Reading from Young & Freedman: For this topic, read the introduction to chapter 25 and sections 25.1 to 25.3 & 25.6.

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1 PHY10 Electricity Topic 6 (Lectures 9 & 10) Electric Current and Resistance n this topic, we will cover: 1) Current in a conductor ) Resistivity 3) Resistance 4) Oh s Law 5) The Drude Model of conduction Reading fro Young & Freedan: For this topic, read the introduction to chapter 5 and sections 5.1 to 5.3 & 5.6. ntroduction The earlier part of this course has been concerned with electrostatics, that is electric charges at rest. We will now exaine charges in otion. We will see that, in a conductor, electrons are always oving within the body of the aterial. However, this rando otion does not result in the transport of charge fro one point to another, and so does not constitute an electric current. When an electric field resulting fro soe potential difference acts on free charges such as electrons, however, there is a net otion superiposed on any rando oveent, and an electric current flows. Electrons in Conductors conductor such as a etal consists of atos arranged in a crystal lattice each of which has one or ore electrons which are very weakly bound to the rest of the ato. They are therefore free to ove around within the body of the conductor. We have already seen that if a steady electric field is applied, they will ove in such a way as to build up a charge density which produces an electric field neutralising the applied external field. n fact, even in the absence of an electric field, the electrons are in constant otion. This is partly due to the theral energy they possess as a consequence of their not being at absolute zero, and partly due to quantu echanical effects that you will eet in your second year courses. s a result, electrons typically have velocities of the order of 10 6 s 1. They are also continually colliding with the stationary ions in the crystal lattice, resulting in scattering of the electrons into different directions. Since their velocities are essentially rando there is no net flow of charge through the aterial, and no current flows. When an electric field is applied to the conductor, the charged electrons experience a force F = q E. s we discussed previously, a copletely free charge, such as an electron, would then be accelerated with a constant acceleration by this force. n a etal, however, the electrons are only accelerated for a short tie before they collide with the ions in the lattice, and have their otion randoised again. The consequence is that the electrons have a sall drift velocity superiposed on their rando otion. lthough this drift velocity is very uch less than the rando otion (perhaps 10 4 s 1 copared with 10 6 s 1 ) because the drift velocity of all the electrons is in the sae direction it constitutes a net transfer of charge in one direction, and this is an electric current. n fact, electric conduction is usually caused by the oveent of negative electrons. The force on the, and hence their acceleration and drift velocity, is in the opposite direction to the electric field. Electric currents were studied long before the constituents of atos were understood, and the conventional direction for a current is that in which positive particles would ove, i.e. fro a positive charge or high potential towards a negative charge or lower potential. Thus the arrow indicating conventional current direction points in the opposite direction fro that taken by the drifting electrons. PHY10 Electricity 35

2 Electric Current and Circuits Current is the rate at which charge flows. n a wire, this could be the rate of charge flow past a given point, or in bulk aterial it could be the flow through a given surface. f in a tie t a charge Q flows, then the average electric current is given by and the instantaneous current is av Q =, t dq =. [1] dt The S.. unit of current is the père or p (). 1 = 1 Coulob per second. For a current to flow along a wire, there ust be a potential difference between its ends. This ay be provided by connecting it to a battery, which raises positive charges fro a low potential (at the negative terinal) to high potential (at the positive terinal). Note that current can only flow continuously through a closed loop or circuit, such as that fored by the wire and battery. s uch charge enters one end of the wire as leaves the other end. The wire does not gain any net charge. (The wire acts soewhat like a hose pipe, where as uch water enters one end as leaves the other.) Current Density The figure alongside shows particles of charge q oving with drift velocity v d along a wire. (Since the rando otion does not contribute to conduction, this is ignored.) f there are n charges per unit volue, the total charge within a cylinder of Q = n q. This charge length and cross sectional area is will take a tie t = /v d to pass through the end of the cylinder. The current, = Q/ t is therefore = nqv d. The average current density J is just the current flowing per unit area at right angles to the current, so J =. [] Note that current is a scalar but current density is a vector with direction given by the local direction of the drift velocity. We can therefore write f the carriers have negative charge, then J is opposite in direction to v d. J = nqv. [3] Resistance and Resistivity The current flowing through a coponent such as a wire depends on the applied potential difference between the terinals of the coponent. We define the resistance R of the coponent as R =. [4] The S.. unit of resistance is the oh (Ω), defined as one volt per ap. less frequently used paraeter is the conductance G which is the reciprocal of resistance, and its units are written as ho (= Ω 1 ) d 1 G = =. R s we will see, the resistance, which relates current to voltage, depends on the geoetry of the aterial through which the current flows. (We have already see how it depends on the cross sectional area.) We can instead consider the relationship between current density and electric field. q v d PHY10 Electricity 36

3 Equation [3] showed that current density is proportional to drift velocity, and for a given aterial at constant teperature, the drift velocity is norally proportional to the electric field. The relationship between J and E can be expressed as 1 J = σ E = E. [5] ρ The constant ρ is known as the resistivity of the ediu, with S.. units oh etre (Ω ), and σ = 1/ρ is the conductivity. The resistivity is a property of the aterial and does not depend on its shape. Consider the section of wire exained previously. f the potential difference across the length is, then E = /. Substituting this and J = / into [5], we arrive at 1 = ρ so R = ρ =. [6] n other words, the resistance of a saple is directly proportional to its length, and inversely proportional to its area. Teperature Coefficient of Resistivity The resistivity of a etal is due to collisions of the conduction electrons with ions in the lattice. s the teperature increases, the aplitude of vibration of the ions increases, and so does the probability of scattering the electrons. The resistivity of the aterial thus increases with teperature. Over odest teperature rises (of the order of 100 C or so), the change in resistivity is approxiately proportional to the teperature difference, and this change can be written as ( T ) 1 ( T T ) ρ ρ + α. [7] 0 0 Here ρ(t) is the resistivity at teperature T and ρ 0 is the resistivity at soe standard teperature T 0, often taken as 0 C or 0 C. α is known as the teperature coefficient of resistivity, and by rearranging [7] we can see this is just ρ ρ α =, [8] ρ 0 0 T i.e. the fractional change in resistivity per degree. ts units are C 1. Since the resistance of a saple is proportional to resistivity, we also have the result 1+ α( ) R T R T T. [9] 0 0 Electrical Power When a current flows as a result of an applied potential difference, the charge carriers give up potential energy. They do not, however, continue to build up their kinetic energy, as the scattering fro the lattice eans that they only acquire a sall drift velocity. nstead, the energy (supplied by the battery) is transferred to the lattice and norally appears as heat. The power P dissipated by the electric current is equal to the rate at which charges lose potential energy. f a charge dq passes through a circuit in tie dt, this is dq/dt. Using [1] this can be written P =. [10] q v d PHY10 Electricity 37

4 Equation [4] allows this to be rewritten in a nuber of useful fors: P = = R =. [11] R Oh s Law Oh s Law states that, for certain conductors under appropriate conditions, the current flowing through the conductor is directly proportional to the potential difference across it. This can be written as constant =. We have already seen, in equation [4], that / = R, so Oh s Law is really a stateent that the resistance of any aterials is not dependent on voltage or current. You should be aware, however, that any aterials do not obey Oh s Law. n a light bulb, for exaple, the filaent becoes very hot as a result of the electrical power dissipated in it. This raises its resistance (according to equation [9]), so voltage certainly does not reain proportional to current. The figures alongside show in (a) a linear - relationship for an Ohic device, such as a copper wire which reains at roo teperature, and in (b) the non-linear - relationship for a non-ohic device in this case a seiconductor diode which allows conduction in one direction uch ore easily than in the other. (a) Ohic conductor (b) Non-Ohic conductor The Drude Model of Conduction classical (non-quantu echanical) odel of electrical conduction was proposed by Paul Drude in s we have already discussed, electrons are oving randoly due to their theral energy. When an electric field E is applied, the electrons experience a force ee, so an acceleration a = ee/. Because of collisions with the ionic lattice, the velocity does not continue to increase. f the ean tie between collisions is τ, at which point the electron s velocity is copletely randoised, the electron achieves a drift velocity of ee v d = τ. [1] The constant τ is a property of the conducting aterial, but does not depend on the drift velocity (since this is so uch saller than the rando theral velocity). Using [3], J = nqv, we have ne τ 1 J = E = E. [13] ρ d so ρ = ne τ. [14] PHY10 Electricity 38

5 Putting What You Have Learnt nto Practice Question 6.1 Copper has a resistivity of wire of diaeter 1.5 and length 5 is connected across a potential difference of 50. Calculate: (a) the resistance of the wire, (b) the current, (c) the power dissipated in the wire. (a) The cross sectional area is (b) The current is (c) The power is = π r. Therefore 8 ρ l R = = = 0.4Ω. π 3 ( ) 50 = = = 07. R P = = = 10.4 kw R 0.4 Question 6. platinu resistance theroeter easures teperature by the change in the electrical resistance of a platinu wire. The coefficient of resistivity for platinu is C 1. t a teperature of 0.0 C, the theroeter has a resistance of 50.0; when iersed in a crucible containing elting indiu its resistance is What is the elting point of indiu? The expression for resistance as a function of teperature is 1+ α( ) R T R T T. 0 0 R R Hence T T0 = = = 137 C α R The elting point of indiu is = 157 C. Question 6.3 n electric fire has a heating eleent rated at 1 kw when operating at 30. (a) What is its resistance? (b) What will be the power dissipation if the ains voltage drops to 10, assuing that the eleent obeys Oh s Law? (a) 30 P = R 5.9 R = P = 1000 = Ω. (b) s we are given that the eleent obeys Oh s Law, its resistance reains constant. Since P =, power is proportional to the square of the voltage. The power at the reduced voltage is R therefore given by PHY10 Electricity 39

6 P 10 = P = = 834 W Question 6.4 Given that copper has a resistivity of and has free electrons per cubic etre, calculate the ean tie between collisions between a conduction electron and the ionic lattice according to the Drude odel. f the copper is exposed to an electric field of 0.5 1, what average drift velocity will the electron achieve? (The ass of an electron is kg.) Fro the lecture notes, we saw ρ = ne τ Therefore Drift velocity τ = = = 8 19 ne ρ v s 19 ee d s = τ = = (The negative sign siply eans that the electron drifts in a direction opposite to that of the electric field.). Probles fro Young & Freedan for Topic 6: Try to do exercises 5.1 to 5.7, 5.39 to 5.45, 5.54 to 5.60, 5.63 to 5.64 and The later probles are ore challenging. (Nuerical answers to odd-nubered questions are available at the back of the book.) PHY10 Electricity 40