Circuit Symbols:

Battery

Resistor

Light-bulb

Switch

Wire

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

There are two ways to put resistors into a circuit.

1. Resistors can be in seriesOR

2. Resistors can be in parallel

Resistors in Series is like a trip to Costco

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.

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

Resistors in Parallel is like a trip to Vons

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.

Combining (adding) Resistors

Parallel Resistors

Current is calculated with Ohm’s Law

Vtotal = V1 = V2 = V3

321

1111

RRRRtotal

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.

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.

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)

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.

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

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

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

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

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

A 25W bulb and a 100W bulb are connected in series. Which bulb will glow brighter?

120V

25W 100W

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.

25W Bulb 100W Bulb

R

VP

2

P

VR

2

25

1202R

144R

100

1202R

576R

Part A.

120V

576 144

B) Calculate the total circuit resistance Rtotal

Rtotal = R1 + R2 Series Resistors

= 576 + 144

= 720

120V

720

C) Calculate the total circuit current (I)

R

VI

720

120voltsamps167.

amps167.

amps167.

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

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

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

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

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

Which bird is in trouble when the switch is closed?

1 21)Bird 1

2) Bird 2

3) Neither

4) Both

Which bird is in trouble when the switch is closed?

1 2

1) Bird 1 2) bird 2 3) neither 4) both

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

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

16

8

120V

Series:R123-4=8+16

R1234=24

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.

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

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

24

120V

I=V/R

I=120v/24

I=5 amps 5am

ps

Another way to do the problem (without the chart)

16

8

120V 5amps

V=IRV=(5)(16)V=80volts

80volts

V=IRV=(5)(8)

V=40volts40volts

120volts

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

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…

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!

6

63

34

122

18 volts

Example: Find the voltage and current for each resistor.

6

63

34

122

18 volts

33

34

122

18 volts

33

34

122

18 volts

6

34

122

18 volts

6

34

122

18 volts

21

111

RRRtotal

12

1

4

11

totalR

3totalR

6

3

3 2

18 volts

21

111

RRRtotal

12

1

4

11

totalR

3totalR

6

3

3 2

18 volts

6

3

5

18 volts

6

3

5

18 volts

5

1

6

11

totalR

73.2totalR

2.73

3

18 volts

2.73

3

18 volts

5.73

18 volts

5.73

18 volts

Now, find the total current flowing

R

VI

73.5

18voltsI

ampsI 14.3

6

63

34

122

18 volts

V=IR

V=(3.14)(3)

V=9.42

9.42volts

3.14

am

p

6

63

34

122

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6

34

122

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6

34

122

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6

3

5

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6

3

5

18 volts

9.42volts

18-9.42

8.57volts

R

VI

6

57.8 voltsI

ampsI 43.1

6

3

5

18 volts

1.43 amps

1.71 amps

9.42volts

18-9.42

8.57volts

3.14 amps

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

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

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

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

6

63

34

122

18 volts

6

63

34

122

18 volts

6

63

34

122

18 volts

Q

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

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

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

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.