Electricity and Magnetism 30 Chapter 3 CURRENT AND RESISTANCE, DIRECT CURRENT CIRCUITS 3.1 Electric Current 1) The electric current in a conductor is defined by dq i = dt [A] (3.1) here dq is the amount of (positive) charge that passes in time dt through a hypothetical surface that cuts across the conductor. By convention, the direction of electric current is taken as the direction in which positive charge carriers would move. The SI unit of electric current is ampere (A) : 1A = 1C/s. 2) The current i (a scalar) is related to the current density J r (a vector) by i = JdA ∫ uuur r (3.2) where dA uuur is a vector perpendicular to a surface element of area dA and the integral is taken over any surface cutting across the conductor. J r has the same direction as the velocity of the moving charges if they are positive and the opposite direction if they are negative. 3.2 A Model for Electrical Conduction When a conductor does not have a current through it, its conduction electrons move randomly, with no net motion in any direction. When the conductor has a current through it, these electrons still move randomly, but now they tend to drift with a drift speed v d in the direction opposite that of the applied electric field that causes the current. The drift speed is tiny compared with the speeds in the random motion. For example, in the copper conductors of house-hold wiring, electron drift speed are perhaps 10 -5 or 10 -4 m/s, where as the random-motion speeds are around 10 6 m/s. (a) (b) (c) Fig. 3.1 : Random motion of an electron from A to F (the electron collides with an atom at B, C, D, E) a: without electric field. b: in presence of an electric field E, the electron drifts rightward. c: superposition of figure a and figure b. Electricity and Magnetism 31 Consider a wire of length L, cross-sectional area A, number of carriers (free electrons) per unit volume n. The total charge of the wire is q = -(nAL)e [C] (3.3) (e = 1.602 x 10 -19 C). Since the free electrons drift along the wire with speed v d (in the direction opposite that of the current i), the total charge q moves through any cross section of the wire in the time interval t = d L v [s] (3.4) and the current i, which is the time rate of transfer of charge across a cross section, is given by i = q t = -neAv d (3.5) The current density J r (current per unit sectional area) is given by J r = -ne d v v [A/m 2 ] (3.6) Note that the minus sign in (3.5) and (3.6) implies that the direction of the current i is opposite to that of the drift of the free electrons in the wire. Example : Consider a copper wire which carries a current i = 17mA Let r = 900 µ m be the radius of the wire. Assume that each copper atom contributes one conduction electron to the current and that the current density is uniform across the wire cross section. The drift speed of the conduction electrons can be determined from (3.5) v d = i - neA = J - ne [m/s] Since each copper atom contributes one conduction electron to the current, the number n of conduction electrons per unit volume is the same as the number of atoms per unit volume n = number of atoms per unit volume = (number of atoms per mole)x(number of moles per unit mass)x(mass per unit volume) number of atoms per mole = Avogadro’s number = N A = 6.02x10 23 number of moles per unit mass = inverse of the mass per mole of copper M = 63.54 g/mol = 63.54x10 -3 kg/mol mass per unit volume = mass density of copper ρ mass = 8.96 g/cm 3 = 8.96x10 3 kg/m 3 n = N A ρ mass /M = 6.02x10 23 x8.96x10 3 /63.54x10 -3 = 0.8489x10 29 electrons/m 3 The current density : J = 17x10 -3 /( π r 2 ) A/m 2 The charge of an electron : e = 1.602 x 10 -19 C Electricity and Magnetism 32 ⇒ v d = J - ne = -4.9x10 -7 m/s 3.3 Resistance and Ohm’s Law Ohm’s law i V R = [ Ω ] (3.7) where V is the potential difference across the conductor and i is the current. Fig. 3.2 Resistivity ρ and conductivity σ of a material J E1 = σ =ρ [ Ω m] (3.8) Vector form JE r r ρ= (3.9) The resistance of a conducting wire of length L and uniform cross section is A L R ρ = [ Ω m] (3.10) where A is the cross-sectional area. Change of ρ with temperature : for many materials, including metals, the relation between ρ and temperature T is approximated by ρ = ρ o [1 + α (T-T o )] (3.11) where ρ o is the resistivity at temperature T o , α is the temperature coefficient of resistivity for the material. Resistivity of a metal τ =ρ n e m 2 (3.12) here n is the number of free electrons per unit volume and τ is the mean time between collisions of an electron with the atoms of the metal. 3.4 Electrical Energy and Power Rate of electrical energy transfer P = Vi [W] (3.13) Resistive dissipation P = Ri 2 [W] (3.14) Electricity and Magnetism 33 In a resistor, electric potential energy is converted to internal thermal energy via collisions between charge carriers and atoms. 3.5 Electromotive Force (EMF) The electromotive force of a device is the work the device does to force a unit positive charge from the negative to the positive terminal dq dW e = [V] (3.15) 3.6 Kirchoff’s Rules Loop rule: The algebraic sum of the changes in potential encountered in a complete traversal of any loop of a circuit must be zero. Junction rule The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction. Single loop circuits (Fig.3.3) : R r + ε = i [A] (3.16) Fig. 3.3 Fig. 3.4 According to the Loop Rule the potential difference caused by the battery ( ) must be compensated for by the potential drops across the two resistors (r and R) in Fig. 3.3. Notice that the potential (V) starts at V a and then returns again to V a after resistor R (Fig. 3.4). Power P = Vi (3.17) P R1 = R 1 i 2 (3.18) 3.7 Resistors in Series and in Parallel Series resistances R eq = Σ R i (3.19) Parallel resistances Electricity and Magnetism 34 ∑ = ieq R 1 R 1 (3.20) 3.8 RC Circuits (Fig. 3.5) 1) Charging a capacitor e = Ri + V = R dt dq + C 1 q ⇒ dt dq + RC 1 q = R 1 e Let x = q - Ce ⇒ dt dx + RC 1 x = 0 ⇒ x dx = - RC dt ⇒ ln(x) = - RC t + const ⇒ x = A RC t e − ⇒ q = Ce + A RC t e − since q(0) = 0 ⇒ A = -Ce ⇒ q = Ce(1- RC t e − ) (3.21) i = R e RC t e − (3.22) Fig. 3.5 2) Discharging a capacitor 0 = Ri + V = R dt dq + C 1 q ⇒ dt dq + RC 1 q = 0 ⇒ q dq = - RC dt ⇒ q = A RC t e − since q(0) = Ce ⇒ A = Ce ⇒ q = Ce RC t e − (3.23) i = - R e RC t e − (3.24) The negative sign indicates that the current flows in the opposite direction. The quantity τ = RC is called the time constant. It dictates the rate of voltage build up on the capacitor, and the rate of current decrease. Problems Electric current 3.1) An isolated conducting sphere has a 10cm radius. One wire carries a current of 1.000.002 A into it. Another wire carries a current of 1.000.000 A out of it. How long would it take for the sphere to increase in potential by 1000 V ? Electricity and Magnetism 35 3.2) A lightning of current I = 100kA strikes the ground at point O (Fig. P3.1). The current spreads through the ground uniformly over a hemisphere centered on the strike point. The resitivity of the ground is ρ = 100 Ω m. Find the potential difference between A and B. The radial distance OA = 60m, OB = 62m Solution : J = 2 r2 I π ⇒ E = ρ J = 2 r2 I π ρ ⇒ V AB = - ∫ OB OA Edr Fig. P3.1 Fig. P3.2 3.3) Consider the circuit in Fig. P3.2 with e(t) = 12sin(120 π t) V, r = 10 Ω . Find the value of R such that the power in R is maximized ? Circuit 3.4) A 9.0 volt battery is connected across a light bulb (R = 3.0 ). How many electrons pass through the resistor in one minute? How many joules of energy are generated in one minute?" 3.5) A battery has an initial internal resistance of 0.75 and an emf of 9 V. It is placed a cross a 5 Ω resistor and a 10 µF capacitor hooked up in parallel. a) After a the capacitor has charged, what is the current through the resistor? b) What is the charge on the capacitor? c) If the battery is disconnected, how long will it take the capacitor to reach one-third of its initial voltage?" 3.6) The capacitor C in Fig. P3.3 is initially uncharged. At t = 0, the switch K is closed. Determine an expression for the potential difference V and the current i of the circuit. Fig. P3.3 Fig. P3.4 Fig. P3.5 3.7) In Fig. P3.4, e 1 = 12V, e 2 = 24V, r 1 = 10 Ω , r 2 = 5 Ω , R = 2 Ω . Determine i 1 , i 2 , i. 3.8) The circuit in Fig. P3.5 has E = 12V, R 1 = 10 Ω , R 2 = 30 Ω , r = 5 Ω . Find the currents i 1 , i 2 , i. 3.9) The circuit in Fig. P3.6 has e 1 = 12V, e 2 = 6V, e 3 = 9V, r 1 = 4 Ω , r 2 = 3 Ω , r 3 = 2 Ω . Find the currents i 1 , i 2 , i 3 . Electricity and Magnetism 36 3.10) The capacitor C in Fig. P3.7 is initially uncharged. At t = 0, the switch K is closed. Determine an expression for the potential difference V and the current i of the circuit. Fig. P3.6 Fig. P3.7 Homeworks 3 H3.1 The capacitor C in Fig. H3.1 is initially uncharged. At t = 0, the switch K is closed. Determine an expression for the potential difference V and the current i of the circuit (e in [V], r and R in [ Ω ], C in [ µ F]) Fig. H3.1 Fig. H3.2 n 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 e 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 r 100 150 200 250 300 100 150 200 250 300 100 150 200 250 300 100 R 300 450 600 750 900 300 450 600 750 900 300 450 600 750 900 300 C 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 n 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 e 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 r 100 150 200 250 300 100 150 200 250 300 100 150 200 250 300 100 R 300 450 600 750 900 300 450 600 750 900 300 450 600 750 900 300 C 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 n 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 e 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 r 100 150 200 250 300 100 150 200 250 300 100 150 200 250 300 100 R 300 450 600 750 900 300 450 600 750 900 300 450 600 750 900 300 C 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 n 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 e 40 42 44 46 48 50 52 54 56 58 60 40 42 44 46 48 r 100 150 200 250 300 100 150 200 250 300 100 150 200 250 300 100 R 300 450 600 750 900 300 450 600 750 900 300 450 600 750 900 300 C 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 25 Electricity and Magnetism 37 H3.2 Determine the currents i, i 1 , i 2 in Fig. H3.2. (e 1 and e 2 in [V], r 1 , r 2 and R in [ Ω ]) n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 e 1 10 15 20 25 30 35 40 45 10 15 20 25 30 35 40 45 e 2 20 25 30 35 40 45 50 20 25 30 35 40 45 50 20 25 r 1 5 6 8 10 12 15 16 18 5 6 8 10 12 15 16 18 r 2 3 4 5 6 7 8 9 3 4 5 6 7 8 9 3 4 R 8 9 10 11 12 13 8 9 10 11 12 13 8 9 10 11 n 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 e 1 10 15 20 25 30 35 40 45 10 15 20 25 30 35 40 45 e 2 20 25 30 35 40 45 50 20 25 30 35 40 45 50 20 25 r 1 3 4 5 6 7 8 9 3 4 5 6 7 8 9 3 4 r 2 5 6 8 10 12 15 16 18 5 6 8 10 12 15 16 18 R 7 8 9 10 11 12 13 14 7 8 9 10 11 12 13 14 n 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 e 1 10 15 20 25 30 35 40 45 10 15 20 25 30 35 40 45 e 2 20 25 30 35 40 45 50 20 25 30 35 40 45 50 20 25 r 1 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 7 r 2 3 4 5 6 7 8 9 3 4 5 6 7 8 9 3 4 R 6 7 8 9 10 11 12 13 6 7 8 9 10 11 12 13 n 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 e 1 10 15 20 25 30 35 40 45 10 15 20 25 30 35 40 45 e 2 20 25 30 35 40 45 50 20 25 30 35 40 45 50 20 25 r 1 6 7 8 9 4 5 6 7 8 9 4 5 6 7 8 9 r 2 5 6 7 8 9 3 4 5 6 7 8 9 3 4 7 8 R 8 9 10 11 12 13 6 7 8 9 10 11 12 13 10 11 . n 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 e 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 r 100 150 200 250 30 0 100 150 200 250 30 0 100 150 200 250 30 0 100 R 30 0 450 600 750 900 30 0. 9 10 11 12 13 14 7 8 9 10 11 12 13 14 n 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 e 1 10 15 20 25 30 35 40 45 10 15 20 25 30 35 40 45 e 2 20 25 30 35 40 45 50 20 25 30 35 40 45 50. 20 21 22 23 24 25 26 27 28 29 30 31 32 e 1 10 15 20 25 30 35 40 45 10 15 20 25 30 35 40 45 e 2 20 25 30 35 40 45 50 20 25 30 35 40 45 50 20 25 r 1 3 4 5 6 7 8 9 3 4 5 6 7 8 9 3 4 r 2