WELCOME BACK CIRCUITS CIRCUIT DIAGRAMS Circuits and Circuit

WELCOME BACK!

CIRCUITS & CIRCUIT DIAGRAMS

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

Ohm’s Law Resistance = Voltage/ Current R=V/I Units Resistance = ohms (Ω) Voltage = volt (V) Current = ampere (A)

Example #1: 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

Series Circuits 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.

Series Circuits 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.

Parallel Circuits Multiple paths for the current to flow; Branches If any Total current equals the sum of currents in branches 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. 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

Combining Parallel Resistors The inverse of the total resistance is the sum of the inverse of the resistors 1/Rtotal = 1/RA + 1/RB

Equations (Make sure to have these in your notes) Series I=V/Rtotal R= R 1 + R 2 + …. . Parallel I=V/R 1/Rtotal= 1/R 1 + 1/R 2 + …

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

Example #2: 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

Practice… Rtotal = 2 Ω + 3 Ω = 5 Ω IA = IB = V/Rtotal…… 9 v / 5 Ω = 1. 8 Amps VA = IA x RA = 1. 8 Amps x 2 Ω = 3. 6 volts VB = IB x RB = 1. 8 Amps x 3 Ω = 5. 4 volts add these and you should get the voltage supplied by the battery, 9 volts

Rtotal= IA = IB = VA = VB =

Rtotal= IA = IB = IC = VA = VB = VC =

Practice… Rtotal = 1/2 + 1/3 = 0. 833, but remember 1/ Rtotal = 0. 833 (rearrange and solve for Rtotal) so Rtotal = 1/0. 833 = 1. 2 Ω I A = V/RA = 9. 0 V / 2 Ω = 4. 5 Amps identical. IB = V/RA = 9. 0 V / 3 Ω = 3 Amps V A the = IA X RA = 4. 5 Amps x 2 Ω = 9 v In a parallel circuit these are NOT these should EACH equal voltage being supplied by battery, 9. 0 volts.

Rtotal = IA = IB = VA = VB =

Rtotal = IA = IB = IC = VA = VB = VC =