2.  Water boy Water boy  Lightning Lightning  Team work Team work  Ac/Dc Charge Ac/Dc Charge

3.  Two different types current:  Alternating current (AC): i.e. Outlets in walls  Direct Current (DC): i.e. Batteries, out the positive, in the negative.

4.  Ohm’s Law: the current between two points is directly proportional to the potential difference or voltage, and inversely proportional to the resistance. ∆V = IR ∆V = Voltage/ Potential difference -Volt (V) I = Current – Amp (A) R = Resistance –Ohm ( Ω )

5.  the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. Refer to page 691 in text book.  Think of gravitational potential difference. ∆E = mgh V = ∆E/Q V = Voltage/ Potential difference -Volt (V) ∆E = change in electrical potential energy- Joule (J) Q = quantity of charge –Coulomb (C)

6.  Ex. A battery has a potential difference of 18.0 V. How much work is done when a charge of 64.0 C moves from the anode to the cathode? V= 18.0 V Q = 64.0C V = ∆E/Q ∆E = VQ ∆E = (18.0V)(64.0C) ∆E = 1152J

7.  Current: a flow of electric charge. From positive to negative.  The current can be calculated with the following equation: I = Q/t  Q = quantity of charge transferred– Coulombs (C)  ∆t = time interval that the charge is transferred – Seconds (s)

8.  Example 1: A toaster runs for 2.5 minutes in which 9.60 A of current were required. Find the amount of charge that passed through the toaster. I = 9.60 A ∆t = 2.5min = 150sI = Q/∆t Q = I ∆t Q = (9.60A) (150s) Q = 1440 C Current cont.

9.  Resistance: is the hindrance to the flow of charge.  3 main variables that contribute to resistance. the total length of the wires will affect the amount of resistance. Longer the wire, more resistance. the wider the wire, the less resistance that there will be to the flow of electric charge the material that a wire is made of

10.  The conducting ability of a material is often indicated by its resistivity. Lower the resistivity the better the conductor.  Kitchen appliances such as electric mixers and light dimmer switches operate by altering the current at the load by increasing or decreasing the resistance of the circuit.

11.  Example 2. A 30 V battery maintains current through a 10 Ω resistance. What is the current? V = 30V R = 10 Ω ∆V = IR I = V/R I = 30V/10 Ω I = 3.0A Resistance cont.

12.  The Ohm's law equation can be rearranged and expressed as: I = ∆V/R The current in a circuit is directly proportional to the electric potential difference.(battery voltage) The current is inversely proportional to the total resistance offered by the circuit.

13. With the formula: I = ∆V/R Battery Voltage Total Resistance Current 1.5V 3 Ω 0.5A 3.0V 3 Ω 1.0A 3.0V 6 Ω 0.5A 3.0V 12 Ω 0.25A

14. Power: is the rate at which electrical energy is supplied to or consumed by a circuit. P = ∆E/t P = rate of change in potential energy – Watt (W) ∆E = change in electrical potential energy- Joule (J) ∆t = time interval that the change occurs over – Seconds (s)

15. To determine the power across a circuit, we must combine the following 3 equations.  P = ∆E/t  ∆V = ∆E/Q  I = Q/t P = (∆V)(Q)/t P = (∆V)(I)

16.  Ex. Determine the power of a saw that draws 12 amps of current when plugged into a 120V outlet. I = 12A ∆V = 120VP = ∆VI P = (120V)(12A) P = 1440W Other formulas for calculating power: P = I 2 R P = V 2 / R

17.  Pg. 692 # 1, 2, 3  Pg. 696 # 4, 5, 6  Pg. 714 # 21, 22