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**Circuits & Circuit Diagrams**

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**Circuits and Circuit Diagrams**

Circuit = Complete path along which electrons can flow When drawing circuits…use the following symbols in your diagrams Wire/Conductor Resistor (light bulbs, fans) Battery Switch

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**Ohm’s Law Resistance = Voltage/ Current R = V / I Units**

Resistance = ohms (Ω) Voltage (Potential Difference) = volt (V) Current = ampere (A)

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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

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**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) Total resistance to current in the circuit is the sum of the individual resistors along the circuit path (Rtotal = RA + RB) 9.0 V 2.0 Ω 3.0 Ω A B

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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 (Rtotal = RA + RB)

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**Parallel Circuits 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

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Combining Resistors Total Resistance or Equivalent resistance in series = sum of all individual resistances Rtotal = RA + RB Parallel – the inverse of the total resistance is the sum of the inverse of the resistors 1/Rtotal = 1/RA + 1/RB

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**Equations 1/Rtotal= 1/R1 + 1/R2 + … Series R= R1 + R2 + ….. Parallel**

I=V/Rtotal R= R1 + R2 + ….. Parallel I=V/R 1/Rtotal= 1/R1 + 1/R2 + …

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**Measures the rate at which energy is transferred **

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

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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

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**Practice… Req = 2 Ω + 3 Ω = 5 Ω IA = V/Rtotal……9 v / 5 Ω = 1.8 Amps**

2.0 Ω 3.0 Ω A B Req = 2 Ω + 3 Ω = 5 Ω IA = V/Rtotal……9 v / 5 Ω = 1.8 Amps IB = VA = IA x RA = 1.8 Amps x 2 Ω = 3.6 volts add these and you should get the voltage supplied by the battery, 9 volts VB = IB x RB = 1.8 Amps x 3 Ω = 5.4 volts

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**Practice… Req => 1/ Req = 1/2 + 1/3 = 0.833,**

9.0 V A B 2.0 Ω 3.0 Ω Req => 1/ Req = 1/2 + 1/3 = 0.833, but remember 1/ Req = (rearrange and solve for Req) so Req = 1/0.833 = 1.2 Ω IA = V/RA = 9.0 V / 2 Ω = 4.5 Amps In a parallel circuit these are NOT identical. IB = V/RA = 9.0 V / 3 Ω = 3 Amps VA = IA X RA = 4.5 Amps x 2 Ω = 9 v these should EACH equal the voltage being supplied by the battery, 9.0 volts. VB = IB X RB = 3 Amps x 3 Ω = 9 v

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