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What we are Doing Today ? 1) Hand in Lab 40 (formal & informal) 2) If you missed the lab 40 formal lab (ie. You were absent with a legit excuse, you will.

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Presentation on theme: "What we are Doing Today ? 1) Hand in Lab 40 (formal & informal) 2) If you missed the lab 40 formal lab (ie. You were absent with a legit excuse, you will."— Presentation transcript:

1 What we are Doing Today ? 1) Hand in Lab 40 (formal & informal) 2) If you missed the lab 40 formal lab (ie. You were absent with a legit excuse, you will be doing lab 41/42 as a formal. 3) Make sure you finished the HW p.173 of the textbook questions 10-18

2 Electrical Power: The amount of work an electrical device can perform per second. P e = W / Δt P e = The electrical Power in Watts (W) W = The Work in Joules (J) Δt = the time interval (in s)

3 Question What is the electrical power an ipod can perform in 2 minutes with 200 J of work? What do we know? P e = ? Δt = 2 min = 120 sec W = 200 J P e = W/Δt P e = 200J / 120sec = 1.67 W Answer: the electrical power is 1.67 W

4 We can also express electrical power as: P e = V x I P e = is the electrical power in Watts (W) V = is the potential difference in Volts (in V) I = is the current intensity in Amps (A)

5 Question What is the potential difference at an ipods terminals if the electrical power is 1000W and the current intensity is 50 Amps ? What do we know? P e = 1000 W I = 50 A V = ? P e = V x I  V = P e / I V = 1000W / 50 A = 20 V Answer: the potential difference is 20 V

6 We can also express the formula for the relationship between electrical power and electrical energy E = P e x Δt E = the electrical energy used (in Joules or kWh, kilowatt hours) P e = is the electrical power in Watts (W) Δt = the time interval in seconds or hours 1 kWh = 3 600 000 Joules

7 Question What is the amount of energy a 10 000 W car (driven by Mr. Garrety) use in 12 minutes?

8 What is the amount of energy a 10 000 W volkswagon beetle (driven by Mr. Garrety) use in 12 minutes? What do we know? P e = 10000 W Δt = 12min x 60sec = 720 seconds E = ? E = P e x Δt E = 10000 W x 720 seconds = 7200000 J or 2 kWh Remember 1 kWh = 3 600 000 Joules

9 EST (Chapter 5) Coulomb`s Law: It explains how strong the force will be between two electrostatic charges. Electrostatic means electric charges without any motion.forceelectrostatic F = the electrical force in Newtons (N) K = Coulomb’s constant, which is 9 x 10 9 Nm 2 /C 2 q1 = the charge of the first particle (in C) q2 = the charge of the second particle (in C) r = the distance between the 2 particles (in metres )

10 Question What is the electrical force between two positively charged particles that are placed 5 cm apart and that each have a charge of 5 x 10 -8 C ?

11 What do we know? k = 9 x 10 9 Nm 2 /C 2 q1 = 5 x 10 -8 C q2 = 5 x 10 -8 C r = 5 cm = 0.05 m F = ? F = 9 x 10 9 Nm 2 /C 2 x 5 x 10 -8 C x 5 x 10 -8 C / (0.05m) 2 F = 9 x 10 -3 N or 0.009 N

12 measuring current Electric current is measured in amps (A) using an ammeter connected in series in the circuit. A

13 measuring current A A This is how we draw an ammeter in a circuit. SERIES CIRCUIT PARALLEL CIRCUIT

14 measuring voltage The ‘electrical push’ which the cell gives to the current is called the voltage. It is measured in volts (V) on a voltmeter V

15 measuring voltage V This is how we draw a voltmeter in a circuit. SERIES CIRCUITPARALLEL CIRCUIT V


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