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106 PHYS - CH17 - Part1 CHAPTER 17 DIRECT CURENTS 106 Phys Dr. Abdallah M. Azzeer 1 INTRODUCTION ¾Most applications of electricity and magnetism involve moving charges or electrical currents in conductors. ¾Direct currents (dc) are produced when a conducting path exists between the terminals of a battery or a dc generator. ¾These devices tend to maintain a constant potential difference between their terminals and convert other kinds of energy, such as chemical or mechanical energy, into electrical energy. 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 2 1 106 PHYS - CH17 - Part1 ¾An alternating current (ac) is produced by an ac generator, which has a terminal potential difference that alternates in sign at some characteristic frequency. 17.1 ELECTRIC CURRENT The electric current in a wire is the rate at which charge moves in the wire A l What causes the charges to move? ¨ applied electric field If ΔQ ≡ net charge crosses the shaded cross-sectional area in time Δt, The average current is 106 Phys I = ΔQ Δt 3 Dr. Abdallah M. Azzeer The instantaneous current is I = lim t →0 ΔQ dQ = Δt dt The S.l. current unit is the ampere (A). 1 A =1 C/s milliampere (mA); 1 mA = 10-3 A. Example 17.1 I=0.5 A, t=1 hr An electrochemical cell consists of two silver electrodes placed in an aqueous solution of silver nitrate. A constant 0.5-A current is passed through the cell for one hour. (a) Find the total charge transported through the cell in coulombs and in multiples of the electronic charge. (b) Each electron reaching the cell discharges one positively charged silver ion, which is then deposited on the negative electrode (cathode). What is the total mass of the deposited silver? (The atomic mass of silver is 107.9 u.) (1 u = 1.66×10-27 kg) 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer - + Ag+ AgNO3 Electrochemical cell 4 2 106 PHYS - CH17 - Part1 (a) Since the current is constant, ΔQ = I Δt =(0.5)(1 hr) =(0.5)(3600 s)=1800 C The number of silver ions transported through the cell and deposited in I hour. N = ΔQ/e = 1.13×1022 ( where e is the electron charge) (b) The mass of the deposited silver is the number of atoms N times the mass of an atom. m = (1.13 x 1022)(107.9 )(1.66 x 10-27 ) = 2.02 X 10-3 kg Conventionally the current in a conductor is assumed to be in the direction of motion of positive charges Metallic conductors ¨ the moving charges are electrons. 106 Phys 5 Dr. Abdallah M. Azzeer In metals, some of the electrons become detached, leaving behind positively charged ions. The heavy ions form a regularly spaced crystalline lattice and vibrate about their equilibrium positions with an energy and an amplitude that increases with the temperature. The detached conduction electrons move randomly among the ions. In the absence of an applied electric field, the average charge flow in any direction is zero. When an electric field is applied, the electrons acquire an average drift velocity opposite to the field, and there is a net current. Relation of the current in a wire to the density of conduction electrons and their drift velocity v. n = number of electrons per unit volume, The total number of electrons in a volume V is n V. A volume V = Al . v Total # of electrons in the wire = nAl Total # of charges in the wire = enAl 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer l 6 3 106 PHYS - CH17 - Part1 The time needed for all of them to pass through the end of the segment shown is; Δt = l / v so the magnitude of the current is; I = ΔQ enAl = = enAv l /v Δt Note that a positive charge flow in one direction is equivalent to a negative flow in the opposite direction. Example 17.2 A copper wire is often used to wire household electrical outlets. Its radius is 1 mm = 10-3 m. If it carries a current of 10 A, what is the drift velocity of the electrons? (Metallic copper has one conduction electron per atom, the atomic mass of copper is 64 u, and the density of copper is 8900 kg m-3.) I , A=π r 2 enA n equals the number of atoms per unit volume. The number of atoms per unit volume × the mass of one atom M = the mass of a unit volume of copper, which is its density d. v = 106 Phys Dr. Abdallah M. Azzeer 7 i.e d = nM , M = mass of one atom n = d/M = 8.38×1028 m-3 ∴ v = I / neA= 2.37×10-4 m/s 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 8 4 106 PHYS - CH17 - Part1 17.2 RESISTANCE The electrical resistance R of a conductor is the potential difference V between its ends divided by the current R= V I The S.l. unit for resistance is the ohm (Ω ) Ohm is a volt per ampere. 106 Phys Dr. Abdallah M. Azzeer 9 For many materials V ∝ I R = constant for a length of copper wire with its ends at an adjustable potential difference V. The current increases linearly with V and reverses direction when V is reversed Materials with a constant resistance are said to obey Ohm's law and are called ohmic conductors. Some conductors R varies with the magnitude or direction of the applied potential difference ¨ nonohmic conductors e.g vacuum tubes and transistors Example 17.3 (READ) 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 10 5 106 PHYS - CH17 - Part1 11 Dr. Abdallah M. Azzeer 106 Phys Resistivity The resistance of a conductor depends on; its (i) geometry (length and cross sectional area) (ii) properties of the material (resistivity) Put two identical wires side by side, the current doubles and hence the resistance is halved ¨ A R ∝ 1/A l If we halve the length l, the potential change and hence the resistance are also halved ¨ R∝l R= ρl A ρ≡ Resistivity, depends only on the properties of the material. The S.I. unit for resistivity is the ohm meter (Ω m). 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 12 6 106 PHYS - CH17 - Part1 1 ρ ≡ σ ≡ conductivity The S.I. unit of conductivity is the ohm-1m-1 . Dr. Abdallah M. Azzeer 106 Phys 13 Example 17.4 Find the room temperature resistance of a copper wire 100 m long with a radius of 1 mm. According to Table 17.1, the resistivity of copper at room temperature (20° C) is 1.72 × 10-8 Ω m. R= 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د ρl A = ρl = 0.547 Ω πr 2 Dr. Abdallah M. Azzeer 14 7 106 PHYS - CH17 - Part1 MORE example: (a) Calculate the resistance per unit length of a nichrome wire with radius 0.321 mm. A = π r2 = π (0.321× 10−3 m )2 = 3.24 × 10−7 m 2 R = (ρ) Ω l R ρ 1.5 × 10−6 Ω⋅ m ⇒ = = = 4.6 A l A 3.24 × 10−7 m 2 m (b) If a potential difference of 10 V is applied across a 1 meter length of this wire, find the current in the wire. Ω ⎞ ⎛ ⋅ ( 1 m ) = 4.60 Ω R = ⎜ 4.60 m ⎟⎠ ⎝ V 10.0 V = = 2.20 A I = R 4.60 Ω 106 Phys 15 Dr. Abdallah M. Azzeer 17.3 ENERGY SOURCES IN CIRCUITS The source that maintains the current in a closed circuit is called a source of electromotive force (emf) Any devices that increase the potential energy of charges circulating in circuits Water flow as analogy are sources of emf for electric current The emf may originate from a chemical reaction as in a battery or from mechanical motion such as in a generator. The EMF E of a battery or generator is defined as the work done per unit charge by the nonelectrical forces. S.l. unit of E is a J/C or volt, which is also the unit of the electric potential. 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 16 8 106 PHYS - CH17 - Part1 The Electric Battery Volta discovered that electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte. This is a simple electric cell. A battery transforms chemical energy into electrical energy. Chemical reactions within the cell create a potential difference between the terminals by slowly dissolving them. This potential difference can be maintained even if a current is kept flowing, until one or the other terminal is completely dissolved. Several cells connected together make a battery, although now we refer to a single cell as a battery as well. 106 Phys Dr. Abdallah M. Azzeer 17 Electric Current By convention, current is defined as flowing from + to -. Electrons actually flow in the opposite direction, but not all currents consist of electrons. 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 18 9 106 PHYS - CH17 - Part1 Potential in a resistor loop V c I b V0 V0 R a d a c b d a As (positive) charge moves from c to d in the resistor it loses electrical potential by an amount ΔV because Vb = Vc and Va = Vd. What goes up, must come down! voltage rises and drops loses must be equal around 12 J/C any closed loop. ε = 12 J/C 19 Dr. Abdallah M. Azzeer 106 Phys From a to b: V increases by E(ΔV = E). From b to c: V is constant (ΔV = 0); R=0. From c to d: V decreases by IR (ΔV = -IR). From d to a: V is constant (ΔV = 0). E When a charge goes around a closed path and returns to the starting point, its potential energy must return to its original value, since the conservative electrical forces do no net work. ΣΔV = 0 ¨ E - IR =0 E = IR 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 20 10 106 PHYS - CH17 - Part1 Circuits (Conceptual) 106 Phys Dr. Abdallah M. Azzeer 21 Example 17.5 AA dry dry cell cell with with an an EMF EMF of of 66 VV isis connected connected to to aa light light bulb bulb with with resistance resistance 44 ohms. ohms. Find Find the the current. current. READ Voltage sources Batteries (chemical reactions) Solar cells (conversion of solar energy to voltage output) Generators (conversion of mechanical energy to voltage output) Real batteries and generators usually have various dissipative effects associated with them that can be thought of as internal resistances This resistance behaves as though it were in series with the emf. 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 22 11 106 PHYS - CH17 - Part1 Real emf’s often have energy losses during operation so that the ideal voltage E is reduced. This energy loss can be modeled by an internal resistance r represented by the equivalent circuit: In this case the current is then determined by E = (r + R)I I= E r+R 106 Phys 23 Dr. Abdallah M. Azzeer Example 17.6 A battery with an EMF of 6 V and an internal resistance r = 2 Ω is connected to an R = 4 Ω light bulb (Fig. 17.7). Find (a)the current; (b)the terminal potential difference of the battery. b E= 6 V, r =2 Ω , R = 4 Ω (a)I = ? , (b) ΔV = ? E + r a R E = IR+Ir ¨ I = E/(R+r) = 1 A Vab = E – Ir = 4 V Vab = IR = 4 V 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 24 12 106 PHYS - CH17 - Part1 Example 17.7 Two Two batteries batteries are are connected connected to to aa resistor resistor as as shown shown in in Fig. Fig. 17.8. 17.8. Find Find (a) (a) the the current; current; (b) (b) the the terminal terminal potential potential difference difference of of each each battery. battery. (a) Since the batteries are connected so that their polarities are opposite, the direction of the current is determined by the larger EMF and is counterclockwise in Fig. 17.8. c E1=6 V + r1=2 Ω d R=3 Ω The potential increases by E2, and in the other EMF the potential decreases by E1. I b E2=18 V + r2=1 Ω a E E22 –– E E11 == I(r I(r11+r +r22+R) +R) ¨ ¨ II == E E22 –– E E11 // (r (r11+r +r22+R) +R) =2 =2 A A 106 Phys Dr. Abdallah M. Azzeer 25 (b) Vab = E2 - Ir2 = 18 V - (2 A)(1 Ω) = 16 V Vdc = E1 + Ir1 = 6 V + (2 A)(2 Ω) = 10 V Note that the terminal potential difference Vdc is greater than the EMF in this case. In calculating the potential difference between points c and d, we are proceeding opposite to the current. Thus we see a potential rise in the resistance as weII as in the EMF. The terminal potential difference exceeds the EMF only when the current is driven "backward" through the battery by another larger EMF. 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 26 13 106 PHYS - CH17 - Part1 1 7.5 SERIES AND PARALLEL RESISTORS; KIRCHHOFF'S RULES There are ways in which resistors can be connected so that the circuits formed can be reduced to a single equivalent resistor Two rules, called Kirchhoff’s rules, can be used instead Statement of Kirchhoff’ Kirchhoff’s Rules • Junction Rule The sum of the currents entering any junction must equal the sum of the currents leaving that junction ¾ A statement of Conservation of Charge • Loop Rule The sum of the potential differences across all the elements around any closed circuit loop must be zero ¾ A statement of Conservation of Energy 27 Dr. Abdallah M. Azzeer 106 Phys Mathematical Statement of Kirchhoff’s Rules • Junction Rule: I 3 Σ Iin = Σ Iout I4 Iout I2 I1 - I2 - I3 - I4 + I5 = 0 I5 Iin I1 I1 = I2 + I3 • Loop Rule: ∑ ΔV =0 closed loop i.e the sum of resistance times the current for every part in the circuit = the sum of emf’s in the circuit 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 28 14 106 PHYS - CH17 - Part1 More about the Loop Rule •• Traveling Traveling around around the the loop loop from from aa to to bb •• In In (a), (a), the the resistor resistor is is traversed traversed in in the the direction direction of of the the current, current, the the potential potential across across the the resistor resistor is is –– IR IR •• In In (b), (b), the the resistor resistor is is traversed traversed in in the the direction opposite of the current, the direction opposite of the current, the potential potential across across the the resistor resistor is is is is ++ IR IR •• In In (c), (c), the the source source of of (electromotive (electromotive force) force) emf emf is is traversed traversed in in the the direction direction of of the the emf emf (from (from –– to to +), +), and and the the change change in in the the electric electric potential potential is is +ε +ε •• In In (d), (d), the the source source of of emf emf is is traversed traversed in in the the direction direction opposite opposite of of the the emf emf (from (from ++ to to -), -), and and the the change change in in the the electric electric potential potential is is -ε -ε Dr. Abdallah M. Azzeer 106 Phys 29 More about the Loop Rule ΔV = ΔVab = Vb −Va 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 30 15 106 PHYS - CH17 - Part1 Loop Rule, final E1 + I2 R2 I1 R1 Σ E = Σ IR R3 R5 I5 E2 + ¾When ¾When apply apply LOOP LOOP RULE, RULE, assume assume specific specific direction for the current , as shown direction for the current , as shown ¾Potential ¾Potential difference difference between between connection connection is is POSITIVE POSITIVE in in this this DIRECTION DIRECTION and and NEGATIVE NEGATIVE in in the the opposite opposite .. Also Also the the same same for for all all EMF. EMF. I3 I4 R4 I1R1+I2R2+I3R3-I4R4-I5R5= E1 – E2 106 Phys ﻋﺒﺪاﷲ ﻣﺤﻤﺪ اﻟﺰﻳﺮ.د Dr. Abdallah M. Azzeer 31 16