Waterfalls and Electricity

Niagara (American side) and Horseshoe (Canadian side) Falls 176 feet high gallons per second

Angel Falls (Venezuela) 3212 feet high 2400 gallons per second

Low current, high voltage High current, low voltage

Current Electricity vocabulary: Voltage: Current: related to the energy of electrons flowing through a circuit; measured in volts the rate of flow of charge; Definition: 1 C/s = 1 Ampere = 1 A measured in amperes

Current Electricity vocabulary: Voltage: Current: related to the energy of electrons flowing through a circuit; measured in volts the rate of flow of charge; Definition: 1 C/s = 1 Ampere = 1 A measured in amperes

Experiment: Vary voltage across a light bulb and measure resulting current Voltage Current General: y = k x Specific: Voltage = k (Current) Voltage Current = k

V I = R Ohm’s Law Georg Ohm (1789 –1854) V = I R V = voltage I = current R = resistance

ex. The current flowing through a light bulb is 0.50 amperes and the voltage is 120 volts. What is the resistance? V = I R V I R = 120 V 0.50 A = R = 240 V/A= 240 ohms= 240 Ω Definition: 1 ohm = 1 Ω = 1 V/A

Resistance x e-e- + - As electrons move down wire, they collide with the particles of the wire ; cause the particles to vibrate ; wire heats up; energy-consuming process So resistance is the result of collisions between conducting electrons and the particles of the conductor

Resistance x e-e- + - Resistance - increases with length more length, more collisions - increases with temperature hotter conductor’s particles are already vibrating, and get in the way more often - decreases with cross-sectional area electrons have more room to move on a thicker wire - depends on the metal silver is the best conductor, followed by copper, then gold, etc.

ex. The voltage across an appliance is 120 volts and the resistance is 20 ohms. What is the current? V = I R I = V R = 120 V 20 Ω I = 6.0 A

Power P = I V ex. Find the power generated in the light bulb in the earlier example ( voltage = 120 V, current = 0.50 A ) P = I V = ( 0.50 A )( 120 V ) P = 60 A V= 60C s J C J s = 60 watts P = 60 W

Cost of Power Power is generated in power plants and transported to our homes by transmission lines We are charged for the energy we use P = W t Energy t = Energy = P t Unit: watt secondkilowatt hour kW h

Schematic Symbols for Drawing Circuits Battery + - Switch Light bulb Resistor Voltmeter V Ammeter A

Series Circuit: (1)Current is the same everywhere I T = I 1 = I 2 = I 3 VTVT V 1 V2V2 V 3 (2) Voltage is divided between resistors ITIT I1I1 I2I2 I3I3 (3) Resistors add to give equivalent resistance only one possible path for current V T = V 1 + V 2 + V 3 R2R2 R 3 R1R1 R eq = R 1 + R 2 + R 3

Parallel Circuit: (1)Current is divided between resistors VTVT V1V1 I T = I 1 + I 2 + I 3 V2V2 V3V3 (2) Voltage is the same everywhere V T = V 1 = V 2 = V 3 (3) Equivalent resistance is a reciprocal relationship I1I1 I2I2 I3I3 ITIT more than one possible path for current R1R1 R2R2 R3R3 1 R eq = 1 R1R1 + 1 R2R2 1 R3R3 +

ex. A 20-Ω resistor, a 10-Ω resistor, and a 30-Ω resistor are connected in series with a 120-V power supply. (a) Draw the schematic diagram. 120 V10 Ω 30 Ω 20 Ω

ex. A 20-Ω resistor, a 10-Ω resistor, and a 30-Ω resistor are connected in series with a 120-V power supply. (b) Find the equivalent resistance. 120 VR eq R eq = R 1 + R 2 + R 3 = (20 Ω) + (10 Ω) + (30 Ω) R eq = 60 Ω 120 V10 Ω 30 Ω 20 Ω

ex. A 20-Ω resistor, a 10-Ω resistor, and a 30-Ω resistor are connected in series with a 120-V power supply. (c) What current comes out of the power source? 120 V60 Ω10 Ω 30 Ω 20 Ω 120 V ITIT ITIT V = I RI = V R eq = 60 Ω 120 V I = 2.0 A Use simpler circuit

ex. A 20-Ω resistor, a 10-Ω resistor, and a 30-Ω resistor are connected in series with a 120-V power supply. (c) What current comes out of the power source? 120 V60 Ω10 Ω 30 Ω 20 Ω 120 V 2.0 A V = I RI = V R eq = 60 Ω 120 V I = 2.0 A ITIT

ex. A 20-Ω resistor, a 10-Ω resistor, and a 30-Ω resistor are connected in series with a 120-V power supply. (c) What current comes out of the power source? 120 V60 Ω10 Ω 30 Ω 20 Ω 120 V 2.0 A V = I RI = V R eq = 60 Ω 120 V I = 2.0 A Circuits are equivalent, so currents are the same in each circuit 2.0 A

120 V V 1 V A I1I1 I2I2 I3I3 10 Ω 30 Ω 20 Ω V2V2 (d) What is the voltage drop across each resistor? 2.0 A V 1 = I 1 R 1 = (2.0 A)(20 Ω) V 1 = 40 V V 2 = I 2 R 2 = (2.0 A)(10 Ω) V 2 = 20 V V 3 = I 3 R 3 = (2.0 A)(30 Ω) V 3 = 60 V Current is the same everywhere = I R 1 = I R 2 = I R 3 Note that V 1 + V 2 + V 3 = 120 V

ex. Three resistors of 40 Ω, 30 Ω, and 40 Ω are connected in parallel across a 120-V potential difference. (a) Draw the schematic diagram. 120 V 40 Ω30 Ω40 Ω

ex. Three resistors of 40 Ω, 30 Ω, and 40 Ω are connected in parallel across a 120-V potential difference. (b) Find the equivalent resistance of the circuit. 120 V 40 Ω30 Ω40 Ω120 VR eq 1 = 1 R1R1 + 1 R2R2 1 R3R3 += = = = 1 R eq = 12 Ω

ex. Three resistors of 40 Ω, 30 Ω, and 40 Ω are connected in parallel across a 120-V potential difference. (c) What current is coming out of the power source? 120 V 40 Ω30 Ω40 Ω120 V12 Ω ITIT ITIT V = I RI = V R eq = 12 Ω 120 V I = 10 A

(d) What is the current through each resistor? 120 V I1I1 I2I2 I3I3 ITIT I 1 = V1V1 R1R1 = V R1R1 = 120 V 40 Ω = I 1 = 3.0 A I 2 = V2V2 R2R2 = V R2R2 = 120 V 30 Ω = I 2 = 4.0 A Voltage is the same everywhere I 3 = V3V3 R3R3 = V R3R3 = 120 V 40 Ω = I 3 = 3.0 A Note that I 1 + I 2 + I 3 = 10 A, the total current from (c)

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (a) Draw the schematic diagram of the circuit. There are a number of equivalent ways to draw this circuit. The following is only one way. Let’s draw it in order of what is written. So, first, the resistors in parallel: 15.0 Ω

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (a) Draw the schematic diagram of the circuit. There are a number of equivalent ways to draw this circuit. The following is only one way Ω Then add the 10.0 Ω resistor in series: 10.0 Ω

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (a) Draw the schematic diagram of the circuit. There are a number of equivalent ways to draw this circuit. The following is only one way Ω Then place the whole circuit across 45.0 V of potential difference: 10.0 Ω 45.0 V

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (b) What is the equivalent resistance of the parallel part of the circuit? 15.0 Ω 10.0 Ω 45.0 VR’ 10.0 Ω 45.0 V 1 R’ = 1 R1R1 + 1 R2R2 1 R3R3 += = 3 1 R’ R’ = 15 3 = R’ = 5 Ω

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (c) What is the equivalent resistance of the circuit? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V R eq 45.0 V R eq = R’ + R 4 = ( 5 Ω ) + ( 10.0 Ω ) R eq = 15.0 Ω

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (d) What is the current coming out of the power source? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V 15.0 Ω45.0 V I T = V R eq = 15.0 Ω 45.0 V I T = 3.00 A ITIT ITIT Use circuit that has only one battery and one resistor

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (e) What is the voltage drop across the 10-Ω resistor? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V 15.0 Ω45.0 V V 1 3 A There are three ways for electrons to go around this circuit (one is highlighted above) V 1 = I 1 R 1 I 1 = ? I1 I1

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (e) What is the voltage drop across the 10-Ω resistor? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V 15.0 Ω45.0 V V 1 3 A In each of the three ways, the electrons go through the 10-Ω resistor V 1 = I 1 R 1 I 1 = ? I1 I1 so I 1 = 3.00 A

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (e) What is the voltage drop across the 10-Ω resistor? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V 15.0 Ω45.0 V V 1 3 A V 1 = I 1 R 1 I 1 = ? I1 I1 = ( 3.00 A )( 10.0 Ω ) V 1 = 30.0 V

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (f) What is the voltage drop across the parallel part of the circuit? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V 30 V 3 A I1 I1 In this case, there are two ways to determine V’ Recall that the electrons go through two resistors on their way around the circuit V’ ; they come out of the battery with 45 V of energy, and drop 30 V in the 10-Ω resistor

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (f) What is the voltage drop across the parallel part of the circuit? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V 30 V 3 A I1 I1 So the electrons that go through any of the other 15-Ω resistors drop the remaining 15 V ( 45 V - 30 V = 15 V ) V’ So V 1 = 15 V This only works when the electrons go through only two resistors; what if they go through three, or more?

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (f) What is the voltage drop across the parallel part of the circuit? 15.0 Ω 10.0 Ω 45.0 V5 Ω 10.0 Ω 45.0 V V’ 3 A I1 I1 V’ 3 A Use the first equivalent circuit that is all in series ; the voltage drop across the 5-Ω resistor is the same as the voltage drop across the parallel part of the original circuit ; and, since the current is the same everywhere in a series circuit, V’ is easy to find V’ = I R’= ( 3 A )( 5 Ω )= V’ = 15 V

ex. Three  resistors are connected in parallel. This arrangement is connected in series with a  resistor. The entire circuit is then placed across a 45.0-V potential difference. (g) What is the current in each branch of the parallel part of the circuit? 15.0 Ω 10.0 Ω 45.0 V V’ =15 V 3 A I2 I2 V = I R = I = V R I 2 = V2V2 R2R2 15 V 15 Ω = I 2 = 1.0 A = I 3 = V3V3 R3R3 15 V 15 Ω = I 3 = 1.0 A = I 4 = V4V4 R4R4 15 V 15 Ω = I 4 = 1.0 A I3 I3 I4 I4

ITIT I1I1 I2I2 I3I3 I4I4 I T = I 1 + I 2 + I 3 + I 4 - When current passes through wires, they wires get hot - As more resistors are added to a parallel circuit, the total current out of the power supply increases; wires could overheat and cause a fire - Use a fuse to prevent this

thin wire Fuses As current passes through the fuse, the thin wire heats up When current reaches a certain value, the wire melts, cutting off the current through the fuse metal contacts insulator

ITIT I1I1 I2I2 I3I3 I4I4 Schematic symbol for fuse: Fuses are in series with the rest of the circuit

; when thin wire melts, current is cut off Schematic symbol for fuse: Schematic symbol for circuit breaker: Circuit breakers protect circuits also, and are re-useable

ITIT I1I1 I2I2 I3I3 I4I4 Fuses are in series with the rest of the circuit ; when thin wire melts, current is cut off Schematic symbol for fuse: Schematic symbol for circuit breaker: Circuit breakers protect circuits also, and are re-useable circuit breaker