3. +
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.
5. +
Schematic Diagrams
Schematic diagrams use
symbols to represent
components.
They show how the parts in
an electrical device are
arranged.
6. +
Electric Circuits
A path through which charges flow
Can have one or more complete paths
Load
An element or group of elements in a circuit
that dissipates energy
Ex: A simple circuit consists of a source of
potential differences and electrical energy,
such as a battery, and a load, such as a
bulb or a group of bulbs
7. +
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
8. +
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).
9. +
Short Circuits
A shortcircuit bypasses the light bulb or
other load.
Itis 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.
10. +
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.
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
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.
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.
15. +
Resistors in Series
Seriesdescribes components of a circuit that
provide a single path for the current.
Thesame electrons must pass through both light
bulbs so the current in each is the same.
16. +
Resistors in Series
Light bulb filaments are resistors
When many resistors are connected in
series, the current in each resistor are the
same
17. +
Resistors in Series
Vbattery= V1 + V2
Conservation of energy
Vbattery= IR1 + IR2
Ohm’s law
Vbattery= I(R1 + R2)
Vbattery= IRequivalent
Requivalent = R1 + R2
18. +
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.
19. +
Equivalent Resistance
The potential difference across the
batter, V, must equal the potential
difference across the load.
20. +
Classroom Practice Problems
A 6.00V lantern battery is connected to
each of the following bulb combinations.
Find the equivalent resistance and current
in each circuit.
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
21. +
Classroom Practice Problems
1. Start by drawing a picture
2. Take your inventory
3. Decide which equations to use
4. Solve
Answers:
7.5 , 0.800 A
15 , 0.400 A
30 , 0.200 A
22. +
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.
23. +
Resistors in Parallel
Ibattery = I1 + I2
Conservation of charge
Vbattery V1 V2
Req R1 R2
Ohm’s law
V V V
VReq VR1 VR2
Vbattery= V1 = V2
Potential energy loss is the same across all parallel resistors.
Because Vbattery= V1 = V2, the equation above reduces as follows:
1 1 1
Req R1 R2
24. +
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.
25. +
Equivalent Resistance
Thesum of currents in parallel resistors
equals the total current
The Req for a parallel arrangement of
resistors must always be the smallest
resistance in the group of resistors
26. +
Classroom Practice Problems
A 9.0V battery is
connected to 4
resistors as shown 2
below. Find the 4
equivalent 5
resistance for the 7
circuit and the total
current in the circuit.
9V
27. +
Classroom Practice Problems
Given:
V= 9V R 1= 2 R 2= 4
R 3= 5 R 4= 7 Req= ?? I= ??
Equations:
1 1 1 1 V=IReq
....
Req R1 R2 R3
28. +
Classroom Practice Problems
1 1 1 1 1
Req 2 4 5 7
0.5 0.25 0.2 0.14 1.09
1 1 1 1 1
1 1
0.915
Req 1.09 V 9V
I 9.84 A
Req 0.915
29. +
Classroom Practice Problems
Findthe 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 , 0.300 A, 0.300 A
10.0 , 0.900 A, 0.300 A
31. +
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?
32. +
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.
34. +
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?
35. +
Complex Resistor Calculations
Forcomplex resistors, you need to follow
a few steps to be successful:
1. Combine series in parallel
2. Combine parallel sets
3. Combine series set
4. Finish problem from there
36. +
Complex Resistor Calculations
1. 6+2= 8
2. 1 1
0.12 0.25 0.37
8 4
1
Tofind the equivalent 3. 2.70
0.37
resistance for the circuit
shown above, follow
the steps shown to the 4. 3 + 9 + 2.70 + 1= 12.7
right:
37. +
Complex Resistor Calculations
Req for 6.0 and 2.0
Answer: 8.0
Req for 8.0 and 4.0
Answer: 2.7
Tofind the equivalent Req
for 3.0 and 6.0 and
resistance for the circuit 2.7 and 1.0
shown above, follow the Answer: 12.7
steps shown to the right: So,the resistance of all 6
resistors is equivalent to a
single 12.7 resistor.
38. +
Complex Resistor Calculations
Findthe total current in the
equivalent circuit.
Answer: 0.71 A
This is the current through the
1.0 , 6.0 (on the left), and
Forthe 2.0 resistor, find 3.0 loads
the current and the Findthe total potential
potential difference. drop across the parallel
To solve this problem, use the combination of three
step-by-step approach resistors.
shown. Answer: 1.9 V
Continued on the next slide
39. +
Complex Resistor Calculations
Findthe current
through the combined
6.0 and 2.0
resistor.
Answer: 0.24 A
Find the potential
difference across the
2.0 resistor.
Answer: 0.48 V
40. +
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
41. +
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?
