Static & dynamic electricty ohm's law

1. Static & dynamic electrictyohm's law

2. Static Electricity • Static electricity describes all the phenomena related to electrical charges at rest.

3. How to charge an object? • An object can be charged in three different ways: • by friction • by conduction • by induction

4. Charging by Friction • Friction pulls electrons away from one of the objects and transfers them to the other. • Before: Two neutral objects • After: Two objects with opposite charges

5. Charging by Friction

6. Charging by Conduction • The charge of one object is shared between two objects when they come into contact. • Before: One charged object and one neutral object • After: Two objects with like charges

7. Charging by Conduction

8. Charging by Induction • The proximity of the charged object causes the charges in the neutral object to separate. • Before: One charged object and one neutral object • After: One charged object and one object carrying a partial positive charge on one side and a partial negative charge on the other side

9. Charging by Induction

10. Practice Problem • The five spheres below, identified A to E, carry an electrical charge. If sphere A carries a positive charge, what is the sign of the charges on each of the other spheres? + + + - + - - -

11. Electroscope • A simple electroscope consists of a glass jar in which two strips of gold leaf are suspended from a metal rod that conducts electricity. • The rod, which enters the jar through a stopper made of a material that does not conduct electricity, has a metal knob on the end outside the jar.

12. Electroscope

13. Electroscope • The strips of gold leaf hang straight down when they are not charged. When a charged body is brought near the metal knob, both strips acquire a like charge (that is, they both become negative or both become positive). As a result, they repel each other and spread apart to form an inverted V. The electroscope is then charged. • If an oppositely charged body is brought close to the knob, the charge on the strips is neutralized, and they again hang straight down. The electroscope is discharged.

14. Electroscope

15. Dynamic Electricity • Dynamic electricity describes all the phenomena related to electrical charges in motion. • Electric current is the orderly flow of negative charges carried by electrons.

16. Ohm’s Law • Ohm’s law states that, for a given resistance, the potential difference in an electrical circuit is directly proportional to the current intensity.

17. Ohm’s Law – Resistance • Resistance: The ability of a material to hinder the flow of electric current. • Symbol: R • Unit of measurement: Ohm (Ω)

18. Ohm’s Law – Potential Difference • Potential difference: The amount of energy transferred between two points in an electrical circuit. • Symbol: V • Unit of measurement: Volt (V)

19. Ohm’s Law – Current Intensity • Current intensity: The number of charges that flow past a given point in an electrical circuit every second. • Symbol: I • Unit of measurement: Ampere (A)

20. Ohm’s Law

22. Practice Problem #1 • If the current in the circuit is 10 amps and the resistance is 3.0 ohms, what is the voltage? STEP 1: STEP 2: STEP 3: I = 10 A R = 3.0 Ω V = ? V = IR V = IR V = (10 A) x (3.0 Ω) V = 30.0 V STEP 4: The voltage of the circuit is 30.0 Volts.

23. Practice Problem #2 • A current of 3.7 amps is running through a circuit with a resistance of 1.5 ohms. What is the voltage? STEP 1: STEP 2: STEP 3: I = 3.7 A R = 1.5 Ω V = ? V = IR V = IR V = (3.7 A) x (1.5 Ω) V = 5.55 V STEP 4: The voltage of the circuit is 5.55 Volts.

24. Practice Problem #3 • If the battery in the circuit is 24 V and the resistance is 12 ohms, what is the current, I? STEP 1: STEP 2: STEP 3: I = ? R = 12 Ω V = 24 V I = V/R I = V/R I = (24 V) / (12 Ω) I = 2 A STEP 4: The current of the circuit is 2 Amperes.

25. Practice Problem #4 • Given a voltage of 120 volts and a current of 5 amps, what is the resistance? STEP 1: STEP 2: STEP 3: I = 57 A R = ? V = 120 V R = V/I R = V/I R = (120 V) / (57 A) R = 2.11 Ω STEP 4: The resistance of the circuit is 2.11 Ohms.

26. Calculating Current Intensity • Current intensity can be determined using the following formula: • I is current intensity [A] • q is charge [C] • Δt is time [s] I = q Δt

27. Calculating Current Intensity q I Δt

28. Practice Problem #5 • The data sheet for a car headlight indicates that the light requires a current intensity of 15 A. What is the charge needed for one minute of operation? I = 15 A Δt = 60 s q = ? q = I Δt q = I Δt q =(15 A)(60 s) q = 900 C ANS: It takes a charge of 900 C to make the headlight work for one minute.

29. Calculating Potential Difference • Potential Difference can be determined using the following formula: • V is potential difference [V] • E is energy transferred [J] • q is charge [C] V = E q

30. Calculating Potential Difference E q V

31. Practice Problem #6 • The electrical circuits in our homes usually supply a potential difference of 120 V. What is the amount of energy transferred by a charge of 200 C? V = 120 V E = ? q = 200 C E = V q E = V q E = (120 V)(200 C) E = 240 000 J ANS: A 200 C charge can transfer 240 000 J of energy.

32. Ohm’s Law

33. Electrical Power • Electrical power (Pe) is the amount of work an electrical device can perform per second. • Unit of measurement: watt • Symbol: W

34. Electrical Power • Pe is electrical power [W] • V is potential difference [V] • I is current intensity [A] Pe = V I

35. Calculating Electrical Power Pe V I

36. Practice Problem #7 • Calculate the power rating of an i-Pod drawing 3.5 A from a 6.0 V battery. Pe = ? I = 3.5 A V = 6.0 V Pe = V I Pe = (6.0 V)(3.5 A) Pe = 21.0 W Pe = V I ANS: The i-Pod has a power rating of 21.0 Watts.

37. Electrical Energy • The amount of Electrical energy (E) used by an electrical device is found by multiplying its electrical power by time. This is a measure of the energy provided by electricity. • Unit of measurement: joule • Symbol: J

38. Electrical Energy E = Pe Δt • E is electrical energy [J] • Pe is electrical power [W or kW] • Δt is time [s or hr]

39. Calculating Electrical Energy E Pe Δt

40. Practice Problem #7 • If a 1000-W microwave oven operates for six minutes, what is the amount of energy it uses? Pe = 1000 W Δt = 6 min = 360 s E = ? E = Pe Δt E = Pe Δt E = (1000 W)(360 s) E = 360 000 J ANS: After six minutes of use, the microwave will consume 360 000 J of energy.

41. Electrical Energy = WORK • HINT: If a problemasksyou to find how muchworkisbeingdone, simplycalculate the elctricalenergy. • Energy and Workcanbetreated as synonyms. • Unit of measurement: joule • Symbol: J

42. Practice Problem #8 • How much work can a 1000 W car engine do in one minute? Pe = 1000 W Δt = 1 min = 60 s W = ? W = Pe Δt W = Pe Δt W = (1000 W)(60 s) W = 60 000 J ANS: A car engine can do 60 000 J of work in one minute.

43. Conversions • 1 kW = 1000 W • Ex: 3.6 kW x 1000 = 3600 W • Ex: 670 W ÷ 1000 = 0.67 kW • 1 hr = 60 min = 3600 s • Ex: 3 hr x 60min/1hr = 180 min • 180 min x 60sec/1min = 10800 sec

44. Work • Pe is electrical power [W] • W is work [J] • Δt is time [s] Pe = W Δt

45. Calculating Work Work [J] and Electrical Energy [J] are interchangeable in the formula. W Pe Δt