Current & ResistancePHY232 – Spring 2008Jon Pumplinhttp://www.pa.msu.edu/~pumplin/PHY232(Ppt courtesy of Remco Zegers)PHY232 - Current & Resistance2Electric current So far we have studied Static Electricity. Now consider the situation where charge can move and hence produce an electric current.A++++Current = amount of charge ∆Q that flows through an area Adivided by the time interval ∆t:PHY232 - Current & Resistance3Electric current IIA matter of convention: The direction of current is the direction in which positive charges flow, even though the flow is often of electrons (negative)+-High VLow VHigh VLow VRemember: positive charge moves from high potential to low potentialPHY232 - Current & Resistance4electric current III: what really happensWhen electrons move through a wire they undergo many collisions and a typical path looks like:Because of the collisions, the velocity is on average constantThe drift velocity of the electrons is actually very slow (less than 1 meter per hour). So why can we have high currents?High VLow VBecause there are so many electrons!!!demo: model of resistancePHY232 - Current & Resistance5electric current IVlet’s assume the average electron speed is vconsider one electron at point xafter time t it will have moved….a distance D = v tin fact all the electrons over the distance D will have movedthe volume of the cylinder V = A D = A v tif n: number of electrons per unit volume, the number of electrons moved is: nV=nAvtthe charge ∆Q that has been moved: nAvtqcurrent I = ∆Q/t = n A v q-PHY232 - Current & Resistance6questionA current of 1 A is running through a Copper wire with cross section 1mm2. Each Copper atom produces 1 free electron. c) How many free charge carriers per unit volume are there? (Given that the molar mass of Cu is 63.5 g and the density of copper is 8.92 g/cm2). b) What is the drift velocity?a) The volume taken by 1 mol of Cu atoms  the number of electrons is also 1 mol in this volume (=NA=6.02x1023) so:PHY232 - Current & Resistance7question…b) Use sowith: n=8.46x1028 m-3, A=1mm2 = 1x10-6m2 q=1.6x10-19 C and I=1A=1C/sso v=2.46x10-5 m/sI.e. this is 0.089 m in one hour.PHY232 - Current & Resistance8wait a second…Wasn’t charge supposed to be collected on the surface of a conductor?That only happens when the conductor has a Net Charge(more electrons than protons or fewer electrons than protons).The conducting wires we are talking about are neutral.PHY232 - Current & Resistance9batteriesA battery can produce a potential difference between the anode (negative) and cathode (positive). When connected(I.e. using a wire or via a device) currentcan flow.The charge is created through chemicalreactions. Once the chemical fuel isused, the battery is emptycommonly used are zinc-carbon batteries.PHY232 - Current & Resistance10A simple circuitPower sources can be DC (Direct Current) or AC (AlternatingCurrent). We will deal with DC circuits first.OFFONA basic electric circuit consists of apower source (e.g. a battery) in which the + and – side are connected via a wire and some device.As long as the circuit is open, nocurrent will flow and hence thedevice not work.PHY232 - Current & Resistance11questionWhich of the following lights will not shine after the switches are closed?1234a) 2b) 2,3c) 2,3,4d) 1,2,3,4lights 2 and 3 will not shine since there isno potential difference over the contactsPHY232 - Current & Resistance12how to measure current?The current anywhere between A and B must be constant, else electrons would accumulate at a certain point in the lineA device to measure current in the light should therefore be placed in line (in series) with the light. Either side!!The device is called an Ampere meter (ammeter)1AB1ABPHY232 - Current & Resistance13how to measure voltage?To measure the voltage to the light, realize that we need to measure the potential difference between A and BA device to measure voltage to the light should therefore be placed in parallel with the lightThe device is called a Volt meter1A B1A BPHY232 - Current & Resistance14Resistance IWhen electrons move through a material, they undergo many collisions which hinders the motion (like friction).Without such collisions, the electrons would accelerate (since there is a force acting on them)The resistive force counterbalances the electric force so the drift velocity is constant When the resistive force is high, the current will go down if the voltage difference that drives the motion remains the same.High VLow VPHY232 - Current & Resistance15Resistance IICompare with water flow through a pipe. If the pipe becomes narrow, flow is reduced. If the length over whichthe pipe is narrow becomes longer, flow is further reduced.so resistance R :high pressurelow pressureflowPHY232 - Current & Resistance16Resistance IIIvoltage is the “equivalent” of pressure and current the equivalent of flowIf pressure (voltage) difference increases, the flow (current) will increaseIf the resistance increases, the flow current will go down if the pressure difference remains the samehigh pressurelow pressureflow+ -VIPHY232 - Current & Resistance17Ohm’s law and resistivityOhm’s lawFor a specific material, the resistance R can be calculated using:where R: resistance (in V/A=Ω (Ohm)), ρ the resistivity (material dependent in Ωm), l the length of the object and A the cross section of the objectPHY232 - Current & Resistance18Ohm’s lawOhm’s law implies that I is proportional to V, which is true for many materials but not for all:Ohmic resistanceNon-ohmic resistancePHY232 - Current & Resistance19questionA voltage of 100V is put over a thick wire of unknown material. The current is measured is 4.5x103 A. The cross section of the wire is 1cm2 and the length is 10m. What material is the cable made of?75x1016Quartz640Silicon22x10-8Lead2.44x10-8Gold1.59x10-8SilverResistivity (Ohm.m)MaterialR = V/I = 0.022 = ρl/Aso: ρ=0.022A/lA=1cm2=0.0001m2l=10 mρ=2.2x10-7 Ohm.mLeadPHY232 - Current & Resistance20a resistor bank..is an adjustable resistorlong wireadjust length of wireVdemoPHY232 - Current & Resistance21questionA person measures the resistance over a 10 m long cable through a measurement of V and I. He finds at V=10 V that I=1 A. A second cable made of the same material and length but with