1. Amps, Volts and Resistance (Ohm’s Law)

2. Coulomb  Recall that one Coulomb has 6.25 X 10 18 electrons.  If the current coming out of the outlet on the wall has 600 C, how many electrons are coming out of the outlet? 1C = 6.25 X 10 18 electrons 600 C e - Cross multiply and you get 3.75 X 10 21 electrons

3. Current Intensity  Current Intensity - the amount of electrons that flow past a given point in a circuit every second (i.e through a wire)  Example: 0.7 amperes is shown by I = 0.7 A  An ammeter is the instrument used to measure current intensity A Symbol: I Units:Amperes, A

4. Current Intensity  The current intensity in a circuit can be determined using the formula: I = q/t I is the current intensity in amps (A) q is the charge in coulombs (C) t is time in seconds (s) NOTE: 1 Amp (A) = 1 Coulombs (C)/1 second (s) A=C/s

5. Example: What is the current flowing through a car headlight if there are 900 C of charge used in 1 minute? I = q/t I = 900 /60 I = 15A

6. Example: How much charge does it take to operate an MP3 player for 15 minutes if the current is 2A? I = q/t 2 = q/900 q = 1800C

7. Potential Difference (Voltage)  Potential Difference is the amount of energy transferred between two points in an electrical circuit.  Example: Energy that is provided by a battery or power supply. It could be a 12 volts battery (also know as 12V)  A voltmeter is used to measure potential difference V Symbol: V Units:Volts, V

8. Potential Difference (Voltage)  The potential difference in a circuit can be determined using the formula: V=E/q V is potential difference in volts (V) E is the energy in joules (J) q is the electric charge in coulombs (C) Note: 1 Volts = 1 Joule / 1 Coulombs V=J/C

9. Example  In a house, how much energy is provided by 120V service providing 200C of charge? V=E/q 120 = E/200 E = 24 000J

10. Symbol: R Units:Ohms, Ω Resistance  Resistance is how difficult it is for current to flow  Resistance is the opposite of conductance!  A resistor is used to slow current down and convert electrical energy into heat energy (e.g. light bulb, stove element). Conductor – low resistance & high conductance Resistor– high resistance & low conductance

11. 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.

12. Formula to Solve Circuits  Resistance, current intensity and potential difference are all related R = V/I R is the resistance in Ohm’s (Ω) V is the potential difference (Volts) I is the current intensity (Ampere) Note: 1 Ohms= 1 Volts / 1 Ampere 1 (Ω) = 1 V / 1 A

13. Example  An stove element 30 Amperes going through it, and carries 120 volts of electricity, how much resistance is in this circuit?  R = V/I  R = 120 V/30 A  R = 4 V/A = 4 Ω

14. Example  A circuit has a potential energy difference of 240 V and offers a resistance of 6Ω, what is the intensity of the current flow?  R=V/I  6 = 240 /I  I = 240 /6  I = 40 A

15. Example  A circuit has a resistance of 10Ω & a current intensity of 5A. What is the potential difference?  R=V/I  10 = V/ 5  10 X 5 = 50V

16. Summary Table ResistanceCurrentPotential Difference Definition Slows current down The flow of electrons Causes electrons to flow SymbolRIV Units Ohms (Ω) Amps (A)Volts (V) Formula (if applicable) R=V/I Ω=V/I I=q/t A=C/s V=E/q V=J/C How measured (if applicable)Circuit Formula Ammeter or formula Voltmeter or formula