2. Electric Current and Resistance

4. Here below an English/ Italian glossary, that we will fill during our activities Current Circuit Charge Potential VoltageVoltaggio = ddp Battery Switch Net Wire Copper Ammeter/Voltmeter Drift Platepiastra Positive/negative terminal electron Electric field Resistor Resistance Series/parallel

5. Which of these connections could work to light the bulb?

6. The Central Concept: Closed Circuit

7. circuit diagram cellswitch lamp wires Scientists usually draw electric circuits using symbols;

8. Remember: Electric Potential Energy- Two Unlike Charges Higher Potential Energy Lower Potential Energy + - To cause movement of a charge, there must be a potential difference.

9. Closing the switch establishes a potential difference (voltage) and an electric field in the circuit. Electrons flow in a net direction away from the (-) terminal. High Potential Low Potential

10. Conventional Current By tradition, direction in which “positive charges” would flow. Direction is opposite of electron flow.

11. Question: What is required in order to have an electric current flow in a circuit? Answers: 1.A voltage source. 2.The circuit must be closed.

12. Battery (Chemical Cell): A device that converts chemical energy to electricity. A battery provides a potential energy difference (voltage source).

13. Electric Current: The flow of electric charges.

14. Electric Current, I I =  q  t Rate Unit: Coulomb / sec = Ampere (A) Andre Ampere (1775-1836)

15. Conventional current has the direction that the (+) charges would have in the circuit.

16. Direct Current DC Provided by batteries Alternating Current AC Provided by power companies TWO TYPES OF CURRENT

17. MEASUREMENT DEVICES :1) Ammeter Measures electric current. Must be placed in series.

18. 2) Voltmeter Measures the voltage between two points in an electric circuit. Must be connected in parallel.

19. Example: What charge flows through a cross sectional area of a wire in 10min, if the ammeter measures a current of 5mA? Answer: 3C

20. RESISTANCE It is known from experiment that the current flowing in a conductor is directly proportional to the potential difference across it. The constant of proportionality, R, is called the RESISTANCE R =  V / I ( resistance equals the ratio of voltage to current ) Unit: Ohm (Ω) 1 Ω = 1 V/A

21. Ohm’s Law I = V / R Georg Simon Ohm (1787-1854) I = Current (Amperes) (amps) V = Voltage (Volts) R = Resistance (ohms) I V Gradient A device that obeys Ohm’s Law is called a ohmic resistor (its resistance does not depend on the voltage)

22. The current through an ohmic conductor is: directly proportional to the voltage across it ( over a limited range of V) (more voltage  more current ) inversely proportional to the resistance of the conductor (the greater the resistance  the less the current)

23. But what is resistance? It is a measure of how much a conductor impedes the flow of current At the molecular level, electrons undergo frequent collisions with the ions of materials. The higher the number of collisions, the higher the resistance of the material is.  It’s a sort of friction, an opposition to the flow of current

24. Resistor An object that has a given resistance.

25. RESISTORS Colour codes are used to identify resistance value Circuit symbol (or the “zig-zag symbol) R The Resistor Colour Code ColourNumber Black0 Brown1 Red2 Orange3 Yellow4 Green5 Blue6 Violet7 Grey8 White9 The four colour code bands are at one end of the component. Counting from the end, the first three (or sometimes four or five) bands give the resistance value and the last the tolerance TOLERANCES BROWN1% RED2% GOLD5% SILVER10% NONE20%

26. Example: Calculate the current through a 3 Ω resistor when a voltage of 12V is applied across it. Answer: 4 A

27. Example: A 6 Ω resistor has a power source of 20V across it. What will happen to the resistance if the voltage doubles? Answer: resistance doesn’t change!

28. Factors that affect resistance: Type of material Size and shape Temperature

29. Resistance R = ρ L (a.k.a. Ohm’s 2 nd law) A L: length of the wire A: cross-sectional area ρ: resistivity (inherent to material) Unit:  m

30. RESISTIVITIES FOR CERTAIN MATERIALS AT 20°C (measured in  m) Silver 1.6 × 10 -8 Copper 1.7 × 10 -8 Aluminium 2.8 × 10 -8 Tungsten 5.6 × 10 -8 Constantan (alloy of copper and nickel) 49 × 10 -8 Nichrome (alloy of nickel, iron & chromium) 100 × 10 -8 Graphite (3 - 60) × 10 -5 Silicon 0.1 – 60( semiconductor ) Germanium (1 - 500) × 10 -3 Pyrex glass 10 12( insulator )

31. The table of resistivity values shows that, although silver has a low resistivity, it is expensive to use in electrical circuits. Copper is the preferred metal although aluminum is commonly used in electricity transmission cables because of its lower density. What is the best material?

32. Example: Determine the resistance of a piece of copper wire that is 10.0 m long and 1.2 mm in diameter. Solution The resistance, R, is given by the formula R = ρL / A, where A = πr 2. This means that R = (1.7 × 10 -8 Ωm) (10.0 m) / π(6.0 × 10 -4 )2 m 2 = 0.150 Ω. The resistance of the copper wire is 0.15 Ω.

33. Question: What happens to the resistance when the length is doubled and the area is quadrupled? Answer: It changes by 1/2

34. Temperature Dependence of Resistance The resistance of a material increases with temperature because of the thermal agitation of the atoms it contains, and this impedes the movement of electrons that make up the current. Resistance increases because resistivity increases, as shown in this formula:  t =  20 [1 +  (t -20°)] where  20 equals the resistivity at 20 °C,  t is the resistivity at some temperature, t °C, above the reference temperature, and  is the temperature coefficient for the material being used.

35. SUPERCONDUCTIVITY One interesting phenomenon of the effect of temperature on resistance is superconductivity. In 1911, H. Kammerlingh Onnes found that mercury loses all its resistance abruptly at a critical temperature of 4.1 K. When a material attains zero resistance at some critical temperature, it is called a superconductor. The possibility of having a material that has an induced electric current that lasts forever has become a topic for research physicists. Just think of the energy saving if the perfect superconductor is found that can give zero resistance at room temperature.

36. POWER DISSIPATION IN RESISTORS Electric power is the rate at which energy is supplied to or used by a device. It is measured in J/ s called watts (W). When a steady current is flowing through a load such as a resistor, it dissipates energy in it. This energy is equal to the potential energy lost by the charge as it moves through the potential difference that exist s between the terminals of the load.

37. Some possible power ratings for household appliances If a vacuum cleaner has a power rating of 1500 W, it means it is converting electrical energy to mechanical, sound and heat energy at the rate of 1500 J/s. A 60 W light globe converts electrical energy to light and heat energy at the rate of 60 J/s.

38. Electric power Basic definition of power: P = W/ t Work / time We know that W = q V and q = I t  P = q V/t = I t V/ t  P = V I (V is the voltage) And applying Ohm’s law: P= R I 2 = V 2 / R

39. UNIT OF ELECTRIC ENERGY The commercial unit of electrical energy is the kilowatthour (kW h). It is the energy consumed when 1 kW of power is used for one hour. The consumer has to pay a certain cost per kilowatt-hour. -------------------------- Question: How many Joules in a kWh? 1 kWh = 1000 W x 3600 s = 3600000 J = 3.6 x 10 6 J

40. HEATING EFFECT OF A CURRENT (Effetto Joule) It was investigated in 1841 by James Joule. He was able to demonstrate that by supplying electrical energy to a high resistance coil of wire this energy could be converted to thermal energy. V × I × t = m× c × ΔT electric energy  heat = infrared radiation

41. Example An electrical appliance is rated as 2.5 kW, 240 V. (a) Determine the current needed for it to operate. (b) Calculate the energy it would consume in 2.0 hours. Solution: (a)Given that P = 2.5 × 10 3 and V = 240 V, we use the formula, P = IV  I = P/W = …10.4 So I = 1.0 × 10 1 A. (b) Next, we use the formula W = VIt, so that W =(240 V) x(10.4 A)x 7.2 10 3 s = = 1.8 × 10 7 J The energy consumed is 1.8 × 10 7 J.

42. Exercises 1) The element of an electric jug has a resistance of 60 Ω and draws a current of 3.0 A. Determine by how much the temperature of 5.0 kg of water will rise if it is on for 6 minutes. 2) Calculate the cost to heat 200 kg of water from 12°C to its boiling point if power costs 14 cents per kilowatt-hour.