1. Ohm’s Law Experiments show that for many materials, including most metals, the resistance remains constant over a wide range of applied voltages or currents Experiments show that for many materials, including most metals, the resistance remains constant over a wide range of applied voltages or currents This statement has become known as Ohm’s Law This statement has become known as Ohm’s Law V = I RV = I R Ohm’s Law is an empirical relationship that is valid only for certain materials Ohm’s Law is an empirical relationship that is valid only for certain materials Materials that obey Ohm’s Law are said to be ohmicMaterials that obey Ohm’s Law are said to be ohmic

2. V = R I Resistance, R = V/I [R] = V/A =  Ohm  For a fixed potential difference across a resistor, the larger R, the smaller current passing through it. R eq

3. Parallel connection Series connection R1R1 R2R2 R3R3 R1R1 R2R2 R3R3 R eq = R 1 + R 2 + R 3 1/R eq = 1/R 1 +1/R 2 +1/R 3

4. Q2. What is the ratio of the current flowing through each resistor (I 1 :I 2 ) in the circuit? 1. 1:1 2. 3:1 3. 1:4 4. Need more info. 6 V R 1 = 10 R 2 = 30

5. Q3. What is the ratio of the current flowing through each resistor (I 1 :I 2 )? R 1 = 10 R 2 = 30 6 V 1. 1:1 2. 3:1 3. 1:4 4. None of above

6. Electrical wires can be bent and/or stretched. A Node point (branching point) can be moved arbitrarily along the wire.

7. There are n identical resistors connected in parallel. R eq ? 1/R eq = 1/R + 1/R + 1/R + … + 1/R = n/R R eq = R/n

8. RaRa RbRb (1) 1/R eq = 1/R a + 1/R b (2) R eq is smaller than R a and R b 20 25 R eq ≈ 10 1000 = 1k 2 R eq < 2 Practically all the current flows Though the bottom one!!

9. Ohm’s law:  = R·I R 1 = 6 6 V I =  /R = (6 V)/(6 Ohm) = 1.0 A What is the electric potential at ? We cannot tell the absolute potential at this point. If  at is +6 V, then 0 V at If  at is +3 V, then -3 V at For both, the potential diff. is 6 V.

10. To be able to specify absolute potential at a given point, we need to specify a reference point “0” potential. GROUND R 1 = 6 6 V  = “0” Then,  at is +6 V.

11. R 1 = 6 6 V R 2 =4R 3 =2  = R 3 I = 2 (V)  = R 2 I = 4 (V)  = 4 + 2 = 6 (V)  = 2 V

12. Resistivity The resistance of an ohmic conductor is proportional to its length, L, and inversely proportional to its cross- sectional area, A The resistance of an ohmic conductor is proportional to its length, L, and inversely proportional to its cross- sectional area, A ρ is the constant of proportionality and is called the resistivity of the materialρ is the constant of proportionality and is called the resistivity of the material

13. L A R = Resistivity: material parameter same for any shape in a given material. [  ] = .m e.g. for copper  = 1.7 x 10 -8 gold  = 2.44 x 10 -8 tungsten  = 5.6 x 10 -8 nickel-chrome  = 150 x 10 -8 L/A 

14. Example A silver wire has a resistance of 2. What would be the resistance of a silver wire twice its length and half its diameter?

15. Electrical Energy and Power In a circuit, as a charge moves through the battery, the electrical potential energy of the system is increased by QV In a circuit, as a charge moves through the battery, the electrical potential energy of the system is increased by QV The chemical potential energy of the battery decreases by the same amountThe chemical potential energy of the battery decreases by the same amount As the charge moves through a resistor, it loses this potential energy during collisions with atoms in the resistor As the charge moves through a resistor, it loses this potential energy during collisions with atoms in the resistor The temperature of the resistor will increaseThe temperature of the resistor will increase

16. Electrical Energy and Power, cont The rate at which the energy is lost is the power The rate at which the energy is lost is the power From Ohm’s Law, alternate forms of power are From Ohm’s Law, alternate forms of power are

17. Electrical Energy and Power, final The SI unit of power is Watt (W) The SI unit of power is Watt (W) I must be in Amperes, R in ohms and V in VoltsI must be in Amperes, R in ohms and V in Volts The unit of energy used by electric companies is the kilowatt-hour The unit of energy used by electric companies is the kilowatt-hour This is defined in terms of the unit of power and the amount of time it is suppliedThis is defined in terms of the unit of power and the amount of time it is supplied 1 kWh = 3.60 x 10 6 J1 kWh = 3.60 x 10 6 J

19. A light bulb with 60 W/120 V When 120 V applied, consume 60 W of power POWER = Energy produced or dissipated per unit time = q High potentialLow potential  tt q./t = (q/t) P = I 2 R =  2 /R [P] = J/s = W (watt)

20. Example Light bulb 60 W, 120 V. Find resistance of the light bulb. Light bulb 60 W, 120 V. Find resistance of the light bulb. Bulbs in series Bulbs in series Bulbs in parallel Bulbs in parallel

21. Spec: 60 W/120 V P =  2 /R = 60 (W) R =  2 /60 = 240 () R measured ≈ 26  WHY are they so different? depends on temperature!!!

22. Example Light bulb 60 W, 120 V. Find resistance of the light bulb. Light bulb 60 W, 120 V. Find resistance of the light bulb. Bulbs in series Bulbs in series Bulbs in parallel Bulbs in parallel

23. Temperature Variation of Resistivity For most metals, resistivity increases with increasing temperature For most metals, resistivity increases with increasing temperature With a higher temperature, the metal’s constituent atoms vibrate with increasing amplitudeWith a higher temperature, the metal’s constituent atoms vibrate with increasing amplitude The electrons find it more difficult to pass the atomsThe electrons find it more difficult to pass the atoms

24. Temperature Variation of Resistivity, cont For most metals, resistivity increases approximately linearly with temperature over a limited temperature range For most metals, resistivity increases approximately linearly with temperature over a limited temperature range ρ o is the resistivity at some reference temperature T oρ o is the resistivity at some reference temperature T o T o is usually taken to be 20° C T o is usually taken to be 20° C  is the temperature coefficient of resistivity  is the temperature coefficient of resistivity

25. Temperature Variation of Resistance Since the resistance of a conductor with uniform cross sectional area is proportional to the resistivity, you can find the effect of temperature on resistance Since the resistance of a conductor with uniform cross sectional area is proportional to the resistivity, you can find the effect of temperature on resistance 3.8 x 10 -3 /C Ag: 3.8 x 10 -3 /C 3.9 x 10 -3 /C 5.0 x 10 -3 /C Cu: 3.9 x 10 -3 /C Fe:5.0 x 10 -3 /C Tungsten: 4.5 x 10 -3 /C

26. Example Light bulb (60 W; 120 V; 240 ) operates at 1800 C. What is the resistance of the filament (tungsten) at 20 C?

27.  Temperature =  o + AT for metals However, insulators have the opposite tendency. The higher temperature is, the lower .

28. Q. A 40 W/120 V light bulb produces 40 W of power when connected to 120 V. How much of power would be produced if the same bulb is connected to 60 V? You can solve this problem by calculating R of the filament and use the formula for power, P = V 2 /R. But we know the resistance of the filament is fixed. Therefore, power is just proportional to V 2. P (for 120 V) = 40 (W), p (for 60 V) = 40*(60/120) 2 = 10 (W)

29. A certain 1400 W space heater is designed to operate on 120 V. How much current flows through it when it is operating? What is Its resistance when operating? Power = IV = I 2 R = V 2 /R Since we know P and V, use P = IV to calculate I. I = P/V = (1400 W)/(120 V) = 11.7 A Now, we can use Ohm’s law to calculate R. When it is operating on 120 V, 11.7 A current flows. R = V/I = (120 V)/(11.7 A) = 10.25  Statement: P = IV If I hook up this heater on 60 V, the power Produced by this heater would drop to ½. No!

30. So far, we have considered circuits which can be reduced to a single loop circuit and applied Ohm’s law to extract I and/or V in the circuit. 6 V 53