2. Parallel Circuit A parallel circuit is one that has two or more paths for the electricity to flow – similar to a fork in a river A parallel circuit is one that has two or more paths for the electricity to flow – similar to a fork in a river In other words, the loads are parallel to each other. In other words, the loads are parallel to each other. If the loads in this circuit were light bulbs and one blew out there is still current flowing to the others. If the loads in this circuit were light bulbs and one blew out there is still current flowing to the others.

3. Electric Current Current is the rate at which electric charges move through a given area. SI unit is the Ampere or Amp. 1 A = 1 C/s I = ΔQ/t Current = charge / time

4. Example problem The current in a light bulb is 0.835 A. How long does it take for a total charge of 1.67 C to pass a point in the wire? ΔQ = 1.6 CI = 0.835 At= ? I = ΔQ/t t = ΔQ/I t= 1.6C/0.835A t= 2.00s

5. Potential Difference The potential difference between the positive and negative ends of batteries: All AA, AAA, C, D Cell Batteries = 1.5 V The only difference is how long they produce the 1.5 V. Car battery = 12 V Positive and Negative slots of an electrical outlet = 120 V

6. Resistance Resistance- The opposition to the flow of current in a conductor R = V/I Resistance = Potential difference/Current SI unit – ohm Symbol-  (omega)

7. Example Problem The resistance of a steam iron is 19.0 Ω. What is the current in the iron when it is connected across a potential difference of 120V? R= 19.0 ΩV= 120VI= ? R=V/I I=V/R I=120V/19.0 Ω I= 6.32 A

8. Ohm’s Law Relates electric current, voltage and resistance. Current = Voltage/Resistance (I=V/R) Voltage (V) push of e- V = PE/Q V = IR Volts V Current (I) flow of e- I = Q/t I = V/R Ampere (Amps) A Resistance (R) opposition to the flow of e- R = V/I Ohms Ω

9. Electric Power Electric power is the rate of conversion of electrical energy Formula for Electric Power: P = IV Electric power = current X potential difference

10. Electric Power Because P= IV and V=IR we can also say; P= IV = I(IR) = I 2 R P = I 2 R Or, because I = V/R, we can also say: P = IV = (V/R)V = V 2 /R P=V 2 /R

11. Electric Power An electric space heater is connected across a 120 V outlet. The heater dissipates 1320 W of power in the form of electromagnetic radiation and heat. Calculate the resistance of the heater. P = V 2 /R R = V 2 /P R = 120 2 /1320 R = 10.9 Ω

12. Electric Power Power companies measure energy not power, using the kilowatt- hour as the unit One kilowatt-hour = the energy delivered in 1 hour at the constant rate of 1 kW. To convert between kWh and the SI unit of Joule: 1 kWh = 3.6 X 10 6 J

13. Energy Usage Cost Power (P) meas. of the rate at which electricity does work. P = V x I Watts (W) Energy (E) meas. By electric meter. E = P x t kilowatt-hour (kWh) Cost ($) cost of electricity Cost = energy x kwh ($.07)

14. Example Problem How much does it cost to operate a 100.0 W light bulb for 24 h if electrical energy costs $0.080 per kWh? P= 100W = 0.100 kW; t= 24 h Energy = Pt = 0.100 kW*24 h = 2.4 kWh Cost = 2.4 kWh*$0.080 = $0.19