CIRCUITS: Series & Parallel...Resistors in Parallel Resistors are considered to be in parallel if...

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CIRCUITS: Series & Parallel

Last Week’s BIG IDEAS:

• Opposite charged objects _____

• Like charged objects ____

attract

repel

• The ______ 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”.

Last Week’s BIG IDEAS:

electrons

• The attractive force between particles/objects depends on the amount of ______ (in Coulombs) the objects have and the ________ (in m) between the two particles/objects.

• The above relationship is called ___________ Law and is given by:

Last Week’s BIG IDEAS:

charge

distance

Coulomb’s

Fe = k𝒒𝟏∙𝒒𝟐

𝒓𝟐

The only difference is that Newton’s Law is only attractiveand Coulomb’s Law can be both ___________ and

______________.

Last Week’s BIG IDEAS:

Coulomb’s Law of attraction:

Is a lot like…

Newton’s Law of Universal ____________

Fe = k𝒒𝟏∙𝒒𝟏

𝒓𝟐

For the eensy-beensyparticles like atoms,

electrons, and protons

Fg = G𝒎𝟏∙𝒎𝟐

𝒓𝟐

for the “big” particles like planets.

Gravitation

attractiverepulsive

Last Week’s BIG IDEAS:Planets behave like _________ to some degree…particles

Both Coulomb’s Law of attraction - Fe = k

𝒒𝟏∙𝒒𝟏

𝒓𝟐

and Newton’s Law of Universal

Gravitation - Fg = G𝑚1∙𝑚2

𝑟2

are examples of forces at a ________________ (far away) where the objects are not touching and can be far away. These are different from ____________ forces such as friction, tension, force of push or pull in which the objects are in direct contact with each other.

Last Week’s BIG IDEAS:

distance

contact

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?

Electric potential energy is like ________ potential

energy and ______________ potential energy

Higher

Potential

Energy

Lower

Potential

Energy

+

-

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

elasticgravitational

EPE

GPE

ELPE

To cause movement of a charge, there must be an

electric potential _______________ between two

points in the electric circuit just as there could be a

difference in __________ pressure between

_______ points in a plumbing system.

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

difference

watertwo

In the image below, there is more electrical “pressure” at the

_____ V side of the battery than the 0 V side. Imagine that the 6

V side is _________________ (-) charged and so electrons are

pushed away (like charges repel) through the circuit towards the

side of the battery which is ____________________ (+) charged

to which the electrons are ________________(opposite charges

attract).

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

6

attractedpositively

+−

negatively

In this circuit the energetic electrons have _________ paths they

can go through. In the middle of each path is something that is

_______________ the flow of electrons and thus they struggle to

get through. In the struggle they ____________some of their

electrical energy which turns into light and ____________ energy.

The electrons are moving slower after the _______________ (in

Joules) they have performed.

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

resistinguse up

+−

two

heatwork

WORD BANK

resisting two heat

Use up work

DIRECTIONS: Use the right

terms below in the blanks above

Suppose there was a U. S. _______ nuclear submarine crew that

just happened to be all guy sailors. They just got off their 3-month

tour on their boat and are ________________ from the sight of

each other. They are seeking out members of the opposite

________and go down to the only two clubs on this small island

they are docked at to party-hardy. Once they’ve ___________ up

their energy they head back tired to their ___________________.

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

usedsex

+−

Navy

repelled

submarine

WORD BANK

sex submarine used

Navy repelled

DIRECTIONS: Use the right

terms below in the blanks above

Potential energy

Potential energy

• For example, to push a positively charged Styrofoam pellet into a positively charged electric field requires __________ (a force through a distance)

• The Styrofoam gains electrical potential energy in the process

• Since like charges _______, the chargedStyrofoam moves away from the Van de Graaf generator converting its electrical potential energy into ________ energy

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

work

repel

kinetic

In an electrical circuit, while the switch is ____________:

• Free electrons (conducting electrons) are always moving in

________________________ motion.

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

WORD BANK

random section

Prom speeds open

DIRECTIONS: Put the correct term in the correct blank below.

• The random ________________ are at an order of 1,000,000 m/s.

• There is no net movement of charge across a cross

________________ of a wire. It is all back and forth and random.

• Its like the random movement of single kids dancing at ___________.

open

random

speeds

sectionProm

What occurs in a wire when the circuit switch is __________?

• An electric field is established instantaneously (at almost the speed of ________, 3x108 m/s).

• _______ electrons, while still randomly moving, immediately begin drifting due to the electric field, resulting in a net flow of charge.

• They are like ___________ 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 ______________ of them, like electrons, the overall effect is large.

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

closedlight

Free

users

billions

WORD BANK

users closed light free billions

http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/imgele/micohm.gif

What occurs in a wire when the circuit switch is

closed and electricity begins to flow?

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

Much like a group of

students going to the

_______ circle after

school, they slowly drift out

at school in a messy but

deliberate pattern. The

electric field is kind of like

__________monitors

telling students to go home

and get out of the building.

bus

hall

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 = 𝑼𝒆𝒍𝒆𝒄

𝒒

An electric field exerts an electrical force FE on a charge q.An electric potential V tells how much electrical energy Uelec

each charge q has.

1 volt = 1 V = 1 Joule per Coulomb, = 1 J/C

Where Uelec is the electric potential energy (in Joules) and q is the charge of an object in Coulombs.

Voltage: Electrical “____________” that

pushes charges along a conducting

pathway.

Metric Unit for Voltage: Volt (V)

VOLTAGE (ELECTRIC POTENTIAL DIFFERENCE)

Another definition….

pressure

Electric Charges always travel from a _________electrical

pressure region to a _________ pressure region.high

low

Electric Current: The flow of electrons through a conducting

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 1018 e-/s

CURRENT

1 AMPERE of current carries about electrons per second

electrons

Resistance: The tendency of a substance to _________________

the flow of e-s.

Metric Unit: The force opposing e- movement is measured in

Ohms (Ω) .

RESISTANCE

resist

All materials have some resistance, insulators have a

large amount of resistance, while conductors have very little resistance . As e-s flow through material and collide with atoms, electrical energy is converted into two forms of energy: ____________ and ____________ .

RESISTANCE

heat light

RESISTANCE

The electricity has two paths to flow through. The 10 Ohm resistor allows ________________ as many electrons/ions to

flow as the 30 Ohm resistor.three times

RESISTANCE

Variable resistor – The knob turning is applying variable resistance to the circuit. This can be seen because the ______________- level of the LED is fluctuating. The

variable resistor is controlling the number of _____________ (e-s) passing through the circuit, or, in

other words, the ______________ (in Amps).

light

electronscurrent

CURRENT, VOLTAGE, & RESISTANCE

Make an analogy for voltage, resistance, and current like the ones below…

Electrical Circuits

Circuit Symbols:

Battery

Resistor

Light-bulb

Switch

Wire

DIRECTIONS: Draw lines from each term to the corresponding item in the circuit diagram

at right

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.

WORD BANK

Power source Short resistance

Closed current Open

DIRECTIONS: Put the correct term in the correct blank below.

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

There are two ways to put

resistors into a circuit.

1. Resistors can be in series

OR

2. Resistors can be in parallel

Resistors in SeriesResistors 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 differentThe rate of electron flow (or current) is determined by which resistor?

Ans. The resistor with the largest amount of

ohms.

R1

R2

R3

Combining (adding) Resistors

To find the total current just add up the currents through each resistor:

Itotal = I1 = I2 = I3

To find the total resistance, just add up the resistances of each resistor:

Req = Rtotal = R1 + R2 + R3

Voltage is calculated with Ohm’s Law

I = 𝑽

𝑹

Series Resistors

R1

R2

R3

Amps Q

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.

R1 R2 R3

Combining (adding) Resistors

Parallel Resistors

Current is calculated with

Ohm’s Law: I = 𝑽

𝑹

Total resistance:

Vtotal = V1 = V2 = V3

321

1111

RRRRtotal

++=

R1 R2 R3

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.

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.

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.

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.

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)

Example 3: What will the current be if an

American-made 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 = I2R , P = IV , P = 𝑽𝟐

𝑹

For this problem we use which one?P = IV

Solving for I… I = 𝑷

𝑽

SOLUTION: I = 𝑷

𝑽=

𝟓𝟓𝑾

𝟐𝟐𝟎 𝑽=0.25 Amps

P.O.D. 3: What will the resistance be if a

European-made 55 W light bulb is plugged in

to a 210 V power source? What about an

American-made 55 W light bulb?

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 …P Q

R

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

…P Q

R

Q

MULTIPLE CHOICE: 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

MULTIPLE CHOICE: 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

EXAMPLE 4: A 25W bulb and a 100W bulb are connected in series. Which bulb will glow brighter?

120V

25W 100W

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.

120V

25W 100W

The Light Bulbs are really Resistors

25W Bulb 100W Bulb

R

VP

2

=

P

VR

2

=

25

1202

=R

W= 144R

100

1202

=R

W= 576R

Part A.

P = 𝐕𝟐

𝐑is the formula for

electrical Power. Another formula for Power is

P = IV

B) The total resistance (Rtotal)

576 W + 144 W

+ =

= 720 W

120V

25W

576 W

100W

144 W

120V

720 W

C) Calculate the total circuit current (I)

R

VI =Use Ohm’s Law:

The current in a series circuit is the same throughout

W=

720

V120amps17.0=

120V

576 W 144 W

D) Calculate the Power used by each resistor.

25 W Bulb 100 W Bulb

P1 = 25 W P2 = 100 W

120V

576 W 144 W

E) Calculate the Voltage across each resistor.

amps167.=

Use Ohm’s Law:

V = IR = (0.167 amps)(576 W) = 96.2 V

100W Bulb

25W Bulb

Use Ohm’s Law:

V = I R = (0.167 amps)(144 W) = 24 V

120V

200 W 300 W

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.

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

AWhen 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

8 W

8 W 4 W16 W

60V

EXAMPLE: Analyze the circuit:

A) Calculate Rtotal

B) Calculate the current through each resistor.

C) Calculate the voltage through each resistor.

16W

16W 32W32W

120V

EXAMPLE: Analyze the circuit:

A) Calculate Rtotal

B) Calculate the current through each resistor.

C) Calculate the voltage through each resistor.

Use the formula for resistors in Parallel:

W==

=++=

++=

−−

84

32R

32

4

32

1

32

1

16

1

R

1

R

1

R

1

R

1

R

1

123

321

321total

8W

16W

16W 32W32W

120V

EXAMPLE: Analyze the circuit:

A) Calculate Rtotal

B) Calculate the current through each resistor.

C) Calculate the voltage through each resistor.

Replace the three resistors with one equivalent resistor

16W

8W

120V

Find the resistance in Series:

Rtotal = R1 + R2 + etc.R123-4=8 + 16

R1234=24

Make chart:

16W

16W 32W32W

120V

R I V

R1 16

R2 16

R3 32

R4 32

R234 8

R1234 24 120

Make a table with the following:

Make chart:

16W

16W 32W32W

120V

R I V

R1 16

R2 16

R3 32

R4 32

R234 8

R1234 24 120

These are in parallel so their voltage is the same along with the total voltage. In Parallel circuits:

V1 = V2 = V3 = etc.

All these numbers will be the same.

Make chart:

16W

120V

R I V

R1 16

R2 16

R3 32

R4 32

R234 8

R1234 24 120

These are in series so their current is the same along with

the total current:I1 = I2 = I3 = etc.

All these numbers will be the same.

8W

Fill out the chart with V=IR

R I V

R1 16 5 80

R2 16

R3 32

R4 32

R234 8 5

R1234 24 5 120

V = IR

120 = I (24)

I = 5 A

V = IR

V = (5) (16)

V = 80 V

Fill out the chart with V=IR

R I V

R1 16 5 80

R2 16 2.5 40

R3 32 40

R4 32 40

R234 8 5 40

R1234 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

Fill out the chart with V=IR

R I V

R1 16 5 80

R2 16 2.5 40

R3 32 1.25 40

R4 32 1.25 40

R234 8 5 40

R1234 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

24W

120V

I=V/R

I=120v/24W

I=5 amps

Another way to do the problem (without the chart)

16W

8W

120V 5amps

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

80volts

V=IRV=(5)(8)

V=40volts

40volts

120volts

16W

16W 32W32W

120V

80volts

40

volts

5 ampsI=V/R

=40volts/16 W=2.5 amps

I=V/R=40volts/32 W

=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 W

6 W

3 W

3 W4 W

12 W2 W

18 volts

P. O. D. 5: Find the voltage and current for each resistor.

6W

6W

3W

3W4W

12W2W

18 volts

3W3W

3W4W

12W2W

18 volts

3W3W

3W4W

12W2W

18 volts

6W

3W4W

12W2W

18 volts

6W

3W4W

12W2W

18 volts

21

111

RRRtotal

+=

12

1

4

11+=

totalR

W=3totalR

6W

3W

3W 2W

18 volts

21

111

RRRtotal

+=

12

1

4

11+=

totalR

W=3totalR

6W

3W

3W 2W

18 volts

6W

3W

5W

18 volts

6W

3W

5W

18 volts

5

1

6

11+=

totalR

W= 73.2totalR

2.73W3W

18 volts

2.73W3W

18 volts

5.73W

18 volts

5.73W

18 volts

Now, find the total current flowing

R

VI =

W=

73.5

18voltsI

ampsI 14.3=

6W

6W

3W

3W4W

12W2W

18 volts

V=IR

V=(3.14)(3W)

V=9.42

9.42volts

6W

6W

3W

3W4W

12W2W

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6W

3W4W

12W2W

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6W

3W4W

12W2W

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6W

3W

5W

18 volts

9.42volts

18-9.42

8.57volts

3.14 amps

6W

3W

5W

18 volts

9.42volts

18-9.42

8.57volts

R

VI =

W=

6

57.8 voltsI

ampsI 43.1=

6W

3W

5W

18 volts

1.43 amps

1.71 amps

9.42volts

18-9.42

8.57volts

3.14 amps

6W

3W4W

12W2W

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

6W

3W4W

12W2W

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

6W

3W4W

12W2W

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/12W

I= 0.43 amps

0.43 amps

1.43 amps

6W

3W4W

12W2W

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/4W

I= 1.28 amps

0.43 amps

Or…1.71 amps – 0.43 =

1.28 amps

6W

6W

3W

3W4W

12W2W

18 volts

6W

6W

3W

3W4W

12W2W

18 volts

6W

6W

3W

3W4W

12W2W

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=1W; R2=2 W; R3=3 W. 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=1W; R2=2 W; R3=3 W. 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.