Circuit Symbols: Battery Resistor Light-bulb Switch Wire.
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Transcript of Circuit Symbols: Battery Resistor Light-bulb Switch Wire.
![Page 1: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/1.jpg)
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Circuit Symbols:
Battery
Resistor
Light-bulb
Switch
Wire
![Page 3: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/3.jpg)
Three general types of circuits:Closed Circuit - There is a complete loop with wires going from one side of the battery through a resistor(s) to the other side of the battery.
Open Circuit - There is not a complete loop.
Short Circuit - There is a complete loop, but it does not contain any resistors.
Only Working Circuit
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There are two ways to put resistors into a circuit.
1. Resistors can be in seriesOR
2. Resistors can be in parallel
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Resistors in Series is like a trip to Costco
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Resistors in Series (like a trip to Costco)
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?
The resistor with the largest amount of ohms.
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Combining (adding) Resistors
Itotal = I1 = I2 = I3
Req = Rtotal = R1 + R2 + R3
Voltage is calculated with Ohm’s Law
Series ResistorsR1
R2
R3
Amps Q
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Resistors in Parallel is like a trip to Vons
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Resistors in Parallel (like a trip to Vons)
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?The resistor with a large resistance will have a smaller current then the resistor with the smaller resistance.
![Page 10: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/10.jpg)
Combining (adding) Resistors
Parallel Resistors
Current is calculated with Ohm’s Law
Vtotal = V1 = V2 = V3
321
1111
RRRRtotal
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Example 1: A circuit has three 8.0 , 5.0 and a 12 resistors 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.
![Page 12: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/12.jpg)
Example 2: A circuit has three resistors: 6.0 , 4.0 and a 12 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.
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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)
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Light bulbs are made to be the only appliance plugged into a socket.The power rating of a light bulb (25
W or 100 W…) is as if that bulb was the only bulb plugged in to a 120 V power source.
The resistance of a light bulb is calculated by knowing the power rating and the voltage (120 V)
Current and actual voltage used by a light bulb depends on the circuit.
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Example 3: What will the power output be if an American-made 45 W light bulb is plugged in to a 310 V power source?
Using 120 V, calculate the resistance of the light bulb.
Using the resistance and the voltage of the new source, calculate the new power
2 212045 320
VP W R
R R
2 22310
3.0 10320
VP P P x W
R
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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 …P Q
R
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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
…P Q
R
Q
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When one bulb is unscrewed, the other bulb will remain lit in which circuit…
Circuit I
Circuit II
1. I
2. II
3. Both
4. Neither
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When one bulb is unscrewed, the other bulb will remain lit in which circuit…
Circuit I
Circuit II
1. I 2. II 3. both 4. neither
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A 25W bulb and a 100W bulb are connected in series. Which bulb will glow brighter?
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120V
25W 100W
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120V
25W 100W
The Light Bulbs are really ResistorsA) 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.
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25W Bulb 100W Bulb
R
VP
2
P
VR
2
25
1202R
144R
100
1202R
576R
Part A.
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120V
576 144
B) Calculate the total circuit resistance Rtotal
Rtotal = R1 + R2 Series Resistors
= 576 + 144
= 720
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120V
720
C) Calculate the total circuit current (I)
R
VI
720
120voltsamps167.
amps167.
amps167.
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120V
576 144
D) Calculate the Power used by each resistor.
amps167.
P = I2R25 W Bulb = .1672 x 576 = 16 watts
100 W Bulb
P = I2R = .1672 x 144 = 4 watts
![Page 27: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/27.jpg)
120V
576 144
E) Calculate the Voltage across each resistor.
amps167.
V = IR
100W Bulb
= .167 x 576 = 96 volts
25W Bulb
V = IR = .167 x 144 = 24 volts
96 volts
24 volts
120 volts
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E) Consider the Percent Power Needed to Light Each Bulb
100 W Bulb
percentwatts
watts4100
100
4
25 W Bulb
percentwatts
watts64100
25
16
Q
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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
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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
A
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.
Q
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Which bird is in trouble when the switch is closed?
1 21)Bird 1
2) Bird 2
3) Neither
4) Both
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Which bird is in trouble when the switch is closed?
1 2
1) Bird 1 2) bird 2 3) neither 4) both
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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
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16
16 3232
120V
Analyze the circuit:A) Calculate Rtotal
B) Calculate the current through each resistor.C) Calculate the voltage through each resistor.
Parallel:
84
32
32
4
32
1
32
1
16
11
123
321
R
R
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16
8
120V
Series:R123-4=8+16
R1234=24
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Make chart:
16
16 3232
120V
R I V
1 16
2 16
3 32
4 32
234 8
1234
24 120
These are in parallel so their voltage is the same
along with the total voltage
All these numbers will be the same.
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Make chart:
16
120V
R I V
1 16
2 16
3 32
4 32
234 8
1234
24 120
These are in series so their current is the same along
with the total current
All these numbers will be the same.
8
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Fill out the chart with V=IR
R I V
1 16 5 80
2 16 2.5 40
3 32 1.25 40
4 32 1.25 40
234 8 5 40
1234
24 5 120
V = IR
120 = I (24)
I = 5 A
V = IR
V = (5) (8)
V = 40 V
V = IR
40 = I (16)
I = 2.5 A
V = IR
V = (5) (16)
V = 80 V
V = IR
40 = I (32)
I = 1.25 A
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24
120V
I=V/R
I=120v/24
I=5 amps 5am
ps
Another way to do the problem (without the chart)
![Page 40: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/40.jpg)
16
8
120V 5amps
V=IRV=(5)(16)V=80volts
80volts
V=IRV=(5)(8)
V=40volts40volts
120volts
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16
16 3232
120V
80volts
40
volts
5 ampsI=V/R
=40volts/16 =2.5 amps
I=V/R=40volts/32 =1.25 amps
5 amps
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1. Bulb A would again be brighter
2. Bulb B would be brighter
3. They would be equal brightness
When the series circuit shown is connected, Bulb A is brighter than Bulb B. If the positions of the bulbs were reversed…
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1. Bulb A would again be brighter
2. Bulb B would be brighter
3. They would be the same
When the series circuit shown is connected, Bulb A is brighter than Bulb B. If the positions of the bulbs were reversed…
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!
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6
63
34
122
18 volts
Example: Find the voltage and current for each resistor.
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6
63
34
122
18 volts
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33
34
122
18 volts
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33
34
122
18 volts
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6
34
122
18 volts
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6
34
122
18 volts
21
111
RRRtotal
12
1
4
11
totalR
3totalR
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6
3
3 2
18 volts
21
111
RRRtotal
12
1
4
11
totalR
3totalR
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6
3
3 2
18 volts
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6
3
5
18 volts
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6
3
5
18 volts
5
1
6
11
totalR
73.2totalR
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2.73
3
18 volts
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2.73
3
18 volts
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5.73
18 volts
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5.73
18 volts
Now, find the total current flowing
R
VI
73.5
18voltsI
ampsI 14.3
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6
63
34
122
18 volts
V=IR
V=(3.14)(3)
V=9.42
9.42volts
3.14
am
p
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6
63
34
122
18 volts
9.42volts
18-9.42
8.57volts
3.14 amps
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6
34
122
18 volts
9.42volts
18-9.42
8.57volts
3.14 amps
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6
34
122
18 volts
9.42volts
18-9.42
8.57volts
3.14 amps
![Page 62: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/62.jpg)
6
3
5
18 volts
9.42volts
18-9.42
8.57volts
3.14 amps
![Page 63: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/63.jpg)
6
3
5
18 volts
9.42volts
18-9.42
8.57volts
R
VI
6
57.8 voltsI
ampsI 43.1
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6
3
5
18 volts
1.43 amps
1.71 amps
9.42volts
18-9.42
8.57volts
3.14 amps
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6
34
122
18 volts
9.42volts
1.71 amps
1.71 amps
V=IRV=(1.71)(2)V=3.42volts
3.42Volts
18-9.42
8.57volts
3.14 amps
1.43 amps
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6
34
122
18 volts
9.42volts
1.71 amps
1.71 amps
3.42Volts
18-9.42
8.57volts
5.15 volts
3.14 amps
1.43 amps
![Page 67: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/67.jpg)
6
34
122
18 volts
9.42volts
1.71 amps
1.71 amps
3.42Volts
18-9.42
8.57volts
5.15 volts
3.14 amps
I=V/RI=5.15volts/
12I= 0.43 amps
0.43 amps
1.43 amps
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6
34
122
18 volts
9.42volts
1.71 amps
1.71 amps
3.42Volts
18-9.42
8.57volts
5.15 volts
3.14 amps
I=V/RI=5.15volts/4I= 1.28 amps
0.43 amps
Or…1.71 amps –
0.43 =1.28 amps
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6
63
34
122
18 volts
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6
63
34
122
18 volts
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6
63
34
122
18 volts
Q
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1. R1 > R2 > R3
2. R1 > R2 = R3
3. R1 = R2 > R3
4. R1 < R2 < R3
5. R1 = R2 = R3
R1 R2 R3
Given: R1=1; R2=2 ; R3=3 . Rank the bulbs according to their relative brightness
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1. R1 > R2 > R3
2. R1 > R2 = R3
3. R1 = R2 > R3
4. R1 < R2 < R3
5. R1 = R2 = R3
R
VRIIVP
22
R1 R2 R3
Given: R1=1; R2=2 ; R3=3 . Rank the bulbs according to their relative brightness
15
Q
![Page 74: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/74.jpg)
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
![Page 75: Circuit Symbols: Battery Resistor Light-bulb Switch Wire.](https://reader031.fdocuments.us/reader031/viewer/2022012323/56649dc95503460f94abf6d5/html5/thumbnails/75.jpg)
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.