© Shannon W. Helzer. All Rights Reserved. Unit 12 Electricity and RC Circuits

© Shannon W. Helzer. All Rights Reserved. Conduction Electrons  In organic compounds, electrons are bound to specific atoms.  In metallic compounds, some of the electrons are not bound to a specific atom.  They are free to move throughout the metal.  These electrons are called conduction electrons.  If a potential difference is placed across the wire (like when you connect the wire to a battery), then the electrons will move.  As they move, the electrons collide with the metallic atoms.  Depending upon the number of collisions an electron has, it may move faster or slower through the metallic structure.  Remember, moving electrons in a wire are known as current. 12-1

© Shannon W. Helzer. All Rights Reserved. Batteries  Batteries come in many shapes and sizes that have a variety of voltage and currents.  Identify the voltages of the four common battery types shown below. 12-1A

© Shannon W. Helzer. All Rights Reserved. Batteries  This slide will explain how a battery provides electricity for use in your small electrical appliances.  Free electrons and “holes,” which are the absences of electrons, are produced within the battery due to electrochemical reactions. 12-2

© Shannon W. Helzer. All Rights Reserved. Capacitors  Like batteries, capacitors come in a wide assortment of shapes.  A capacitor acts as a buffer to temporally store excess charges.  At other times a capacitor acts as a short lived battery in order to provide additional current (electrons) when the need is larger than a battery can provide.  Basically, a capacitor consists of two parallel plates.  One of these has a net positive charge while the other has a net negative charge.  In the next slide we will take a closer look at the operation of a capacitor. 12-3

© Shannon W. Helzer. All Rights Reserved. Capacitors  Observe the capacitor acting in the capacity of a short lived battery. 12-4

© Shannon W. Helzer. All Rights Reserved. Resistors  Resistors impede the flow of electrons (current).  They impede this flow because certain electrical items have maximum limitations on the current they can handle.  Consider the Light Emitting Diode (LED) in the figure to the right.  The battery supplies too high a current to the LED.  As a result, the LED is damaged by the current.  When a resistor is placed into the circuit, the current is reduced to a level appropriate for use with the LED, and the LED is not damaged. 12-5

© Shannon W. Helzer. All Rights Reserved. Batteries and Current  A complete circuit is one that connects a battery to an electrical component back to a battery.  Remember, current flows from the negative end of a battery through the light bulb (resistor) and back into the positive end of the battery.  The symbol “I” is used to denote current.  Current is the number of electrons passing a certain point in a circuit per unit of time.  We draw an arrow in the direction that the current would flow through the circuit. 12-6

© Shannon W. Helzer. All Rights Reserved. Multiple Batteries and Current  Depending on how they are placed, multiple batteries in a circuit can enhance or impede the current in a circuit.  Consider the circuit below. The current flowing through both batteries travels in the same direction.  As a result, the net current is enhanced in the circuit.  Now, let us flip one of the batteries.  When we flip the battery, its current now acts in the opposite direction.  As a result, it impedes the current.  Since both batteries are the same, no net current flows, and the bulb goes out. 12-7

© Shannon W. Helzer. All Rights Reserved. Multiple Batteries and Current  In this figure the batteries are of different voltages  As in the previous figure, when the batteries are connected in the same way, the net current is enhanced.  When one battery is flipped, the net current is impeded.  Determine the net current direction and the net voltage of the circuit below.  The key to determining the net current direction lies in considering the voltages of the batteries.  The net current will flow in the direction of the current belonging to the battery with the highest voltage. 12-8

© Shannon W. Helzer. All Rights Reserved. WS ???? #2  Determine the voltage and the net current direction in the circuits below Volt

© Shannon W. Helzer. All Rights Reserved. Circuit Configurations: Series Resistors  Look at the drawing animation below.  The skateboarders leaving the school only have one place to go.  They also only have one place to return to.  Their round trip occurs in a series circuit

© Shannon W. Helzer. All Rights Reserved. Circuit Configurations: Series Resistors  Now lets replace the school building with a battery.  Turn the house and restaurant into a light bulb (resistor).  Change the roads into wires.  Now let some electrons flow through the circuit.  They have only one destination: through the resistor and back the battery

© Shannon W. Helzer. All Rights Reserved. Circuit Configurations: Multiple Series Resistors  The lights below are in series with each other because the same current flows through all three of them.  Notice how the first bulb is brighter than each consecutive bulb thereafter.  This is because the electric potential (voltage) drops as it passes through the resistor.  As a result, there is less voltage available for the next resistor

© Shannon W. Helzer. All Rights Reserved. Circuit Configurations: Parallel Resistors  In a parallel circuit, the skateboarders have a choice of going home or to the Big Burger for lunch.  They make this choice at the intersection.  When there is an intersection resulting in a choice between two or more directions, the circuit is a parallel circuit

© Shannon W. Helzer. All Rights Reserved. Circuit Configurations: Parallel Resistors  Now let us make the appropriate replacements in this circuit and turn it into an electrical circuit.  Now allow the current to flow.  Notice how more electrons flow through the small bulb.  This action happens because the small bulb has a lower resistance than the large bulb.  We will look at resistance more in the next slide

© Shannon W. Helzer. All Rights Reserved. Resistance  Lets consider the circuit to now have two ramps.  Most unskilled skateboarders would choose the lower ramp.  As a result, we could say that the lower ramp has a lower resistance.  Now let us make the appropriate replacements.  In this replacement, we replaced the ramps with actual resistors instead of with light bulbs.  As you can see, more electrons flow through the orange resistor because it has a lower resistance

© Shannon W. Helzer. All Rights Reserved. + - Circuit Configurations: Multiple Parallel Resistors  The circuit to the right shows three parallel resistors (bulbs) that are in series with a fourth resistor.  When we close the top switch, the two lighted bulbs are in series, and the top bulb is slightly brighter than the bottom bulb.  They share equal currents.  What do you think will happen when we close the middle switch?  Notice that the top two bulbs were equally dimmed while the bottom bulb remained the same.  The currents through the top bulbs are equal but half that through the bottom bulb.  A similar result is observed when the bottom switch is closed

© Shannon W. Helzer. All Rights Reserved. A Closer Look – Series Resistors  When the switch to the right is closed, which bulb will be brighter?  Why?  The voltage drops across R 1 making R 2 less bright than R 1.  If we remove R 2, then what will happen to R 1 ?  Why?  The same current flows through both light bulbs (resistors).  If you remove one of these series resistors, then the current can not flow, and the bulbs go out R1R1 R2R2

© Shannon W. Helzer. All Rights Reserved. A Closer Look – Parallel Resistors  Note: R 2 = 2.0  and R 1 = 1.0   When the switch to the right is closed, which bulb will be brighter?  Why?  R 1 will be brighter because more current will flow through it due to the fact that it has less resistance.  If we remove R 1, then what will happen to R 2 ?  Why?  R 2 will get brighter because all of the current now passes through R 2 instead of being split between R 1 and R Volt R1R1 R2R2

© Shannon W. Helzer. All Rights Reserved. A Closer Look – Combination Resistors  Note: R 4 = 4.0 , R 3 = 3.0 , R 2 = 2.0 , and R 1 = 1.0 .  When the switch to the right is closed, which bulb will be brighter?  Why?  All of the current flows through R 1 before it splits after R 1.  What will happen to R 4 if we remove R 1 ?  Why?  All of the bulbs go out because the electrons can not flow through R 1.  What would happen to R 1, R 3, and R 4 if we removed R 2 ?  Why?  What would happen to R 2 and R 1 if we removed R 4 ?  Why? R1R1 R2R2 R3R3 R4R4

© Shannon W. Helzer. All Rights Reserved. Schematic Symbols  When engineers design electrical circuits, they replace actual pictures of electrical components with schematic symbols.  Here are five electrical components we will use frequently.  The schematic symbol for each of these electrical components is as follows.  Here are some other symbols you must be familiar with VA

© Shannon W. Helzer. All Rights Reserved. Parallel and Series Electrical Configurations  There are two basic electrical configurations: series and parallel.  In a series connection, all electrical components share the same current.  In a parallel connection, the current through each component varies depending upon the components resistance.  Let us take a look at the schematic diagrams for the circuits below.  In the lower left picture, the two resistors are series.  Note that we can move one of the resistors any where in the circuit while maintaining the same current through each.  In the lower right picture the resistors are parallel.  Again, we may move one resistor and still have a parallel circuit.  The resistors do not need to be geometrically parallel in order to be electrically parallel

© Shannon W. Helzer. All Rights Reserved. Equivalent Circuits  It is often desirable to reduce numerous electrical resistors in a circuit to an equivalent circuit with fewer resistors.  In a series circuit, the series resistors may be replaced with a single resistor with the equivalent resistance to that of the ones it replaced.  On paper, simply redraw the circuit with only one resistor in the place of the two (or more) you are replacing.  The same concept holds true for parallel resistors.  The exact same procedure is followed when doing equivalent circuits with capacitors instead of resistors

© Shannon W. Helzer. All Rights Reserved. Series Test  Series Component Test – must be able to go from only one side of a component to only one side of an equivalent component without passing an intersection or a nonequivalent component.  An intersection kills a series possibility.  Are R 4 and R 2 Series?  Are R 5 and R 3 Series?  Are R 4 and R 5 Series?  Are C 2 and C 1 Series?  Are C 3 and C 4 Series? R3R3 R1R1 R2R2 C2C2 C1C1 R4R4 C4C4 C3C3 R5R5

© Shannon W. Helzer. All Rights Reserved. Parallel Test  Parallel Component Test – must be able to go from both sides of a component to both sides of an equivalent component without passing through a nonequivalent component.  An intersection has no impact on a parallel circuit possibility.  Are R 3 and R 1 Parallel?  Are R 4 and R 2 Parallel?  Are C 2 and C 1 Parallel? R3R3 R1R1 R2R2 C2C2 C1C1 R4R4 C3C3

© Shannon W. Helzer. All Rights Reserved. Series & Parallel Test  The circuit to the right is a very complicated RC (Resistor-Capacitor) circuit.  Let us apply the tests for series and parallel circuits in order to reduce the circuit to its simplest form.  Always replace series components with their equivalent series component before attempting to replace parallel components.  Do you see any series components?  Once all series components are replaced, proceed to replace any parallel components that may remain.  Are the remaining resistors series?  Why or why not?  Once there are no longer any series or parallel components, the circuit is reduced as far as possible

© Shannon W. Helzer. All Rights Reserved. Equivalent Resistance for Series Circuits  When you replace series resistors with an equivalent resistance, you must calculate the value of the new resistor.  When replacing two series resistors with an equivalent resistance, use the following formula in calculating the equivalent resistance.  If you are replacing many resistors, use the following formula in order to calculate the equivalent resistance

© Shannon W. Helzer. All Rights Reserved. Equivalent Resistance for Parallel Circuits  When you replace parallel resistors with an equivalent resistance, you must calculate the value of the new resistor.  When replacing two parallel resistors with an equivalent resistance, use the following formula in calculating the equivalent resistance.  If you are replacing many resistors, use the following formula in order to calculate the equivalent resistance

© Shannon W. Helzer. All Rights Reserved.  When you replace series capacitors with an equivalent capacitance, you must calculate the value of the new capacitor.  When replacing two series capacitors with an equivalent capacitance, use the following formula in calculating the equivalent capacitance.  If you are replacing many capacitors, use the following formula in order to calculate the equivalent capacitance. Equivalent Capacitance for Series Circuits 12-28

© Shannon W. Helzer. All Rights Reserved. Equivalent Capacitance for Parallel Circuits  When you replace parallel capacitors with an equivalent capacitance, you must calculate the value of the new capacitor.  When replacing two parallel capacitors with an equivalent capacitance, use the following formula in calculating the equivalent capacitance.  If you are replacing many capacitors, use the following formula in order to calculate the equivalent capacitance

© Shannon W. Helzer. All Rights Reserved. WS ??? # 1 and WS ??? #1  Find the equivalent capacitance or resistance for the circuits in the following problems. WS ??? # 1. If C 1 = 5.0 F, C 2 = 25.0 F, and C 3 = 9.5 F, then C eq = ____? WS ??? # 1. If R 1 = 5.0 , R 2 = 25.0 , and R 3 = 9.5 , then R eq = ____? R1R1 R2R2 R3R3 C1C1 C2C2 C3C3

© Shannon W. Helzer. All Rights Reserved. WS ??? # 1 and WS ??? #1  Find the equivalent capacitance or resistance for the circuits in the following problems. WS ??? # 1. If C 1 = 5.0 F, C 2 = 25.0 F, and C 3 = 9.5 F, then C eq = ____? WS ??? # 1. If R 1 = 5.0 , R 2 = 25.0 , and R 3 = 9.5 , then R eq = ____? C3C3 C2C2 C1C1 R1R1 R2R2 R2R2

© Shannon W. Helzer. All Rights Reserved. WS ??? # 4  Find the equivalent capacitance or resistance for the circuits in the following problems.  If R 1 = 5.0 , R 2 = 25.0 , and R 3 = 9.5 , then R eq = ____? R1R1 R2R2 R3R3 R1R1 R2R2

© Shannon W. Helzer. All Rights Reserved. WS ??? # 4  Find the equivalent capacitance or resistance for the circuits in the following problems.  If C 1 = 12.0 F, C 2 = 25.0 F, C 3 = 5.0 F, and C 4 = 1.5 F, then C eq = ____? C3C3 C2C2 C1C1 C4C4 C4C4

© Shannon W. Helzer. All Rights Reserved. Resistor/Capacitor Circuits WS ??? # 2  Reduce the circuit below as far as possible by finding the equivalent capacitance and resistance. Draw the final circuit for each problem. C 1 = 22.8 F, C 2 = 2.3 F, C 3 = 5.9 F, C 4 = 5.0 F, R 1 = 2.2 , R 2 = 14.8 , R 3 = 9.5 , and R 4 = 12.0 . R2R2 R3R3 R1R1 C3C3 C2C2 C1C

© Shannon W. Helzer. All Rights Reserved. Ohm’s Law  Ohm’s Law gives us a mathematical expression relating the voltage (V), Current (I), and Equivalent Resistance (R) of a circuit.  Previously, we reduced the circuit below to its equivalent resistance.  If we do not know the voltage but we do know the current and the resistance, then we can use the equation to find the voltage.  If we know the voltage of the battery and the resistance, then we can find the current flowing through the resistor

© Shannon W. Helzer. All Rights Reserved. Ohm’s Law  You can use Ohm’s Law to calculate the current through a resistor if you know the voltage across the resistor and the resistance of the resistor.  Consider the parallel circuit below.  Suppose the voltage (V) of the battery in both circuits below is 10.0 V.  Since both sides of the battery are connected to both sides of both resistors, the voltage across both resistors would be 10 volts.  However, they would not have the same current because the current splits before it reaches the resistors.  You can use Ohm’s law in order to find the current through these resistors

© Shannon W. Helzer. All Rights Reserved. Ohm’s Law  Both sides of the resistors below are not connected to both sides of the battery.  As a result, they do not have the same voltage across them.  However, as these resistors are in series, they share the same current.  The voltage drop across R 1 (from A to B) is given by the equation below right (Ohm’s Law).  The voltage would also drop across R 2 and can be calculated with the equation below left

© Shannon W. Helzer. All Rights Reserved. Ohm’s Law Example 1  R 1 = 10.0 , R 2 = 20.0 , and R 3 = 30 .  What is the current through the circuit below?  What are the voltage drops through R 1, R 2, and R 3 ? R1R1 R2R2 R3R3

© Shannon W. Helzer. All Rights Reserved. + - Ohm’s Law Example 2  R 1 = 10.0 , R 2 = 20.0 , and R 3 = 30 .  What is the current through each resistor in the circuit below?  What are the voltage drops through R 1, R 2, and R 3 ? R1R1 R2R2 R3R3

© Shannon W. Helzer. All Rights Reserved. Power  Once we know both the current and the voltage across a resistor, we can determine the power consumed by that resistor.  The power consumed may be determined using the following equation.  The power consumed in R 1 in both circuits below may be determined as follows

© Shannon W. Helzer. All Rights Reserved. Batteries and emf  So far we have considered batteries as perfect sources of electrons meaning that all the electrons produced by the electrochemical reactions inside the battery are delivered to the electrical circuit.  However, this statement is not true because of the fact that the materials the battery is made of resist the flow of electrons produced by the battery itself.  We call this resistance the internal resistance (r) of the battery.  This internal resistance, when multiplied by the current flowing through the battery, reduces the electric potential (voltage) the battery can deliver to the circuit.  Consider the D-Cell battery below that provides a voltage (V) of 1.5 V to a circuit.  The actual electrical potential produced by the battery known as the electromotive force (emf or  ) is larger than the voltage V delivered.  The relationship between the emf and the battery voltage is given in the below equation where I is the current produced by the battery.  The schematic symbols for a battery with and without internal resistance is given below

© Shannon W. Helzer. All Rights Reserved. Electro Motive Force (emf)  The V below is the ideal value of the voltage across the battery (i.e. 9 Volts).  Although the potential difference across the terminals of a battery is V when no current is flowing, the actual potential difference is reduced when a current is flowing through the battery due to internal resistance (r)in the battery.  The actual potential difference is known as the electromotive force, .  When we consider the internal battery resistance, the figure below would change to look as follows.  The electromotive force may be found by using the following equation.  The internal battery resistance is treated like a regular resistor, R, when doing calculations BA V R BA R  r

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