Download Chapter 21-21

Chapter Twenty One: Electrical
Systems
21.1 Series Circuits
21.2 Parallel Circuits
21.3 Electrical Power
21.1 Electrical Systems
In a series circuit,
current can only
take one path, so
the current is the
same at all points
in the circuit.
21.1 Electrical Systems
Inexpensive strings
of holiday lights are
wired with the bulbs
in series.
If you remove one of
the bulbs from its
socket, the whole
string of mini bulbs
will go out.
21.1 Current and resistance in
series circuits
If you know the resistance of each device,
you can find the total resistance of the
circuit by adding up the resistance of each
device.
21.1 Current and resistance in
series circuits
Think of adding
resistances like
adding pinches to
a hose.
 Each pinch adds
some resistance.
21.1 Current and resistance in
series circuits
Everything has some resistance, even
wires.
A series circuit contains
a 12-V battery and three
bulbs with resistances
of1Ω, 2 Ω, and 3 Ω.
What is the current in
the circuit?
1. Looking for:
 …current (amps)
2. Given
 …Voltage = 12V; resistances = 1Ω, 2 Ω, 3 Ω.
3. Relationships:
 Rtot = R1+R2+R3
 Ohm’s Law I = V ÷ R
4. Solution
 Rtot = 6 Ω
 I = 12 V ÷ 6 Ω = 2 amps
Total = 6 Ω
21.1 Voltage drop
As each device in
series uses power,
the power carried by
the current is
reduced.
As a result, the
voltage is lower after
each device that uses
power.
This is known as the
voltage drop.
21.1 Voltage drop
The law of conservation
of energy also applies to
a circuit.
In this circuit, each bulb
has a resistance of 1
ohm, so each has a
voltage drop of 1 volt
when 1 amp flows
through the circuit.
21.1 Kirchhoff’s Voltage Law
Kirchhoff’s voltage law states that the
total of all the voltage drops must add up
to the battery’s voltage.
The circuit shown
contains a 9-volt battery,
a 1-ohm bulb, and a 2ohm bulb.
Calculate the circuit’s
total resistance and
current.
Then find each bulb’s
voltage drop.
1. Looking for:
 …total resistance; voltage drop each bulb
2. Given
 …Voltage = 9V; resistances = 1Ω, 2 Ω
3. Relationships:
 Rtot = R1+R2+R3
 Ohm’s Law I = V ÷ R
4. Solution- part 1
 Rtot = 3 Ω
 I = 9 V ÷ 3 Ω = 3 amps
=3Ω
4. Solution- part 2
 Use resistance to find current
I = 9 V ÷ 3 ohms = 3 amps
 Solution- part 3
 Rearrange Ohm’s law to solve for voltage
 Use current to find each voltage drop
V=IxR
V1 = (3 A) x (1 ohms) = 3 volts
V2 = (3 A) x (2 ohms) = 6 volts
Voltage total = (3v + 6v ) = 9 V
21.2 Parallel Circuits
In parallel circuits the current can
take more than one path.
21.2 Kirchhoff’s Current Law
All of the current
entering a branch
point must exit
again.
This is known as
Kirchhoff’s
current law.
21.2 Voltage and parallel circuits
If the voltage is
the same along
a wire, then the
same voltage
appears across
each branch of a
parallel circuit.
21.2 Voltage and parallel circuits
 Parallel circuits have two
advantages over series circuits.
1. Each device in the circuit has a voltage
drop equal to the full battery voltage.
2. Each device in the circuit may be turned
off independently without stopping the
current in the other devices in the circuit.
21.2 Current and parallel circuits
Each branch
works
independently
so the total
current in a
parallel circuit
is the sum of
the currents in
each branch.
21.2 Calculating in circuits
In a series circuit,
adding an extra
resistor increases
the total resistance
of the circuit.
In a parallel circuit,
more current flows
so the total
resistance decreases.
RESISTORS IN PARALLEL
• If no other resistance is present, the potential difference
across each resistor equals the potential difference across
the terminals of the battery.
• The equivalent resistance (R) of a parallel combination is
always less than the smallest of the individual resistors. The
formula for the equivalent resistance is as follows:
•
1/R = 1/RI + 1/R2 + 1/R3
The potential difference
across each resistor in
the arrangement is the
same, i. e.

V = VI = V2 = V3
21.2 Parallel vs. Series
 Remember: series/same/current;
parallel/same/voltage.
 Use Ohm’s law for both.
 All of the electrical outlets
in Jonah’s living room are
on one parallel circuit.
 The circuit breaker cuts off
the current if it exceeds 15
amps.
 Will the breaker trip if he
uses a light (240 Ω), stereo
(150 Ω), and an air
conditioner (10 Ω)?
1. Looking for:
 whether current exceeds 15 amps
2. Given:
 ……resistances = 240 W150 W; 10 W
3. Relationships:
 Assume voltage for each branch = 120 V
 Ohm’s Law I = V ÷ W
 Kirchhoff’s Current Law Itotal = I1 +I2 +I3
4. Solution:
 Ilight = 120 V ÷ 240 W = 0.5 amps
 Istereo = 120 V ÷ 150 W = 0.8 amps
 Ia/c = 120 V ÷ 10 W = 12 amps
0.5
0.8
+12.0
13.3
Breaker will
not trip
Solving Problems
1. Given:





……resistances = 240 W 150 W; 10 W
1/R = 1/ 240 W + 1/ 150 W + 1/ 10 W
1/ R = 240/36000 + 150/36000 + 3600/36000
1/R = 3990/36000
R = 36000/3990
R = 9.02 W
 Ohm’s Law


I = V ÷R
I = 120 v / 9.02 W
I = 13.3 A
Breaker will
not trip
21.2 Short circuits
A short circuit is a parallel path in a circuit
with very low resistance.
A short circuit can be created accidentally
by making a parallel branch with a wire.
21.2 Short circuits
Each circuit has its own fuse or circuit
breaker that stops the current if it exceeds
the safe amount, usually 15 or 20 amps
If you turn on too many appliances in one
circuit at the same time, the circuit breaker
or fuse cuts off the current.
To restore the current, you must FIRST
disconnect some or all of the appliances.
21.2 Fuses
In newer homes, flip the
tripped circuit breaker.
In older homes you must
replace the blown fuse (in
older homes).
Fuses are also used in car
electrical systems and in
electrical devices such as
televisions or in electrical
meters used to test circuits.
Chapter Twenty One: Electrical
Systems
21.1 Series Circuits
21.2 Parallel Circuits
21.3 Electrical Power
21.3 Electrical Power
Electrical power is measured in watts,
just like mechanical power.
Power is the rate at which energy is
changed into other forms of energy
such as heat, sound, or light.
Anything that “uses” electricity is
actually converting electrical energy
into some other type of energy.
21.3 Power
Power is a “rate” and is measured using
current and voltage .
21.3 Different forms of the Power
Equation
21.3 Important review
Watts
Used to measure
power of light bulbs
and small
appliances
An electric bill is
measured in kW/hrs.
1 kilowatt = 1000 W
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Horsepower (hp) = about 746 watts
Traditionally associated with engines.
(car,motorcycle,lawn-mower)
The term horsepower was developed to
quantify power. A strong horse could
move a 746 N object one meter in one
38
second.
What is the SI unit of power?
Watt
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21.3 Electrical Power
The watt is an
abbreviation for
one joule per
second.
A 100-watt light
bulb uses 100
joules of energy
every second.
21.3 Kilowatt
Most electrical
appliances have a
label that lists the
power in watts (W) or
kilowatts (kW).
The kilowatt is used
for large amounts of
power.
Solving Problems
 A 12-volt battery is
connected in series to
two identical light
bulbs.
 The current in the
circuit is 3 amps.
 Calculate the power
output of the battery.
Solving Problems
1. Looking for:
 …power of battery
2. Given:
 …voltage = 12 V; current = 3 amps
3. Relationships:
 Power:
P=IxV
4. Solution:
 P = 3 A x 12 V = 36 watts
21.3 Buying
Electricity
Utility companies charge customers for the
number of kilowatt-hours (kWh) used each
month.
A kilowatt-hour is a unit of energy.
The number of kilowatt-hours used equals
the number of kilowatts multiplied by the
number of hours the appliance was turned
on.
21.3 Buying Electricity
There are many
simple things you
can do to use less
electricity.
When added up,
these simple
things can mean
many dollars of
savings each
month.
Solving Problems
 How much does it cost to run a
3,000 W electric stove for 2 hours?
 Use an electricity cost of $0.15 per
kilowatt-hour.
1. Looking for:
 …cost to run stove for 2h
2. Given:
 … P = 3,000W; T = 2h; price $0.15/kW
Solving Problems
3. Relationships:
 1000 watts = 1 kW
 Charge in kWh
4. Solution:
 3000 W x 1 kW = 3 kW
1000 W
 Charge = 3 kW x 2 h = 6 kWh
 Cost = 6 kWh x $ 0.15
1 kWh
= $ 0.90
How much power does a 100 watt light
bulb use if it is turned on for 30
seconds?
One more duh!
100 watts!!!!!!!!!!!!!!!!!!!
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