1. Preview Section 1 Schematic Diagrams and Circuits
Section 2 Resistors in Series or in Parallel
Section 3 Complex Resistor Combination

2. What do you think?
Scientists often use symbols to represent electrical components, such as batteries, bulbs, and wires. On the next slide, you will see the symbols for eight common electrical components that you have seen and discussed previously.
Predict the component shown by looking at each symbol.
Briefly explain why you think each symbol represents that particular electrical component.
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting students’ ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.

3. What do you think? 1 2 3 4 5 6 7 8
Allow students a few moments to try to analyze the symbols and determine what they might represent. You might point out the similarities between 2 and 3, as well as between 6 and 7, to help students.
The next slide has a Visual Concept that is Table 1 from the text. This Visual Concept slide will allow discussion about each of students’ predictions one at a time.

4. Schematic Diagram and Common Symbols
Click below to watch the Visual Concept.
Visual Concept
Place the pointer on each image and then discuss the explanation in the right hand column. This should help students recall the symbols.

5. Schematic Diagrams
Schematic diagrams use symbols to represent components.
They show how the parts in an electrical device are arranged.
Point out the right angles used for connecting wires.

6. Electric Circuits
An electric circuit is a set of components providing a complete, closed-loop path for the movement of electrons.
Called a closed circuit
If the path is broken, the electrons do not flow.
Called an open circuit
The circuit shown is closed.

7. Inside a Light bulb
A complete conducting path is established inside the light bulb.
The tip of the bulb (a) is connected to one side of the filament (see the black line).
The threads on the side of the bulb (c) are connected to the other side of the filament (see the white line).
Hold up a light bulb and point out the tip of the bulb and the threads. Show students that they are separated from each other by a non-conducting material.

8. Short Circuits A short circuit bypasses the light bulb or other load.
It is a closed circuit.
Electrons flow directly from - to + without passing through the bulb.
The current is large and the wire becomes hot.
Short circuits in homes can cause fires.
Fuses or circuit breakers are designed to turn off the electron flow if short circuits occur.
Try using the PhET simulation at:
Choose “Simulations,” then choose “Electricity, Magnets and Circuits,” and then choose “Circuit Construction Kit -DC version.”
With this simulator, you can build a a closed circuit (bulb, battery, switch, and wires) and see the electrons flow. You can then open the switch to see an open circuit, and you can add wires to go around the bulb and see a short circuit and the flames that ensue.
Other aspects of this simulator can be used later with series and parallel circuits. It has many features that allow the study of simple DC circuits, including a voltmeter and ammeter.

9. Potential Difference in Circuits
A device that increases the PE of the electrons, such as a battery, is a source of emf (electromotive force).
Not really a force, but a PE difference
Energy is conserved in electric circuits.
The potential difference (V) for the battery equals the energy converted into heat as the electrons move through the bulb.
Electrons gain energy (battery) and lose energy (bulb) as they make a complete trip.

10. Internal Resistance, EMF, and Terminal Voltage
Click below to watch the Visual Concept.
Visual Concept

11. Now what do you think?
Draw schematic diagrams showing each of the following circuits:
An open circuit including a battery, open switch, and bulb
A closed circuit including a battery, closed switch, and resistor
A short circuit including a battery, bulb, and closed switch
After allowing some time, show students the correct schematic drawings for each of the circuits. The short circuit should show a wire connecting the two sides of the bulb or it could show both connecting wires going to the same side of the bulb.

12. The student is expected to:
TEKS
5F design, construct, and calculate in terms of current through, potential difference across, resistance of, and power used by electric circuit elements connected in both series and parallel combinations

13. What do you think?
Figure (a) shows a single bulb and battery as seen before. Figures (b) and (c) each show two bulbs connected to the battery. The batteries and bulbs are all identical. Answer the three questions on the next slide and explain your reasoning.
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting students’ ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.

14. What do you think?
How will the brightness of (b) and (c) compare to each other and how does each compare to (a)? Explain.
How will the brightness of (d) and (e) compare to each other and how does each compare to (a)? Explain.
Compare the total current leaving the battery in each of the three circuits. Explain.
Be sure students explain WHY they believe the bulbs will be brighter or dimmer or the same. Do not bother to label the circuits as series or parallel yet. See if students can deduce the relative brightness based on their knowledge of electron flow and potential energy. It might be helpful to ask students if all three batteries provide the same potential energy to the electrons. Since they are identical, the V is the same for all three batteries. The third question refers to the total current, not the current in each bulb. The current in each bulb will be addressed later but it could also be used as a follow-up question for this slide.

15. Resistors in Series
Series describes components of a circuit that provide a single path for the current.
The same electrons must pass through both light bulbs so the current in each is the same.

16. Resistors in Series Vbattery= V1 + V2 Vbattery= IR1 + IR2
Conservation of energy
Vbattery= IR1 + IR2
Ohm’s law
Vbattery= I(R1 + R2)
Vbattery= IRequivalent
Requivalent = R1 + R2
Point out that the current is the same in each so there is not an I1 and an I2, but just an I.

17. Equivalent Resistance
Solving problems with series resistors:
Find the equivalent resistance.
Use Req with Ohm’s law to find V or I.
Use I and R1, R2, etc. to find V1, V2, etc.

18. Classroom Practice Problems
A 6.00 V lantern battery is connected to each of the following bulb combinations. Find the current in each circuit and the potential difference across each bulb.
One bulb with a resistance of 7.50 
Two bulbs in series, each with a resistance of 7.50 
Four bulbs in series, each with a resistance of 7.50 
Answers:
0.800 A, 6.00 V
0.400 A, 3.00 V each
0.200 A, 1.50 V each
For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.
Point out that the energy lost in each bulb decreases as bulbs are added, from 6.00 V to 3.00 V to 1.50 V.
Ask students how this would affect the brightness of the bulbs in each base.
It would be a good follow-up to this problem to change the resistance of the bulbs so that some have a resistance of 7.50  and others a resistance of 2.50 . This will help students see how the potential difference across the bulbs depends on the resistance. Both bulbs will have the same current. The energy drop is greater in the bulb with greater resistance.
You could also have students calculate the potential difference across each bulb for Sample Problem A in the text.

19. Resistors in Parallel
Parallel describes components providing separate conducting paths with common connecting points.
The potential difference is the same for parallel components.
Electrons lose the same amount of energy with either path.
Point out to students that the battery provides the same PE to all electrons. When they reach the junction point, they split, and on either path, they lose that energy before returning to the battery with zero PE.
Also point out that the current is not the same in each path because all electrons do not go through both bulbs. (This is also addressed on the next slide.) Depending on the resistance of the bulbs, more current may flow through one than the other. If the bulbs are identical, each will have half of the total current.

20. Resistors in Parallel Ibattery = I1 + I2 Vbattery= V1 = V2
Conservation of charge
Ohm’s law
Vbattery= V1 = V2
Potential energy loss is the same across all parallel resistors.
Because Vbattery= V1 = V2, the equation above reduces as follows:
Point out that the total current splits into two parts, so the total is the sum of the two.

21. Equivalent Resistance
Solving problems with parallel resistors:
Find the equivalent resistance.
Use Req with Ohm’s law to find V or Itotal.
Use V to find I1, I2, etc.

22. Classroom Practice Problems
Find the equivalent resistance, the total current drawn by the circuit, and the current in each resistor for a 9.00 V battery connected to:
One 30.0  resistor
Three 30.0  resistors connected in parallel
Answers:
30.0 , A, A
10.0 , A, A
For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.
Point out that the current in each bulb is the same for one bulb as it is for three bulbs. This is a consequence of the PE lost being the same for each bulb. The electrons’ potential difference drop is 9.00 V for every bulb.
Ask students how the brightness of the three bulbs would compare to that of the single bulb. Since current and potential difference are the same, all three bulbs would be as bright as the single bulb.

23. Comparing Resistors in Series and in Parallel
Click below to watch the Visual Concept.
Visual Concept

24. Summary Try using the PhET simulation at:
Choose “Simulations,” then choose “Electricity, Magnets and Circuits,” and then choose “Circuit Construction Kit -DC version.”
With this simulator, you can build a a closed circuit (bulb, battery, switch, and wires) and see the electrons flow.
You can build series and parallel circuits. When making a parallel circuit, make one resistance double the other, and you can see the electrons splitting off (two through the smaller resistor for every one going through the larger resistor). You also have the option of adding voltmeters and ammeters to the circuit so you can measure the potential difference and the current. When using the meters, be aware that this simulator does not treat the connecting wires as zero resistance components. It shows a slight potential difference drop as the electrons move through the wire.

25. Wiring Lights The series circuit shows a bulb burned out.
What will happen to the other bulbs?
Would this also happen in the parallel circuit?
Assuming the bulbs are identical:
Which circuit will draw more current?
In which circuit are the bulbs brighter?
All bulbs go out in the series circuit, but not in the parallel circuit. Some decorative lighting is wired in series using lower resistance bulbs. Adding bulbs in series has the advantage of reducing the current flowing into the circuit.
In newer lighting systems for the holidays, when one light goes out, the others remain lit. However, when a light is removed, they all go out.
Ask students how this circuit is designed. (The bulbs have both a filament and a higher resistance by-pass wire. So, if the filament goes out, the current can go through the parallel wire and the lights remain lit. Under normal operation, most current goes through the filament because the parallel wire has a higher resistance than the filament.)
Parallel circuits maintain the brightness of each bulb as they are added.

26. Now what do you think?
How will the brightness of (b) and (c) compare to each other and how does each compare to (a)? Explain.
How will the brightness of (d) and (e) compare to each other and how does each compare to (a)? Explain.
Compare the total current leaving the battery in each of the three circuits. Explain.
Bulbs (b) and (c) will be equally bright and both will be dimmer than (a) because the current through the (b) and (c) series circuit is less, and therefore the potential difference in each bulb is less.
Bulbs (d) and (e) will be equally bright and the same as (a) because the potential difference through parallel resistors connected to a battery is the same as that of a single resistor. Each bulb has its own path when connected to the battery.
See the two Classroom Practice Problems (slides 6 and 10) to remind students how potential difference and current change in series and parallel circuits.

27. The student is expected to:
TEKS
5F design, construct, and calculate in terms of current through, potential difference across, resistance of, and power used by electric circuit elements connected in both series and parallel combinations

28. What do you think?
Household circuits typically have many outlets and permanent fixtures such as hanging light fixtures on each circuit.
Are these wired in series or in parallel?
Why do you believe one of these methods has an advantage over the other method?
What disadvantages would the other method of wiring have for household circuits?
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting students’ ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.
Ask students to think about the advantages and disadvantages of each system.

29. Analysis of Complex Circuits
Click below to watch the Visual Concept.
Visual Concept

30. Complex Resistor Calculations
Req for 6.0  and 2.0 
Answer: 8.0 
Req for 8.0  and 4.0 
Answer: 2.7 
Req for 3.0  and 6.0  and 2.7  and 1.0 
Answer: 12.7 
So, the resistance of all 6 resistors is equivalent to a single 12.7  resistor.
To find the equivalent resistance for the circuit shown above, follow the steps shown to the right:
For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.
These problems are complex and are very difficult to solve without redrawing the circuit as each Req is calculated, and then using all of the equivalent circuits drawn to solve the problem.

31. Complex Resistor Calculations
Find the total current in the equivalent circuit.
Answer: 0.71 A
This is the current through the 1.0 , 6.0  (on the left), and 3.0  loads
Find the total potential drop across the parallel combination of three resistors.
Answer: 1.9 V
Continued on the next slide
For the 2.0  resistor, find the current and the potential difference.
To solve this problem, use the step-by-step approach shown.
For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.
It is a good idea to make a LARGE drawing of the circuit to start. Then, draw the equivalent circuits as you keep reducing the circuit to a single resistor and battery.
Then, find the current for the equivalent circuit. Work your way back up through the circuits by labeling the current and potential difference in each resistor as it is calculated (using V = IR or I = V/R).
SOLUTION:
Draw the equivalent circuit on the board (9.0 V battery and a single 12.7  resistor).
I = V/R = 0.71 A
Draw the equivalent circuit showing the parallel combination (2.7  ) and the other 3 resistors (1.0 , 6.0 , and 3.0 ).
For the 2.7  resistor, the V = (0.71 A)(2.7 ) = 1.9 V . Therefore the V is 1.9 V for each portion of the parallel segment (the upper path and the lower path).

32. Complex Resistor Calculations
Find the current through the combined 6.0  and 2.0  resistor.
Answer: A
Find the potential difference across the 2.0  resistor.
Answer: 0.48 V
The current through the 6.0  and 2.0  resistor depends on the total resistance (8.0 ) and the potential difference 1.9 V.
I = V/R = (1.9 V) / (8.0 ) = 0.24 A
Finally, find the potential difference for the 2.0  resistor.
V = IR = (0.24 A)(2.0 ) = 0.48 V

33. Classroom Practice Problems
For the circuit shown, find the:
Equivalent resistance
Current through the 3.0  resistor
Potential difference across the 6.0  resistor
Answers:
6.6 , 1.8 A, 6.5 V
Draw the circuits showing the battery, the 3.0  resistor, and the 3.6  resistor (Req for the 6.0  and 9.0  resistors).
Then draw the equivalent circuit showing just the 12 V battery and the single 6.6  resistor.
Now work backwards to get:
I = V/R = (12 V)/(6.6 ) = 1.8 A through the 3.0  and the 3.6  equivalent resistor
V = IR = (1.8 A) (3.6 ) = 6.5 V across the parallel combination. (It is the same potential difference across both the 6.0 and 9.0  resistors.)

34. Now what do you think?
Household circuits typically have many outlets and permanent fixtures such as hanging light fixtures on each circuit.
Are these wired in series or in parallel?
Why do you believe one of these methods has an advantage over the other method?
What disadvantages would the other method of wiring have for household circuits?
They are wired in parallel so that each load is connected directly to the 120 V source. Adding loads does not have any effect on other loads. Turning one load off does not affect other loads.