Transcript Chapter 22 Current Electricity

Chapter 22
Current Electricity
Electric Current The flow of electrons
Electric Circuit A closed loop in which
electrons can move
The flow is due to a potential difference
which is maintained by a pump,
Chapter 22
Current Electricity
Forms of electrical energy
Chemical
Solar
Hydroelectric
Wind
Nuclear
Chapter 22
Current Electricity
Simple Electric Cell
Carbon
Electrode
(+)
•Two dissimilar metals or carbon rods in acid
•Zn+ ions enter acid leaving terminal negative
•Electrons leave carbon leaving it positive
•Terminals connected to external circuit
•‘Battery’ referred to several cells originally
_
_
_
+
+
+
Zn
Electrode
(-)
Zn+
Zn+
Zn+
Zn+
Sulfuric acid
Chapter 22
Current Electricity
Electric Current
The flow of electrons
The flow is due to a
potential difference
which is maintained by
a pump,
Chapter 22
Current Electricity
An Electric Circuit has three components
1..A source of electrical potential difference or voltage.
(typically a battery or electrical outlet)
2. A conductive path which would allow for the
movement of charges. (typically made of wire)
3. An electrical resistance (resistor) which is loosely
defined as any object that uses electricity to do work.
(a light bulb, electric motor, heating element, speaker,
etc.)
Chapter 22
Current Electricity
Chapter 22
Current Electricity
Generator A device using available energy to produce a
potential difference.
Electric potential energy can be changed to kinetic
energy.
Electric current is measured in amperes
(coulomb/second)
Chapter 22
Current Electricity
•Electric Power The product of the current and the
potential difference
• P = VI (measured in watts)
Chapter 22
Current Electricity
Ohm’ Law
 For wires and other circuit devices, the current is proportional to the voltage applied
to its ends:
IV
 The current also depends on the amount of resistance that the wire offers to the
electrons for a given voltage V. We define a quantity called resistance R such that
V = I R (Ohm’s Law)
 The unit of resistance is the ohm which is represented by the Greek capital omega
().
 Thus
V
1 
A
Chapter 22
Current Electricity
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SIMP
Chapter 22
Current Electricity
Chapter 22
Current Electricity
Find the current and the power in the circuit
4Ω
I = V/R
I = 6V / 4Ω = 1.5 A
P = VI = 6V * 1.5 A
6 Volts
P = 9 watts
Chapter 22
Current Electricity
Draw a circuit with a 9 volt battery and a 6 Ω
resistor. Calculate the current and the power in the circuit.
6Ω
I = V/R
I = 9V / 6Ω = 1.5 A
P = VI = 9V * 1.5 A
P = 13.5 watts
9 Volts
1&2
1&5
1,3,4
1,3,6
1,2,3,4
1&6
1,3,5
Chapter 22
Current Electricity
goes out
increases
increases
increases
increases
Chapter 22
Current Electricity
Heating effects:
P = VI and V = IR so P = I2R
E = Pt = I2Rt
Q = I2Rt measured in joules
Chapter 22
Current Electricity
Kilowatt Hour The rate of energy consumption
(power) multiplied
by one hour
1 Kwh = (Kj/s)(3600 s) = 3.6 x 106 J
Chapter 22
Current Electricity
A clock has a resistance of 12000  and is plugged into a 115 V
outlet.
a. How much current does it draw?
b. How much power does it use?
c. If it costs 9¢ per kWh, how much does it cost to operate the
clock for 30 days?
a. I = V/R = 115V/12000  = .0096 A
b. P = VI = (115 V)(.0096 A) = 1.1 W
c. Cost = (1.1 x 10-3 kW)($0.09/kWh)(30 days)(24h/day) = $0.07
Chapter 23
Series and Parallel Circuits
Series Circuit:
1. Electric current has a single pathway through the
electrical circuit; therefore the current passing through
the electrical devices is the same everywhere.
2. The current is resisted by the first device, the second,
third, etc. The total resistance is the sum of the
individual resistors.
Rt = R1+ R2+ … + Rt
3. The current (I) is equal to the voltage divided by the
total resistance I = V/RT.
4. The total voltage across a series circuit divides among
the individual resistors so that the sum of the voltage
drops is equal to the total voltage.
5. Voltage is directly proportional to the resistance.
It= V/R = 2 Amps
V1= IR = (2A)(30Ω) = 60 V
V1 +V2 + V3 = 120 V
V2= IR = (2A)(15Ω) = 30 V
V3= IR = (2A)(15Ω) = 30 V
Chapter 23
Series and Parallel Circuits
Three resistors of 3 , 4 , and 5  are
connected in series across a 12 V battery.
a. What is the equivalent resistance?
b. What is the current through each resistor?
c. What is the voltage drop across each resistor?
d. Find the total voltage drop across each
resistor.
3
4
5
It= V/R = 12/12 = 1 A
RT = 3 + 4 + 5 = 12 
12 V
V1= IR = (1A)(3Ω) = 3 V
V1 +V2 + V3 = 12 V
V2= IR = (1A)(4Ω) = 4 V
V3= IR = (1A)(5Ω) = 5 V
Chapter 23
Series and Parallel Circuits
Parallel Circuit:
1. Every device connects to the same two points of the
circuit. Therefore the voltage is the same.
2. Current divides among the parallel branches. It follows
the path of least resistance.
3. The current (I) is equal to the sum of the current in the
parallel branches.
4. As the number of parallel branches increases, the
overall resistance of the circuit decreases.
1/Rt = 1/R1+1/R2+ … + 1/Rt
1
1
1
1
6
1





RT 60 30 20 60 10
RT = 10 Ω
I = V/R = 90/10 = 9 Amps
I1 = V/R1 = 90 V/60Ω = 1.5 A
I2 = V/R2 = 90 V/ 30Ω = 3 A
I = 1.5 A + 3 A + 4.5 A = 9 A
I3 = V/R3 = 90 V/20Ω = 4.5 A
1
1
1
1



Rt 120 60 40
12 V
120
60
40
Rt  20
I 
V 12V

 .6 A
R 20 
V
12V

 .1 A
R 120 
V 12V
I2  
 .2 A
R 60 
I1 
Find the equivalent resistance
and the readings in each meter
V
12V
I3  
 .3 A
R 400 
V = 20 V
V = 10 V
I = V/RT = 90/45 = 2A
Rt = 10 + 5 + 6 + 24 = 45 Ω
I = V/R = 90/45 = 2A
5.0 Ω
V = 12 V
24.0 Ω
20 +10 + 12 + 48 = 90 V
V = 48 V
I = V/R = 10/15 = 2/3 A
I = V/R = 90/45 = 2A
I = V/R = 48/60 = .8 A
Chapter 23 & 24
Series and Parallel Circuits
A 30 Ω resistor is connected in parallel with a 20 Ω resistor. The
parallel connection is placed in series with a 8 Ω resistor, and the
entire circuit is placed across a 60 V difference of potential.
a.
b.
c.
d.
e.
Draw the circuit.
What is the effective resistance in the entire circuit?
What is the voltage drop across the 8 Ω resistor?
What is the voltage drop across the parallel branch?
What is the current through each resistor?
Chapter 23 & 24
Series and Parallel Circuits
Chapter 23 & 24
Series and Parallel Circuits
Chapter 23 & 24
Series and Parallel Circuits
b. RT = 20 Ω
I=3A
c. V8Ω = IR = (3A)(8 Ω) = 24 V
d. Vpar = IR = (3A)(12 Ω) = 36 V
e. I8Ω = V/R = 24/8 = 3 A
I30Ω = V/R = 1.2 A I20Ω = V/R = 1.8 A
Chapter 23 & 24
Series and Parallel Circuits
Does the fuse melt?
Chapter 23 & 24
Series and Parallel Circuits
Does the fuse melt?
I = V/R = 120/6 = 20 A
YES
6Ω
Chapter 23 & 24
Series and Parallel Circuits
Voltmeters: Connected in parallel, high resistance.
Ammeters: Connected in series, low resistance.
Measuring in Circuits
How to use a multimeter!
Taking Measurements
Placement of meter depends on what you
want to measure.
#1 Goal: the meter should not alter the
behavior of the circuit.
Measuring Current
The current must flow
through the meter.
Circuit must be
broken to place the
meter in series.
Ammeters must have
very low resistance!
Measuring Potential
 The voltmeter is
connected in parallel
between the two points
where the measurement
is to be made.
 The voltmeter provides a
parallel pathway and
should take very little
current.
 A voltmeter must have
very high resistance!
Measuring Resistance
The component must
be removed from the
circuit.
Ohmmeters work by
running a current
through the
component being
tested!
Volts
DC
Volts
AC
Amps
The Meter
Always start HIGH!
If you are unsure of the
measurement range, begin
with larger values and reduce
multiplier until you get a
reading.
CURRENT
Red lead
Ohms
Volts and Ohms
Red lead
Ground
Black lead