back V1 = IR1 Next > V2 = IR2 V3 = IR3 c. V = V1 + V2 + V3 d. Effective resistance, 14 End CashPlants R = R1 + R2 + R3 I R1 R2 R3 V1 V2 V V3 HolisticTuition Parallel Circuit Home When resistors are connected in parallel: a. Same potential differences across all resistors, V b. Current in the resistors, < Back Next > I1 = I2 = I3 = V R1 V R2 V R3 I R1 I1 R2 I2 R3 I3 c. I = I1 + I2 + I3 d. Effective resistance, 15 End CashPlants 1 1 1 1 + = + R R2 R3 R1 V HolisticTuition Home Electromotive Force Electromotive force (e.m.f.), E Work done to drive a unit charge (1 C) around circuit – where the unit is volt, V = J C-1 < Back Using a high resistance voltmeter Next > E = 1.5 V Potential difference V < e.m.f. E because work is done to drive a charge through a cell with internal resistance, r. E = V + Ir = I(R + r) 16 End CashPlants E r R+r =1+ = R R V r I V R V I HolisticTuition Home Electrical Energy The potential difference V across a conductor is the work done in moving a charge of 1 C across the conductor. The work done is transformed into heat which is dissipated from the conductor. < Back From volt, V = J C-1 = Next > Energy dissipated, E Charge, Q substitutions Energy dissipated, E = QV = IVt = I2Rt 17 End CashPlants V2t E= R Q = It V = IR I = V/R HolisticTuition Home Electrical Power Electrical power, P = < Back Next > Energy dissipated Time, t E = IVt = IV V = IR substitutions = I2R I = V/R V2 P= R Power rating of an electrical appliance is the power consumed by it when the stated voltage is applied. V2 Resistance of the appliance, R = P 1 unit of electrical energy consumed = 1 kW h = (1000 Js-1)(3600 s) = 3.6 x 106 J 18 End CashPlants Cost of electrical energy = units x cost per unit HolisticTuition Summary Home What you have learned: < Back 19 End CashPlants 1. Electric fields & charge flow 2. Electric current & potential difference 3. Series & parallel circuits 4. Electromotive force & internal resistance 5. Electrical energy & power Thank You