1 Handout 5: Current and resistance Electric current and current density Figure 1 shows a flow of positive charge. Electric current is caused by the flow of electric charge and is defined to be equal to the rate of change of the electric charge: I = dq. The unit of current I is Cs -1 or ampere (A). The direction of current follows the direction of flow of positive charge but is opposite to the direction of flow of negative charge (Figure 2). Figure 1: Flow of electric charge through a surface area In Figure 1, the current passes through an area A. The current density J is a vector quantity in the direction of charge flow. The magnitude J is equal to the current per unit (perpendicular) area. The electric I can be obtained by integration over the surface: I = J da. If the current is uniform and perpendicular to the crosssection area, I = J da = J da = JA J = I A. Figure 2: Directions of electric currents The unit of current density is Am -2. Drift velocity Consider motion of free electron in metal under electric field in Figure 3. The electron moves along zigzag path because it is subject to many collisions with atoms and other electrons. However, the electron tend to drift with a drift velocity v d in the direction opposite to electric field. Figure 3: Drift velocity Assume that the number density (number per unit volume) of free electron is n. Consider a section of the metal in Figure 3 whose length is x. The number of electron in this section is nax. Each electron has charge e. Therefore the amount of charge in this section is
Q = enax. Differentiation gives us an expression for electric current 2 I = dq = ena dx = enav d. Example 1 Electric charges q coulombs flow through a substance. At time t seconds, the charge is given by q t = 3t 2 + t + 3. a) Calculate the electric charge at t = 1.3 s. b) Calculate the electric current at t = 1.3 s. Example 2 Lightning strikes the ground and spread over a hemisphere under the ground with current I = 5.2 10 4 A. Determine the current density at distance r = 1.5 m from the strike point. *Example 3 Estimate the average drift speed v d of the electrons in a copper wire of radius r = 2 mm carrying a current I = 25 A. Assume that each copper atom contributes one electron to the drift current, that the relative atomic mass of copper is m A = 63.5 and that its density is ρ = 8.9 10 3 kgm -3.
3 Resistivity and conductivity In Figure 3, free electron in metals experiences many collisions with atoms under the applied electric field. These collisions lead to energy lost and hence resistance of the material. Consider a conducting wire in Figure 4. Given the length L and cross-section area A, the resistance R is given by Figure 4: A conducting wire R = ρl A, where ρ is the resistivity which is a property of the material. Unit of resistance is ohm (Ω). The reciprocal of the resistivity is called conductivity ς = 1 ρ. At higher temperature, atoms vibrate more violently and thus further impede motion of free electrons. As a result, the rate of collisions of the electrons is higher and the resistivity increases. If the temperature change T is not too high, the change in resistivity ρ is given by ρ = αρ 0 T, where ρ 0 is the initial value of the resistivity and α is the temperature coefficient of resistivity for the material. Figure 5 shows numerical values of ρ and α for some materials. Figure 5: Resistivity (at 20 C) and its temperature coefficient Example 4 Calculate the resistance of a copper rod of length 30 cm and radius 0.5 cm. The resistivity of copper is 1.7 10 8 Ωm. *Example 5 A hollow conducting cylinder of length L has inner and outer radii a and b respectively and is made from a material with resistivity ρ. The current flows from the inner curved surface to the outer in the radial direction. Show that the resistance is given by ρ R = 2πL ln b a.
4 Ohm s law When a potential difference V is applied to a device with resistance R, the current I passes through it. Ohm s law states that V = IR. The graph of current against voltage is a straight line as shown in Figure 6. The slope of the graph is 1 R (constant). However, some materials are not ohmic and this can be seen in the I-V characteristics which is not a straight line. Figure 7 shows the variation of current and voltage of a semiconducting diode. In Figure 8, the rod is given a voltage V between both ends, causing the electric field E = V L inside the rod. Substituting V = EL and R = ρl A into V = IR, we can rewrite Ohm s law as Figure 6: I-V characteristic of ohmic device Figure 7: I-V characteristic of a semiconducting diode EL = I ρl A = ρl I A E = ρj, where J = I A is the current density. In terms of conductivity, Ohm s law becomes J = ςe. The microscopic picture is that drifting free electron (charge e, mass m) experiences an electric force F = ee and hence having the acceleration a = ee m. The speed of the electron increases until it collides with atom and loses the speed. Let the time between collisions be τ, the electron will acquire a drift speed of v d = aτ = eeτ m. From the relation I = enav d or J = env d, one obtain Figure 8: Electric field inside a rod as a result of applied voltage J = en eeτ m = ne 2 τ m E. Comparing with J = ςe, we have the formula for the conductivity ς = ne2 τ m. For Ohm s law, ς has to be constant. Hence, we need to assume that τ is constant, independent of applied field.
Example 6 A Nichrome wire consists of a nickelchromium-iron alloy, is commonly used in heating elements, and has conductivity 2.0 10 6 Ω 1 m 1. The wire with cross-section area of 2.3 mm 2 carries current of 5.5 A when a 1.4 V voltage is applied between its ends. What is the wire s length? 5 Power In Figure 9, a device (lamp) is connected to a voltage source (battery). The current flows in the circuit. Consider a small charge dq moving through a constant potential difference V. The electric potential energy decreases by du = VdQ. Therefore, the rate of decrease in the energy, power, is P = du = V dq P = VI. Figure 9: Battery connected to a lamp One can use Ohm s law, V = IR, to show that power can be expressed as P = I 2 R and P = V 2 R. Example 7 A 850 W radiant heater is operated at 115 V. a) What is the current? b) What is the resistance in the heating coil? c) How much thermal energy is produced in 5 hr? Example 8 A copper wire, having cross-section area 2.4 10 6 m 2 and length 4.0 m, has a current of 2.0 A uniformly distributed across that area. a) Find the magnitude of electric field along the wire. b) How much electrical energy is transferred in 30 minutes?