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I. Current I is _________________________________________ Units: 1 coulomb/second = 1...

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Current I is ______________________________________ ts: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ the answer can be written I = ________ 22 C of charge pass a point in 4.0 seconds. Fin I =
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Page 1: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

I. Current I is _________________________________________

Units: 1 coulomb/second = 1 _______________

1 C/s = 1 ____

So the answer can be written I = ________

Ex. 22 C of charge pass a point in 4.0 seconds. Find I.

I =

Page 2: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. How much charge passes a point if the currentat the point is __________ ampere for ________seconds?

Amperes are ______________________ units. (m___sk)

Coulombs are ______________________ units

What is a coulomb written in terms of fundamental units?

1 C =

Page 3: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. How many electrons pass a point in 2.5 seconds if

the current at that point is 0.50 A?

Don't forget that the amount of charge q passing a

point can be written with units of __________________

or __________________________________ . The conversion

from one unit to the other is found on pg 1 of PhysRT:

1 C = _______________________ e

or 1 e = ________________________ C

Page 4: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

To have current, you need _________________ 2 things:

1.___________________________ (aka ___________ source)

a/ a ______:

wire wire

two _________________

The amount of ____________ depends on the metals used.

b/ a battery = _______________cell:

_________side or _______ potential

c/ a ______________ :

more cells _________________

_________side or _______ potential

Page 5: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

A voltage source V supplies ____________ to a circuit by setting up ___________________________ within the circuit.

_______ pot. _______ pot.

V flow of___________charge

flow of________( )

Electrons flow out of the ________________ side of V.This is equivalent to flow of ___ charge out of ___ side of V.The e-'s collide with _____________ of wire. This absorbs electrical __________________ and ______________ the wire.The speed of e-'s ______________ collisions is ____________, but the drift (________________) velocity of e- is __________ .

magnified wire cross section

.

Page 6: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Batteries supply a constant V _____________ current _______

I

I

t

Generators supply a varying V ________________ current ______

charge moves in ___________________

charge moves in _________________

t

charge moves in __________________

In Regents Physics, we will mostly study _____ ,

but the basic ideas are also valid for ______.

Page 7: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

2. A ________________ circuit

Can you have voltage without current?

Can you have current without voltage?

Ex:

Switch up ___________ circuit ________ current

Close the switch ____________ circuit _____________ __________ flows ______________

Using circuit symbols:

+ -

Page 8: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Current I is measured with an _________________, which

is often part of a DMM (______________________________ ).

Its symbol is:

In an electrical circuit, ammeters are connected __________________ . This means the circuit must be ________________ and the _______________ must be _____________ into it:

circuitpart 2

circuitpart 3

circuitpart 2

circuitpart 3

Circuit _______ ammeter:

circuitpart 1

circuitpart 1

Circuit _________ ammeter:

Page 9: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Voltage V is measured with an _________________, which

is often part of a _________. Its symbol is:

In an electrical circuit, voltmeters are connected __________________ . This means the circuit is _______opened up , but the _______________ must be connected _____________ two points in the circuit:

Circuit:

circuitpart 3

circuitpart 2

circuitpart 1

circuitpart 1

circuitpart 2

circuitpart 3

Page 10: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ammeters measure the current that passes _________ a part of the circuit, in other words, the amount of ___________ each second that passes ____________ it.

Voltmeters measure the ________________________ from one side of a part of the circuit to the other side. This is called the voltage _____________ the circuit part. It represents the ____________ or _____________ needed to force each ________________ of charge through.

Ideally, neither ammeter nor voltmeters ______________________________________________ , but in reality they do.

In symbols:

Page 11: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

II. Resistance R - ___________________________________________________________________________________________________________________________________________

2. R is a __________________ . It has _________________ .

units of R: ___________________

It is a ______________________ unit.

1. Resistance occurs as a result of ________________

colliding with ___________________ and with the

__________________________ , resulting in ____________ .

This converts __________________energy to ___________.

3. Any factor that makes it more _________________for

_______________ to move will through a material will

__________________________________ of the material:

Page 12: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

R

L

R

A

A. __________________

B. __________________

For _____________, there are four factors that affecthow much resistance it has:

Page 13: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

temperature:R

T

Higher T atoms of the metal _________________

________________________ for e-'s

to move through the metal

more ____________________

D. ______________________ : Different metals have different numbers of ____________________ .

______ electrons ______ current _______ resistance

R

# of free electrons

Page 14: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

These 4 factors are summed up in:

(rho) is called the _________________ of a material. depends on the ___________________ of a metal

and is different for different _____________ .

units of :

_________________

_________________

Lowest = _______________Highest = _______________Metals that have more free_____________ will have a_________ and _________ R.

Page 15: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. Calculate the resistance of 100 meters of copper wire that has a cross-sectional area of 3.44 x 10-6 m2.

R =

=

=

Page 16: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

A _________________is a device that is designed to

have a definite amount of _________________.

Resistors are used to

1. control _____________ flow; and

2. provide a _____________________

of a certain amount.

Symbols:

1. resistor:

2. variable resistor:

Page 17: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Semiconductors (like ___________ and ______________ )

have ____________ resistance at higher temperatures.

Here’s why:

___________ silicon (Si) is

an _______________________ .

It _____________ its outer e-’s

with 4 other silicon atoms in

a ___________________ bond,

so that its own electrons

_______________________

electricity.

Two materials that do not follow these rules for metals

are _____________________ and ______________________ .

= a ________ of shared e-s

= Si atom

Page 18: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Phosphorus P and arsenic Ashave __________ outer e- than Si.

Boron B and gallium Ga have __________ outer e- than Si.

If you add _________________ of P, As, B or Ga to pure Si, it creates extra charge carriers. This is called _____________ . Higher temps “free up” more of these extra charges and allows for more __________ and so less _____ . And because of the extra charge carriers, semiconductors have _________________ resistancesthat can be ______________ . They are now used in making almost all _______________________________ .

outer

e-’s3 4 5

a

t

o

m

C N

Al Si

Ge

Page 19: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Superconductors:

The resistance R of superconductors is _________

as long as the material is _____________________________.

Because they have no _____ , electrons can travel through

them __________ , and so they can carry ________ currents

for _________________ without producing large amounts

of ___________ . This in useful in the ___________________

___________ and _________________________________________

Originally (around 1911), only certain ____________

were found to be superconducting. But they had to be

cooled to near ___________________ using liquid helium

(boiling point about _______ ) for this to happen.

This is very expensive.

Page 20: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Materialmetal=mceramic=

c

critical temp.

(K)

absolute zero 0

Zinc 0.88

Aluminum 1.19

Tin 3.72

Mercury 4.15

liquid nitrogen

YBa2Cu3O7 90

TlBaCaCuO 125

room temp. 293

In _______, a new type of superconductor was discovered whose makeup is similar to ________________ . These become superconductors at higher temperatures. This makes them much more ____________________.

much ___________to use liquid N

Page 21: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Using R = L/A, R can be found using the ____________________________of a metal wire.

In a circuit, R is defined for any device as the ratio of __________________ the device to the________________________the device:

A simple circuit:

III. Ohm's Law

Page 22: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. If the potential difference across a resistor is _______

and the current through it is __________, find R.

units: [ ] = [ ]/[ ]

=

=

Page 23: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

R = V / I Solve this for

To remember all 3 equations, use:

V = ?

I = ?

units: [V] = [I][R] =

[I] = [V]/[R] =

Page 24: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex: What is the potential difference across a 25- resistor when it carries a current of 3.0 A?

Ex: What is the potential difference across a wire that has no resistance?

Ex: How much energy is required to make each coulombof charge pass through the above resistor?

Page 25: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

3. I = charge flowing _______________________________ . The charge going __________ any circuit element must_________ the charge __________ that element. Assume ____ charge flows out of the ____ side of the source.

4. V = potential difference __________________ = ____________________________ available to do work = energy converted to _______________________ by R = energy is __________________ by passing through R = _________________________ across R = _______ if there is no resistance, e.g. in a __________

1. Assume the connecting wires have _________________ resistance. (They usually have ________________ R than the circuit elements.)

2. For simple devices such as _______________________ ,we often replace the device with the symbol for__________________ : and assume that it has all of the ______________________.

Page 26: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

distance around the circuit

V

Vsource

R

Ex: A simple circuit has 1 _______.All of the __________________ is dropped across the ___________, because it is the only element in the circuit that requires ____________ (voltage).

Graph the voltage drops as you follow ____________ charge from the _________ potential side of source, throughthe _____________ , back to the _______ side of the source.

no V dropped in wireb/c V = IR = ________

Page 27: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Other ____________________ways to hook up the meters:

ammeter – measures current passing ____________ R - Ideally, it has no ____, so no ________ across it

voltmeter – measures potential difference _________ R - Ideally, it does not allow any ____ to enter it

R

A simple circuit with ____________:

The voltmeter must be connected across _______________ sides of Rto measure potential _____________.

Page 28: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ohm’s Law: For __________________ conductors at _______________ temp., I is ___________ prop. to V.

V

I

V

I

Case A: a device obeys Ohm’s Law _____________

Case B: ________________ devices

slope = ΔV/ΔI = constantso the ratio V/I = ____ is ___________

slope = V/I = R is ________________

Remember:V ____________ I So changing V _______________ I.

(Traditionally, V is plotted on the ____ axis)

In the case shown, R _______________

Page 29: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. If R is _____________ , then I is _____________ .

V

Ex. If R is ___________ , then I is _____________ .

= V

As R ____, I ____ . This is an ___________ circuit.

= V R =

V R =

This situation can be _______________________________ .Body resistance can be lowered by getting __________ .

As R ____, I ____ . This is a ___________ circuit.

Page 30: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Currents and the harm they can cause:

"It's _______________ that jolts, (shocks you)

But it's ___________ (milliamps of current) that kills."

AC tends to send heart nerves into _______________, which can be harder to fix than simply ________________________.

Page 31: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

A__________, short for 'fusible link', is a type of

overcurrent protection device. Its essential

component is a

__________________________________________________

____________________ . Fuses usually are rated in

_______________ . If the current exceeds the rating, the

metal strip melts, and it _________ the circuit. This

protects the circuit from __________________ which may

damage other circuit parts or ________________ .

A _______________________ is an automatically-

operated electrical ______________ . Like a fuse, it is

designed to protect an electrical circuit from damage

caused by excess_________. Unlike a fuse, which

operates once and then must be replaced, a circuit

breaker ________________ once the problem that caused

the excess current is fixed.

Page 32: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

A downed power linecan set up a _________through the ground.

Since the cables have _______ R, most voltage will be dropped along_________________ .

If the distance between thedowned line and the sourceis _________, there can be large _________ between2 nearby points along theground between your feet.___________ or ____________!

Page 33: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

IV. Series circuits - _______________________________________________________________________________________

Assume:

1. _____________________________________________________

2. _____________________________________________________

3. _____________________________________________________

4. _____________________________________________________

circuitelement ___

voltage source circuit

element ___

circuitelement ___

wire________potential

________potential

wire

wirewire

Page 34: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

_______________ Conservation: V =

_______________ Conservation: I =

_______________ (Total) Resistance: Req =

__________ Law applies to the total: V = and to each individual element: V1 =

V2 =

For a circuitwith 2 resistors:

Page 35: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. Find all the voltages and currents in the circuit below:

20. V 40

120

V1 = I1 = R1 =

V2 = I2 = R2 =

V = I = Req =

Page 36: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

V = 20. V

40

120

•V “divides up” ______________________________ as the R’s•This is because ___________ R requires _________ energy.•Series circuits are _______________________________.

Form the __________ of each resistance to Req = ________ ,

and then multiply by the ___________ voltage V.

R1

Req

R2

Req

Page 37: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Plot V vs. “distance around circuit.”

20

15

V dropped across the ______ resistor

0

back to ____side of thebattery

________ drop across wires because we assume ________

distance around circuit

potentialdifference

(V)

____ side ofbattery

V dropped across the ______ R at the ___

side of thebattery

Page 38: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Important:

“I is ______________ everywhere in ___________ circuit” does

NOT mean that I is ___________ in _________________ circuit!

10. V R1=

R2 =

I =

I1 =

I2 =

10. V R2=

R3=

I =

I1 =

I2 =

I3 =

I is still the _______________ in all parts of the second circuit, but it is a ________________ I than the first one!

R1=

Page 39: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Equivalent resistance: _________________________________

________________________________________________________

The total I =

20. V40

120

Replacing this part of thecircuit with a single_______________ resistor:

Req = R1 + R2 =

=

…gives you this circuit:

20. V

This is the ____________ as before.

Page 40: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

V = 20 V

V = 12 V

All _______________ circuits can be ___________________ in this way.

This can be done even if the ______________________ is not shown.

Req =

_____ V = 20 V

V = 12 V

Req results in the _____________ as the _________________ circuit.

A.

B.

C.

D.

Req =

_____

Req =

_____

Req =

_____

Page 41: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Series Circuits__________ Hookups:

V R1

R2

Original circuit:

To measure I1, the current through R1, _________________ the circuit and ____________ an ________________ next to R1.

V R1

R2

V R1

R2

Other possibilities: V R1

R2

V R1

R2

___ is the same everywhere, so _________________________

Page 42: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

To measure V1, the voltage across R1, __________disconnect the circuit. Simply connect the ______________ across R1

Other possibilities:

V R1

R2

V R1

R2

Original circuit:

V R1

R2

V R1

R2

Similarly, to measure the _________ voltage V or V2:

V R1

R2

V R1

R2

Page 43: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

V. Parallel circuits - ___________________________________ ___________________________________________

Assume:

1. _____________________________________________________

2. _____________________________________________________

3. _____________________________________________________

4. _____________________________________________________

voltage source

wire________potential

________potential

wirewire

wire

circuitelement ___

circuitelement ___

circuitelement ___

circuitelement ___

wire

Page 44: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

_______________ Conservation: V =

_______________ Conservation: I =

_______________ (Total) R: 1/Req =

__________ Law applies to the total: V = and to each individual element: V1 =

V2 =

For a circuitwith 2 resistors:

Page 45: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

V1 = I1 = R1 =

V2 = I2 = R2 =

V = I = Req =

Ex. Find all the voltages and currents in the circuit below:

Page 46: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

To find Req without using V and I:

1 Req

= 1 R1

+ 1 R2

=

=

=

=

NOTE: In a ________________ circuit,

Req is __________________ either R1 or R2.

Page 47: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

•I “divides up” ______________________________ to the R’s

•R1 has ______ the current b/c it has __________resistance.

•Parallel circuits are _______________________________.

Compare the _________ of the resistances R1/R2 to the __________ of the currents: I1/I2.

I1

I2

R1

R2

But…

Page 48: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

20. V 50 100

Notice what happens if one branch is __________________ :

I1= V1/R1

I2= V2/R2

20. V 50 100

These are ____________________ answers as before.

Each branch is_______________________of the others.

This is why __________________ circuits are used.

Page 49: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Plot V vs. “distance around circuit.”

20

____ resistor

0

back to ____side of thebattery

______ drop across wires because we assume _________

distance around circuit

potentialdifference

(V)

____ side ofbattery

_____ resistor at the ___side of thebattery

Page 50: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Important:

“V is ______________ everywhere in ___________ circuit” does

NOT mean that V is ___________ in ______________ circuit!

10. V R2=R1 =V=

V1=

V2=

20. V R2=

V =

V1 =

V2 =

V is the _______________ across all parts of the second circuit, but it is a ________________ V than the first one!

R1 =

Page 51: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Equivalent resistance: _________________________________

________________________________________________________

The total I =

Replacing this part of thecircuit with a single_______________ resistor:

Req = 1/(1/R1 + 1/R2)

= = =

…gives you this circuit:

20. V

This is the ____________ as before.

Page 52: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

V = 3.0 V

All _______________ circuits can be ___________________ in this way.

This can be done even if the ______________________ is not shown.

Req = ____

V = 3.0 V

Req results in the _____________ as the _________________ circuit.

A.

B.

C.

Req =

_____

Req =

_____

Page 53: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Alternative ways to draw parallel circuits:

V R1 R2

The circuitat left can also be drawn:

or:

NOTE: The diagram below is _________________________

because there is ________________

__________ for the current.

V

or:

V V

V

Page 54: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex 1. Draw two 10- resistors in parallel between points A and B.

What is the equivalent resistance between points Aand B in each of the examples above? (Hint: For identicalparallel resistors, divide 1 R by the ___________ of resistors.)

A B

Ex 2. Draw three 60- resistors in parallel between points A and B.

A

B

Ex 1: Divide _______ by ____ Req = __________

Ex 2: Divide _______ by ____ Req = __________

Page 55: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Parallel Circuit__________ Hookups:

Original circuit:

V R1 R2

A. To measure, V1, the voltage across R1, connect the ______________ across R1.

V R2

To measure I1, the current through R1, _________________ the circuit and _____________ an ________________ next to R1

____ = where the ammeter could also be placed.

Page 56: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

B. To measure the voltage _____________ and the current __________ resistor R2:

V R1

C. To measure the _________ voltage and current :

R1 R2

___ = optional ammeter position

___ = optional ammeter position

Page 57: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

In an ideal parallel circuit, all of the ______________

are equal, so placing the ________________ across any

element gives ___________________________ .

V R1R2

In reality, each voltage will _______________________ .

This is because the wires have a small amount of ______ ,

and so by Ohm's law: __________ , there is a small amount

of _________________ dropped along each wire.

Page 58: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

VI. Electrical Power P is the _________ at which electrical _____________________________ light, heat, mechanical, etc, energy or vice versa.

P

P is a _____________

The units of P are watts, W ( _____________ )

Electrical work: W =

1 watt = 1 W =

=

=

Since W = , you can write:

Page 59: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. At what rate is electrical energy converted to heatand light in a 75 W bulb?

Ex: A 55-W toaster oven is used for 25 seconds. How much electrical energy is converted to heat?

Page 60: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex. A 12- resistor carries a current of 3.0 A for 5.0 s. a/ At what rate does the resistor convert energy?

b/ How much energy does the resistor convert in the 5.0 s?

Page 61: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex: How much power is developed in the circuit at right?

Ex. What quantity does P “·” represent?

Given:

At what rate is energy converted to heat?

How much energy is converted in one minute?

Page 62: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

VR2

R1

Ex.

R1 R2V

Power of resistor R1: P1 =

or sum up the powers: P =

Power of resistor R2: P2 =

All of these equations work ___________________________ .

Total power: P =

Page 63: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex: Find the power in each resistor and the total power.

V1 = I1 = R1 = P1 =

V2 = I2= R2 = P2 =

V = I= R = P =

Page 64: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Ex: Find the power in each resistor and the total power.

V1 = I1 = R1 = P1 =

V2 = I2= R2 = P2 =V = I= R = P =

Page 65: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

Which resistor develops more power in the series circuit?

Notice both circuits above have ____________ V, R1 and R2.

Which of the two circuits would drain a battery faster?

Which develops more power in the parallel circuit?

___ b/c both have same ___ but it has more ____________

The ____________ circuit has a ______________ Req.

This means it will have ____________ I. Because

P = ______ and both circuits have the _____ , the

_____________ circuit will develop more power.

___ b/c both have same ___ but it has more ____________

Page 66: I. Current I is _________________________________________ Units: 1 coulomb/second = 1 _______________ 1 C/s = 1 ____ So the answer can be written I = ________.

__________________ passed ___________ through a wire that was submerged in __________ and found that the amountof ________ produced was proportional to ____ and _____.This is called_________ or _____ heating. Less ___ means ______________ electrical energy is converted to _______.Because P = ____ , the current can be reduced if ____ is increased. For this reason, power is transmitted at high_____________ .

Power plant:~_____

At your house:_________

The power istransmitted at_____________

A substation cutsthe V to __________


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