Current and Resistance. February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 1

Transcription

1 Current and Resistance February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 1

2 Helproom hours! Strosacker learning center, BPS 1248! Mo: 10am noon, 1pm 9pm! Tue: noon 6pm! We: noon 2pm! Th: 10am 1pm, 2pm 7pm! Fri: noon to 2pm February 11, 2014 Physics for Scientists & Engineers 2, Chapter 21 2

3 Electric Current! Total charge is conserved, which means that charge flowing in a conductor is never lost! The unit of current is Coulombs per second, which has been given the unit Ampere (abbreviated A), named after French physicist André-Marie Ampère, ( )! Some typical currents are February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 3

4 Current! Current is defined as the flow of positive charge! Current is i = dq dt = nev da! Current flowing through a cross section area A is i = J d A! And the current density is J = i A = nev d! A current that flows in only one direction, which does not change with time, is called direct current February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 4

5 Current through portion of a wire! The current density in a cylindrical wire of radius R = 2.0 mm is uniform across a cross section of the wire and has the value of J= A/m 2. PROBLEM What is the current i through the outer portion of the wire between radial distances R/2 and R? THINK The current density is uniform across the cross section, the current density, J, the current i and the cross section area A are related by J = i A However, we only want the current through the reduced cross sectional are A. February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 A 5

6 Current through portion of a wire The current density in a cylindrical wire of radius R = 2.0 mm is uniform across a cross section of the wire and has the value of J= A/m 2. PROBLEM What is the current i through the outer portion of the wire between radial distances R/2 and R? CALCULATE Cross sectional area A (outer portion) A = πr 2 π( R / 2) 2 = π 3R2 4 = m 2 Current through A A i = J A = ( A/m 2 )( m 2 )= 1.9 A February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 6

7 Resistivity and Resistance! Some materials conduct electricity better than others! If we apply a given voltage across a conductor, we get a large current! If we apply the same voltage across an insulator, we get very little current! The property of a material that describes its ability to conduct electric currents is called the resistivity, ρ! The property of a particular device or object that describes its ability to conduct electric currents is called the resistance, R! Resistivity is a property of the material! Resistance is a property of a particular object made from that material February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 9

8 Resistance! If we apply an electric potential difference ΔV across a conductor and measure the resulting current i in the conductor, we define the resistance R of that conductor as R = ΔV i! The unit of resistance is volt per ampere! In honor of Georg Simon Ohm ( ), resistance has been given the unit ohm, Ω 1 Ω = 1 V 1 A! Rearrange the equation to get Ohm s Law i = ΔV R or ΔV = ir February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 10

9 Resistivity! We will assume that the resistance of the device is uniform for all directions of the current; e.g., uniform metals! The resistance R of a device depends on the material from which the device is constructed as well as the geometry of the device! The conducting properties of a material are characterized in terms of its resistivity! We define the resistivity of a material by the ratio ρ = E J! The units of resistivity are [ ρ] = E J [ ] [ ] = V/m A/m = V m 2 A = Ω m February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 11

10 Typical Resistivities! The resistivities of some representative conductors at 20 C are listed below (more in Table 25.1)! Typical values for the resistivity of metals used in wires are on the order of 10-8 Ω m. February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 12

11 Resistance! Knowing the resistivity of the material, we can then calculate the resistance of a conductor given its geometry! Consider a homogeneous wire of length L and constant cross sectional area A, we can relate the electric field and potential difference as ΔV = E ds E = ΔV L! The magnitude of the current density is J = i A! Combining our definitions gives us ρ = E J = ΔV / L i / A = ΔV i A L = ir i A L = R A L R = ρ L A February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 13

12 Resistance of a Copper Wire! Standard wires that electricians put into residential housing have fairly low resistance PROBLEM! What is the resistance of a length of m of standard 12-gauge copper wire, typically used in household wiring for electrical outlets? SOLUTION! The American Wire Gauge (AWG) size convention specifies wire cross sectional area on a logarithmic scale! A lower gauge number corresponds to a thicker wire! Every reduction by 3 gauges doubles the cross-sectional area! See Table 25.2 February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 16

13 Resistance of a Copper Wire! The formula to convert from the AWG size to the wire diameter is d = (36 AWG )/39 mm! So a 12-gauge copper wire has a diameter of 2.05 mm! Its cross sectional area is then A = 1 4 πd 2 = 3.31 mm 2! Look up the resistivity of copper in Table 25.1 R = ρ L A = ( Ω m) m m 2 = Ω February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 17

14 Resistors! In many electronics applications one needs a range of resistances in various parts of the circuit! For this purpose one can use commercially available resistors! Resistors are commonly made from carbon, inside a plastic cover with two wires sticking out at the two ends for electrical connection! The value of the resistance is indicated by four color-bands on the plastic capsule! The first two bands are numbers for the mantissa, the third is a power of ten, and the fourth is a tolerance for the range of values February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 20

15 Resistor Color Codes Band 1 Digit 1 - Brown 1 Band 2 Digit 2 - Green 5 Band 3 Power of 10 - Brown 1 Band 4 Tolerance - Gold Ω = 150 Ω 5% tolerance Color Tolerance Brown 1% Red 2% Gold 5% Silver 10% None 20% February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 21

16 Temperature Dependence of Resistivity! The resistivity and resistance vary with temperature! For metals, this dependence on temperature is linear over a broad range of temperatures! An empirical relationship for the temperature dependence of the resistivity of metals is given by! where ( ) ρ ρ 0 = ρ 0 α T T 0 ρ is the resistivity at temperature T ρ 0 is the resistivity at temperature T 0 α is the temperature coefficient of electric resistivity February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 22

17 Temperature Dependence of Resistance! In everyday applications we are interested in the temperature dependence of the resistance of various devices! The resistance of a device depends on the length and the cross sectional area! These quantities depend on temperature! However, the temperature dependence of linear expansion is much smaller than the temperature dependence of resistivity of a particular conductor! The temperature dependence of the resistance of a conductor is, to a good approximation, ( ) R R 0 = R 0 α T T 0 February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 23

18 Temperature Dependence! Our equations for temperature dependence deal with temperature differences so that one can use C as well as K (T-T 0 in K is equal to T-T 0 in C)! Values of α for representative metals are shown below (more can be found in Table 25.1) February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 24

19 Example: Which Metal is it?! A metal wire of 2mm diameter and length 300m has a resistance of Ω at 20 o C and Ω at 150 o C. Find the values of α, R(0 o C), ρ(0 o C) and identify the metal!! Solution: R(150 o C)=2.415Ω=R(0 o C)(1+α(150-0)) R(20 o C)=1.6424Ω=R(0 o C)(1+α(20-0)) First equation: R(0 o C)=2.451Ω/(1+α(150-0)), substitute in second equation, solve for α: α= o C -1 With α known, use any of the above equations to get R(0 o C)=1.5236Ω. Then use R(0 o C)=ρ(0 o C)L/A Ω = ρ(0 C) 300m 0.25π( m) 2 ρ(0 C) = Ωm ρ(20 C) = ρ(0 C)(1+ α 20) = Ωm Table 25.1: Copper! February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 25