Note 5: Current and Resistance

Transcription

1 Note 5: Current and Resistance In conductors, a large number of conduction electrons carry electricity. If current flows, electrostatics does not apply anymore (it is a dynamic phenomenon) and there can be electric field in conductors (metals). Otherwise, current cannot be driven. The direction of current is defined in terms of motion of positive charge. In conductors, current flow is against the flow of electrons. In some metals, positive carriers (holes) exist. Electrons collide with lattice ions and lose momentum (origin of resistivity). current Electron drift

2 Current Density J (A/m ) 3 If the number density of electrons is (m ) and the average electron velocity is v (m/s), the current density is given by J = n e v = nev ( )( ) (A/m ) The electron density is a large number and is of 8 wave c = m/ 0 0 n order 10 / m. Then the electron 9 3 velocity is expected to be very small. The electron velocity should not be confused with the velocity of electromagnetic 1 = εµ s. J (A/m ) I (Amp) S v E

3 Example 1 6. A 1 gauge copper wire (cross-section area m ) is carrying 10 A current. Find the average electron drift velocity. Assume each copper atom contributes one coduction electron. The copper density 3 is 8.9 g/cm and the molar mass of copper is 63.5 g/mol. Sol. One mole contains atoms. Then the number density of electrons is g/cm / mol /c n = = 63.5 g/mol The current density is I 10 A J = = = A/m 6 A m The electron drift velocity is 6 6 J A/m v = = = ne /m C m = /m m/s.

4 Origin of Resistance In coductors, electrons are accelerated by electric field. However, electrons frequently collide with lattice ions and lose momentum. In steady state, electrons acquire a consatnt (terminal) velocity according to dv ee m = 0 = ee mν cv v = dt mν where ν c is the collision frequency (1/sec). The current density is ne J = E J = σ E mν c ne σ = : conductivity (S/m) (Not surface charge density) mν c E = ρ 1 mν ρ = c = : resistivity Ohm m σ ne ( ) c

5 Example 8 : The ressitivity of copper at room temperature is m. Estimate the lectron collision frequency. η = Ω m ν Sol. The resistivity is η = ne m e e c 31 (electron mass) = kg n (electron density) = /m Then e 8 3 where ηne ( ) ν c = = = 31 m / sec

6 Resistivity and Resistance The microscopic Ohm's law: E = ρj. ρ is resistivity ( Ω m) In a resistive rod of cross-section Aand length l, the voltage is V = El I and current density is J = A V I l E = ρj = ρ V = ρ I = RI l A A l Resistnce of the rod is R = ρ ( Ω) A

7 Example 3 6. Nichrome has a resistivity m at room temperature 0 C. Find the resistance of -gauge Nichrome wire 1 m long. The radius of the wire is 0.3 mm. Sol. l R = ρ = Ω π r Supplement m 1 m 3 π ( m) ( ) R( ) α ( ) ρ = Ω = 3.1 Ω 3 The temperature coefficient of Nichrome is = / C. Find the resistance of the wire at 500 C. Sol. R Example 4. Find the resistance of a conical conductor shown C = 0 C = = 8.9 Ω. Sol. The radius of the cross section is r r h 1 r = r1 + z The resistance is R h dz ρ = ρ = πr π 0 0 z= h ρ h 1 ρ h = = π r r r r π rr r + z h z= 0 h r 1 + r dz r 1 h z α h

8 Energy Dissipation in Resistor In a simple closed circuit consisting of power supply (e.g., battery) and a resistor, the rate of charge flow (current) is constant. In unit time one second, a charge I (C) is transfered everywhere along the circuit. In the power supply, the charge gains energy being lifted by the electromotive force (EMF) at the rate C J I V ( Volts = = Watts) sec sec In the resistor, the energy is converted into heat at the same rate. V Power = IV = RI = R Note on Energy Dissipation. (Serway warned about the use of energy dissipation in resistors and instead he suggested energy delivered to resistors. I disagree. In resistors electric energy is consumed irreversibly into heat.)

9 Energy Transfer from EMF to Resistor The amount of charge transferred in unit time (1 second) is I 1 sec = I C. At the EMF, the charge gains energy at the rate resistor, the charge gives away energy at the same rate I (voltage in R) = I RI = RI I EMF (J/sec). At the EMF + - Current Resistor

10 Quiz. Find the current in each light bulb. The voltage is 10 V. 30W The current in the 30 W bulb is I = = 0.5 A 10 V 60 W and that in the 80 W bulb is I ' = = 0.5 A. 10 V Quiz. Determine the curents at positions a through f. a: 0.75 A; b: 0.75 A; c and d: 0.5 A; e and f: 0.5 A. 0.5 A 0.5 A 0.75 A

11 Example 5. An electric heater has a resistance of 8 Ω. When it is connected to 10 V line, what is the power? How long does it take the heater to boil water of 1.5 liter from 10 C? Assume 80% efficiency. V 10 Sol. The power is P = = = 1800 W. R 8 Amount of energy needed to heat 1.5 L water from 10 to 100 degree C is U ( ) = = / 0.8 J J (Recall that to raise the temperature of 1 g water by 1 degree C, 4. J = 1 cal of energy is needed.) It takes t J = = 350 seconds =5.8 Min J/s 3 6 If the rate is $ 0.1/kWh (1 kwh = = J), the cost is about 1.75 cents.

12 Example 6: Explain why high voltage power transmission is more beneficial than low voltage. Ans. For a given power P = VI, the loss in transmission line of a resistance R is P Loss rate = RI = R V Therefore, the loss decreases as the voltage increases. Power line voltage of 100 kv to 1 MV is commonly employed in long distance power transmission. Consider a power line having a resistance of 50 Ω over 100 miles. At 700 kv, 1000 A, the power is transmitted at 700 MW and the loss rate in the line is RI = 50 MW. At 1 MV, the current is 700 A for the same power. But the loss decreases to 4.5 MW.

13 Example 7. Using 1 g of copper, design a wire having a resistance of 0.5 Ω. The ρ = Ω 8 3 copper resistivity is m and copper density is m = 8900 kg/m. ρ Sol. Let the wire radius be r and length be l. The resistance is l R = ρ = 0.5 Ω π r The volume of the wire is V= πrl. Then ρπ m rl= kg ρρml Solutions are l = 1.8 m and r = 0.8 mm. 4 4 = 5 10 l = = 3.31 m 8

14 Superconductors In 1911, Onnes (Dutch physicist) found that the resistivity of mercury totally disappeared at cryogenic temperatures lower than 4. K (liquid He temperature). In 1990 s, high temperature superconductors were developed. The ultimate goal is to develop superconductors that work at the room temperature. Applications: high magnetic field (NMI, fusion, high energy accelerators, etc.), loss-less power transmission.

15 500 km/hr Superconducting Maglev Train in Japan (commercial line under construction)