CIRCUITS: Series & Parallel


 Emerald Ramsey
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1 CIRCUITS: Series & Parallel
2 Last Week s BIG IDEAS: Opposite charged objects attract Like charged objects repel
3 Last Week s BIG IDEAS: The electrons are the loose particles that move to make things charged not the protons. Protons are locked in more solid relationships in the nucleus and don t get around.
4 Last Week s BIG IDEAS: The attractive force between particles/objects depends on the amount of charge (in Coulombs) the objects have and the distance (in m) between the two particles/objects. The above relationship is called Coulomb s Law and is given by: F e = k q 1 q 2 r 2
5 Last Week s BIG IDEAS: Coulomb s Law of attraction: F e = k q 1 q 1 r 2 For the eensybeensy particles like atoms, electrons, and protons Is a lot like Newton s Law of Universal Gravitation F g = G m 1 m 2 r 2 for the big particles like planets. The only difference is that Newton s Law is only attractive and Coulomb s Law can be both attractive and. repulsive
6 Last Week s BIG IDEAS: Planets behave like particles to some degree
7 Last Week s BIG IDEAS: Both Coulomb s Law of attraction  F e = k q 1 q 1 r 2 and Newton s Law of Universal Gravitation  F g = G m 1 m 2 r 2 are examples of forces at a distance (far away) where the objects are not touching and can be far away. These are different from contact forces such as friction, tension, force of push or pull in which the objects are in direct contact with each other.
8 Voltage (Electric Potential Difference), Current, and Resistance What is this illustration showing? What does voltage do? What does current (ampere) represent? And what does resistance do?
9 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) elastic Electric potential energy is like potential energy and gravitational potential energy Higher Potential Energy + GPE Lower Potential Energy  ELPE EPE
10 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) To cause movement of a charge, there must be an electric potential difference between two points in the electric circuit just as there could be a difference in water pressure between two points in a plumbing system.
11 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) In the image below, there is more electrical pressure at the 6 V side of the battery than the 0 V side. Imagine that the 6 V side is negatively () charged and so electrons are pushed away (like charges repel) through the circuit towards the side of the battery which is positively (+) charged to which the electrons are (opposite attracted charges attract). +
12 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) In this circuit the energetic electrons have two paths they can go through. In the middle of each path is something that is resisting the flow of electrons and thus they struggle to get through. In the struggle they some use up of their electrical energy which turns into light and heat energy. The electrons are moving slower after the work (in Joules) they have performed. DIRECTIONS: Use the right terms below in the blanks above WORD BANK resisting two heat Use up work +
13 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) Suppose there was a U. S. Navy nuclear submarine crew that just happened to be all guy sailors. They just got off their 3month tour on their boat and are repelled from the sight of each other. They are seeking out members of the opposite and sex go down to the only two clubs on this small island they are docked at to partyhardy. Once they ve used up their energy they head back tired to their. submarine DIRECTIONS: Use the right terms below in the blanks above WORD BANK sex submarine used Navy repelled +
14 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) Potential energy Potential energy For example, to push a positively charged Styrofoam pellet into a positively charged electric field requires work (a force through a distance) The Styrofoam gains electrical potential energy in the process Since like charges, repel the charged Styrofoam moves away from the Van de Graaf generator converting its electrical potential energy into kinetic energy
15 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) DIRECTIONS: Put the correct term in the correct blank below. WORD BANK random section Prom speeds open In an electrical circuit, while the switch is : open Free electrons (conducting electrons) are always moving in random motion. speeds The random are at an order of 1,000,000 m/s. There is no net movement of charge across a cross section of a wire. It is all back and forth and random. Its like the random movement of single kids dancing at. Prom
16 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) closed What occurs in a wire when the circuit switch is? An electric field is established instantaneously (at almost the speed of, light 3x10 8 m/s). Free electrons, while still randomly moving, immediately begin drifting due to the electric field, resulting in a net flow of charge. They are like users who think have free will but are in fact being manipulated to move in certain directions. Average drift velocity is about 0.01cm/s. The effect of the manipulation on FB users is small but since there are billions of them, like electrons, the overall effect is large. WORD BANK users closed light free billions
17 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) What occurs in a wire when the circuit switch is closed and electricity begins to flow? Much like a group of students going to the bus circle after school, they slowly drift out at school in a messy but deliberate pattern. The electric field is kind of like monitors hall telling students to go home and get out of the building.
18 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) The electric potential V (or voltage) is the potential for creating electric potential energy if a charge is placed at a given point. It is defined as the energy per unit charge: V = U elec q Where U elec is the electric potential energy (in Joules) and q is the charge of an object in Coulombs. 1 volt = 1 V = 1 Joule per Coulomb, = 1 J/C An electric field exerts an electrical force F E on a charge q. An electric potential V tells how much electrical energy U elec each charge q has.
19 VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE) Another definition. Voltage: Electrical pressure that pushes charges along a conducting pathway. Metric Unit for Voltage: Volt (V)
20 high Electric Charges always travel from a electrical pressure region to a pressure region. low
21 1 AMPERE of current carries about electrons per second CURRENT Electric Current: The flow of electrons through a conducting electrons pathway. It relates to how many pass a given place in a wire in one second. Metric Unit for Current: 1 ampere (A) = 6.24 x e  /s
22 RESISTANCE resist Resistance: The tendency of a substance to the flow of e  s. Metric Unit: The force opposing e  movement is measured in Ohms (Ω).
23 RESISTANCE All materials have some resistance, insulators have a large amount of resistance, while conductors have very little resistance. As es flow through material and collide with atoms, electrical energy is converted into two forms of energy: and. heat light
24 RESISTANCE The electricity has two paths to flow through. The 10 Ohm resistor allows three times as many electrons/ions to flow as the 30 Ohm resistor.
25 RESISTANCE Variable resistor The knob turning is applying variable resistance to the circuit. This can be seen because the  light level of the LED is fluctuating. The variable resistor is controlling the number of electrons (e  s) passing through the circuit, or, in other words, the current (in Amps).
26 CURRENT, VOLTAGE, & RESISTANCE Make an analogy for voltage, resistance, and current like the ones below
27 Electrical Circuits
28 Circuit Symbols: Battery Resistor Lightbulb Switch Wire DIRECTIONS: Draw lines from each term to the corresponding item in the circuit diagram at right
29 Three general types of circuits: DIRECTIONS: Put the correct term in the correct blank below. WORD BANK Power source Short resistance Closed current Open Closed Circuit  There is a complete loop with wires going from one side of the power source through a resistor(s) to the other side of the. Open Circuit  There is not a complete loop for to flow. Short Circuit  There is a complete loop, but it does not contain any to the flow of electricity.
30 Three general types of circuits: Closed Circuit  There is a complete loop with wires going from one side of the power source through a resistor(s) to the other side of the. Open Circuit  There is not a complete loop for to flow. Short Circuit  There is a complete loop, but it does not contain any to the flow of electricity. Only Working Circuit WORD BANK Power source Short resistance Closed current Open
31 There are two ways to put resistors into a circuit. 1. Resistors can be in series OR 2. Resistors can be in parallel
32 Resistors in Series Resistors are considered to be in series if the current must go through all of the resistors in order. The current (amps) through all resistors in series is the same. The voltage across resistors in series may be different The rate of electron flow (or current) is determined by which resistor? Ans. The resistor with the largest amount of ohms. R 1 R 2 R 3
33 Combining (adding) Resistors Series Resistors R 1 To find the total current just add up the currents through each resistor: I total = I 1 = I 2 = I 3 R 2 To find the total resistance, just add up the resistances of each resistor: R eq = R total = R 1 + R 2 + R 3 R 3 Voltage is calculated with Ohm s Law Amps I = V R Q
34 Resistors in Parallel Resistors are considered to be in parallel if the current is shared between multiple resistors. The current (amps) through all resistors in parallel may be different. The voltage across all parallel resistors is the same. Will a resistor with a large resistance have more or less current through it then a resistor with a small resistance? Ans. The resistor with a large resistance will have a smaller current then the resistor with the smaller resistance. R 1 R 2 R 3
35 Combining (adding) Resistors Parallel Resistors Current is calculated with Ohm s Law: I = V R R 1 R 2 R 3 Total resistance: = + + R R R total 1 V total = V 1 = V 2 = V R 3
36 Example 1: A circuit has three resistors  an 8.0 W, 5.0 W and a 12 W resistor  in series along with a 24 V battery. Draw the circuit. Calculate the total resistance of the circuit. Calculate the total current through the circuit. What is the current through each resistor? Calculate the voltage across each resistor.
37 P.O.D. 1: A circuit has three resistors  a 6.0 W, a 4.0 W and a 9 W resistor  in series along with a 36 V battery. Draw the circuit. Calculate the total resistance of the circuit. Calculate the total current through the circuit. What is the current through each resistor? Calculate the voltage across each resistor.
38 Example 2: A circuit has three resistors: 6.0 W, 4.0 W and a 12 W resistors in parallel along with a 24 V battery. Draw the circuit. Calculate the total resistance of the circuit. Calculate the total current through the circuit. What is the voltage across each resistor? Calculate the current across each resistor.
39 P.O.D. 2: A circuit has three resistors: 4.0 W, 6.0 W and a 8 W resistors in parallel along with a 30 V battery. Draw the circuit. Calculate the total resistance of the circuit. Calculate the total current through the circuit. What is the voltage across each resistor? Calculate the current across each resistor.
40 Electrical Outlets Electrical outlets provide electric potential (or the voltage) for any appliance plugged in to it. In the United States ALL outlets provide 120 V (in Europe it is 240 V)
41 Example 3: What will the current be if an Americanmade 55 W light bulb is plugged in to a 220 V power source? We have worked with Power before in our Energy Unit. It is measured in Watts It was given by the formula P = W t For electricity we have three alternative shortcut formulas for Power based on what is given in the problem: P = I 2 R, P = I V, P = V2 For this problem we use which one? P = I V Solving for I I = P V SOLUTION: I = P 55 W = =0.25 Amps V 220 V R
42 P.O.D. 3: What will the resistance be if a Europeanmade 55 W light bulb is plugged in to a 210 V power source? What about an Americanmade 55 W light bulb?
43 MULTIPLE CHOICE: As more identical resistors R are added to the parallel circuit shown, the total resistance between points P and Q R 1. Increases 2. Remains the same 3. Decreases P Q
44 MULTIPLE CHOICE: As more identical resistors R are added to the parallel circuit shown, the total resistance between points P and Q 1. Increases 2. Remains the same 3. decreases R P Q Q
45 MULTIPLE CHOICE: When one bulb is unscrewed, the other bulb will remain lit in which circuit 1. I 2. II 3. Both 4. Neither Circuit II Circuit I
46 MULTIPLE CHOICE: When one bulb is unscrewed, the other bulb will remain lit in which circuit 1. I 2. II 3. both 4. neither Circuit I Circuit II
47 EXAMPLE 4: A 25W bulb and a 100W bulb are connected in series. Which bulb will glow brighter?
48 25W 100W 120V
49 The Light Bulbs are really Resistors A) Calculate the resistance for each resistor shown. B) Calculate the total resistance of the circuit. C) Calculate the current through each resistor. D) Calculate the power used by each resistor. E) Calculate the voltage across each resistor. 25W 100W 120V
50 25W Bulb Part A. 100W Bulb V P = R 2 V R = P R = 25 R = 576W P = V2 R is the formula for electrical Power. Another formula for Power is P = I V R = R = 144W
51 B) The total resistance (R total ) 25W 576 W 100W 144 W 120V + = 576 W W = 720 W
52 C) Calculate the total circuit current (I) 720 W 120V The current in a series circuit is the same throughout Use Ohm s Law: I = V R 120V = 720 W = 0.17amps
53 D) Calculate the Power used by each resistor. 576 W 144 W 25 W Bulb 100 W Bulb 120V P 1 = 25 W P 2 = 100 W
54 E) Calculate the Voltage across each resistor. 576 W 144 W =.167 amps 120V 25W Bulb 100W Bulb Use Ohm s Law: V = I R = (0.167 amps)(576 W) = 96.2 V Use Ohm s Law: V = I R = (0.167 amps)(144 W) = 24 V
55 P.O.D. 4: B) Calculate the total resistance of the circuit. A) Calculate the total current in the circuit. C) Calculate the current through each resistor. D) Calculate the power used by each resistor. E) Calculate the voltage across each resistor. 200 W 300 W 120V
56 The circuit below consists of two identical light bulbs burning with equal brightness and a single 12V battery. When the switch is closed, the brightness of bulb A A 1. Increases 2. Decreases 3. Remains unchanged
57 The circuit below consists of two identical light bulbs burning with equal brightness and a single 12V battery. When the switch is closed, the brightness of bulb A 1. Increases 2. decreases 3. remains unchanged When the switch is closed, bulb B goes out because all of the current goes through the wire parallel to the bulb. Thus, the total resistance of the circuit decreases, the current through bulb increases, and it burns brighter. A Q
58 Which bird is in trouble when the switch is closed? 1) Bird 1 2) Bird 2 3) Neither 4) Both 1 2
59 Which bird is in trouble when the switch is closed? 1) Bird 1 2) bird 2 3) neither 4) both 1 2
60 Charge flows through a light bulb. Suppose a wire is connected across the bulb as shown. When the wire is connected 1. All the charge continues to flow through the bulb, and the bulb stays lit. 2. Half the charge flows through the wire, the other half continues through the bulb. 3. Essentially all the charge flows through the wire and the bulb goes out. 4. None of these. Q
61 EXAMPLE: Analyze the circuit: A) Calculate R total B) Calculate the current through each resistor. C) Calculate the voltage through each resistor. 8 W 60V 8 W 16 W 4 W
62 EXAMPLE: Analyze the circuit: A) Calculate R total B) Calculate the current through each resistor. C) Calculate the voltage through each resistor. Use the formula for resistors in Parallel: 1 R R 120V 1 2 R total = = = 1 R = R 8W R = W 16W 32W 32W
63 EXAMPLE: Analyze the circuit: A) Calculate R total B) Calculate the current through each resistor. C) Calculate the voltage through each resistor. 16W 120V Replace the three resistors with one equivalent resistor 16W 32W 8W32W
64 120V Find the resistance in Series: R total = R 1 + R 2 + etc. R = R 1234 =24 16W 8W
65 Make a table with the following: Make chart: R I V 16W R V R W 32W 32W R 3 32 R 4 32 R R
66 Make chart: These are in parallel so their voltage is the same along with the total voltage. In Parallel circuits: V 1 = V 2 = V 3 = etc. 120V 16W 16W 32W 32W All these numbers will be the same. R I V R 1 16 R 2 16 R 3 32 R 4 32 R R
67 120V Make chart: 16W 8W All these numbers will be the same. These are in series so their current is the same along with the total current: I 1 = I 2 = I 3 = etc. R I V R 1 16 R 2 16 R 3 32 R 4 32 R R
68 Fill out the chart with V=IR V = IR 120 = I (24) I = 5 A V = IR V = (5) (16) V = 80 V R I V R R 2 16 R 3 32 R 4 32 R R
69 Fill out the chart with V=IR V = IR 120 = I (24) I = 5 A V = IR V = (5) (8) V = 40 V V = IR V = (5) (16) V = 80 V V = IR 40 = I (16) I = 2.5 A V = IR 40 = I (32) I = 1.25 A R I V R R R R R R
70 Fill out the chart with V=IR V = IR 120 = I (24) I = 5 A V = IR V = (5) (8) V = 40 V V = IR V = (5) (16) V = 80 V V = IR 40 = I (16) I = 2.5 A V = IR 40 = I (32) I = 1.25 A R I V R R R R R R
71 Another way to do the problem (without the chart) I=V/R 120V I=120v/24W I=5 amps 24W
72 V=IR V=(5)(16) V=80volts 16W 80volts 120V V=IR V=(5)(8) V=40volts 5amps 8W 40volts 120volts
73 I=V/R =40volts/16 W =2.5 amps 5 amps 16W 80volts 120V I=V/R =40volts/32 W =1.25 amps 16W 32W 32W 40 volts 5 amps
74 When the series circuit shown is connected, Bulb A is brighter than Bulb B. If the positions of the bulbs were reversed 1. Bulb A would again be brighter 2. Bulb B would be brighter 3. They would be equal brightness
75 When the series circuit shown is connected, Bulb A is brighter than Bulb B. If the positions of the bulbs were reversed 1. Bulb A would again be brighter 2. Bulb B would be brighter 3. They would be the same The bulbs are connected in series, so the same current passes through both of them. Different brightnesses indicate different filament resistances. Bulb A is NOT brighter because it is first in line for the current of the battery! After all, electrons deliver the energy, and they flow from negative to positive  in the opposite direction!
76 P. O. D. 5: Find the voltage and current for each resistor. 3 W 3 W 6 W 6 W 4 W 12 W 2 W 18 volts
77 3W 3W 4W 12W 6W 6W 2W 18 volts
78 3W 3W 3W 4W 12W 2W 18 volts
79 3W 3W 3W 4W 12W 2W 18 volts
80 6W 3W 4W 12W 2W 18 volts
81 6W 3W 1 1 = + R total R 1 1 R 2 4W 12W 2W 1 R total = volts R total =3W
82 6W 3W 1 1 = + R total R 1 1 R 2 3W 2W 1 R total = volts R total =3W
83 6W 3W 3W 2W 18 volts
84 6W 3W 5W 18 volts
85 6W 1 R total R total 3W 1 = = 2. 73W 5W 18 volts
86 3W 2.73W 18 volts
87 3W 2.73W 18 volts
88 5.73W 18 volts
89 Now, find the total current flowing I V I = R 18volts = 5.73W I = amps 5.73W 18 volts
90 9.42volts 3W 3W V=IR V=(3.14)(3W) 4W V= W 6W 6W 2W 18 volts
91 volts 9.42volts 3W 3.14 amps 3W 4W 12W 6W 6W 2W 18 volts
92 volts 9.42volts 6W 3W 3.14 amps 4W 12W 2W 18 volts
93 volts 9.42volts 6W 3W 3.14 amps 4W 12W 2W 18 volts
94 volts 9.42volts 6W 3W 3.14 amps 5W 18 volts
95 volts 9.42volts I = 3W I = V R 8.57volts 6W I =1. 43amps 6W 5W 18 volts
96 volts 9.42volts 3W 3.14 amps 6W 1.43 amps 5W 1.71 amps 18 volts
97 volts 9.42volts 3W 3.14 amps 1.71 amps V=IR V=(1.71)(2) V=3.42volts 6W 1.43 amps 4W 12W 18 volts 3.42Volts 2W 1.71 amps
98 volts 9.42volts 6W 1.43 amps 3W 3.14 amps 1.71 amps 5.15 volts 4W 12W 3.42Volts 2W 1.71 amps 18 volts
99 I=V/R I=5.15volts/12W 9.42volts I= 0.43 amps 3W 3.14 amps 1.71 amps 6W 4W volts 8.57volts 12W 0.43 amps 1.43 amps 3.42Volts 2W 1.71 amps 18 volts
100 I=V/R I=5.15volts/4W 9.42volts I= 1.28 amps 3W 3.14 amps 1.71 amps 6W 4W Or 8.57volts 1.71 amps 0.43 = 1.28 amps 5.15 volts 12W 0.43 amps 3.42Volts 2W 1.71 amps 18 volts
101 3W 3W 4W 12W 6W 6W 2W 18 volts
102 3W 3W 4W 12W 6W 6W 2W 18 volts
103 3W 3W 4W 12W 6W 6W 2W 18 volts Q
104 Given: R1=1W; R2=2 W; R3=3 W. Rank the bulbs according to their relative brightness 1. R 1 > R 2 > R 3 2. R 1 > R 2 = R 3 3. R 1 = R 2 > R 3 4. R 1 < R 2 < R 3 R 1 R 2 R 3 5. R 1 = R 2 = R 3
105 Given: R1=1W; R2=2 W; R3=3 W. Rank the bulbs according to their relative brightness R 1 > R 2 > R 3 2. R 1 > R 2 = R 3 3. R 1 = R 2 > R 3 4. R 1 < R 2 < R 3 R 1 R 2 R 3 5. R 1 = R 2 = R 3 P = IV = I 2 R = V 2 R Q
106 If the four light bulbs in the figure below are identical, which circuit puts out more total light? 1. I 2. II 3. Same Circuit II
107 If the four light bulbs in the figure below are identical, which circuit puts out more total light? 1. I 2. II 3. Same Circuit II The resistance of two light bulbs in parallel in smaller than that of two bulbs in series. Thus the current through the battery is greater for circuit I than for circuit II. Since the power dissipated is the product of current and voltage, it follows that more is dissipated in circuit I.
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