 Circuit = Complete path where electrons can flow  Circuit diagram symbols:  Wire/Conductor  Resistor (light bulbs, fans)  Battery  Switch

 Resistance = Voltage/ Current  R = V / I  Units  Resistance = ohms ( Ω )  Voltage = volt (V)  Current = ampere (A)

 A 30.0 V battery is connected to a 10.0 Ω resistor. What is the current in the circuit?  I = V/R  I = 30.0 V / 10.0 Ω  = 3.00 A

 Single path for electrons to flow  If any resistor (bulb) in the circuit is removed or burnt out, then NO BULBS will light – the circuit is not complete.  Electric current is the same through each device / resistor / bulb. (I = V/Rtotal)  Current stays the same across all resistors.

 Voltage drop across each device depends directly on its resistance (V = I x R)  Total voltage divides among the individual electric devices in the circuit.

 Multiple paths for the current to flow; Branches  If any resistor (bulb) in the circuit is removed or burnt out, then the other bulbs will light as long as there is an unbroken path from the battery through that bulb and BACK to the battery.  Total current equals the sum of currents in branches  As the number of branches is increased, overall resistance of the circuit is decreased  think about driving on a 4 lane highway – little resistance to the flow of traffic  now consider an accident that blocks three of the lanes…a reduction to only one lane INCREASED the resistance  opening all lanes DECREASED the resistance

The inverse of the total resistance is the sum of the inverse of the resistors 1/R total = 1/R A + 1/R B

 Series I=V/R total R= R 1 + R 2 + ….. Parallel I=V/R 1/R total = 1/R 1 + 1/R 2 + …

 Measures the rate at which energy is transferred  Power = Current x Voltage  P = IV  The unit of power is the Watt

 A 6.0 V battery delivers a 0.50 A current to an electrical motor. What power is consumed by the motor?  P = IV  P = (0.50 A)(6.0 V)  P = 3.0 W

R total = 2 Ω + 3 Ω = 5 Ω I A = V/R total ……9 v / 5 Ω = 1.8 Amps I B = V A = I A x R A = 1.8 Amps x 2 Ω = 3.6 volts add these and you should get the voltage supplied by the battery, 9 volts V B = I B x R B = 1.8 Amps x 3 Ω = 5.4 volts

R total = I A = I B = V A = V B =

R total = I A = I B = I C = V A = V B = V C =

R total = 1/ R total = 1/2 + 1/3 = 0.833,  but remember 1/ R total = (rearrange and solve for R total)  so R total = 1/0.833 = 1.2 Ω  I A = V/R A = 9.0 V / 2 Ω = 4.5 Amps In a parallel circuit these are NOT identical.  I B = V/R A = 9.0 V / 3 Ω = 3 Amps  V A = I A X R A = 4.5 Amps x 2 Ω = 9 v these should EACH equal the voltage being supplied by the battery, 9.0 volts.  V B = I B X R B = 3 Amps x 3 Ω = 9 v

R total = I A = I B = V A = V B =

R total = I A = I B = I C = V A = V B = V C =