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

• Electrical Power: The amount of work an electrical device can perform per second.

Pe = W / Δt

Pe = The electrical Power in Watts (W)

W = The Work in Joules (J)

Δt = the time interval (in s)

• Question

What is the electrical power an ipod can perform in 2 minutes with 200 J of work?

What do we know?Pe = ?Δt = 2 min = 120 secW = 200 J

Pe = W/Δt

Pe = 200J / 120sec = 1.67 W

Answer: the electrical power is 1.67 W

We can also express electrical power as:

Pe = V x I

Pe = is the electrical power in Watts (W)

V = is the potential difference in Volts (in V)

I = is the current intensity in Amps (A)

• 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?Pe = 1000 W I = 50 A V = ?

Pe = V x I V = Pe / I

V = 1000W / 50 A = 20 V

Answer: the potential difference is 20 V

We can also express the formula for the relationship between electrical power and electrical energy

E = Pe x Δt

E = the electrical energy used (in Joules or kWh, kilowatt hours)

Pe = is the electrical power in Watts (W)

Δt = the time interval in seconds or hours

1 kWh = 3 600 000 Joules

• Question

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

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

What do we know?Pe = 10000 W Δt = 12min x 60sec = 720 seconds E = ?

E = Pe x Δt

E = 10000 W x 720 seconds = 7200000 J or 2 kWh

Remember

1 kWh = 3 600 000 Joules

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.

F = the electrical force in Newtons (N)K = Coulomb’s constant, which is 9 x 109 Nm2/C2

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)

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 ?

What do we know?k = 9 x 109 Nm2/C2

q1 = 5 x 10-8 C

q2 = 5 x 10-8 C

r = 5 cm = 0.05 m

F = ?

F = 9 x 109 Nm2/C2 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

measuring current

Electric current is measured in amps (A)

using an ammeter connected in series in

the circuit.

A

measuring current

A A

This is how we draw an ammeter in a circuit.

SERIES CIRCUIT PARALLEL CIRCUIT

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

measuring voltage

V

This is how we draw a voltmeter in a circuit.

SERIES CIRCUIT PARALLEL CIRCUIT

V