CHAPTER 19
 How does the energy generated by wind farms get to
people’s houses to power their appliances?
 Current is the rate of change of electric charge
 A current exits whenever there is a net movement of
electric charge through a medium
 The unit for current is the ampere
 1 ampere= 1 Coulomb
second
Electric Charge 
Q
I
t
Charge passing through a given area
time interval
 In a particular television tube, the beam current is 60
µA. How long does it take for 3.75 x 1014 electrons to
strike the screen?
 First calculate the electric charge of 3.75 x 1014 electrons.
 1 electron has a charge of 1.60 x 10-19 C
 (3.75 x 1014 ) (1.60 x 10-19 C)= 6 x 10-5 C
5
Q 6 x10 C
t 


1
.
0
sec
6
I
60 x10 A
 Batteries and generators work by converting other
forms of energy into electrical potential energy
 Batteries convert chemical energy into electrical
potential energy
 Generators convert mechanical energy (KE and PE) into
electrical potential energy
 Potential Difference, ΔV, is the driving force behind
current
 Increasing potential difference results in a greater
current
 i.e. using a 9.0 V battery generates a greater current than a 6.0
V battery
 V is measured in volts
 1 volt= 1 Joule/Coulomb
 Some conductors allow charges to move through them
more easily than others
 The opposition to the motion of charge through a
conductor is the conductor’s resistance
 The unit for resistance is the ohm (Ω)
 Ohm’s Law:
V
R
I
Potential Difference
Resistance 
Current
 Resistance is inversely proportional to current
 As the resistance increases, the current decreases
 For most materials, resistance is independent of V.
 Therefore, changing V affects the current, not the
resistance
 The current in a certain resistor is 0.50 A
when it is connected to a potential
difference of 110 v. What is the current in
this same resistor if
 a. The operating potential difference is
90.0 V?
 b. The operating potential difference is 130
V?
 I= 0.50 A, V = 110 V
 We’re looking for the new current if the potential
difference is changed
 According to Ohm’s Law:
V
I
R
 We’re missing R. Let’s find it 
V 110V
R

 220
I
0.50 A
 Let’s find the new current for each potential difference
 A.
 B.
V
90V
I

 0.41 A
R
220
V 130V
I

 0.59 A
R
220
 Superconductors have zero resistance below a certain
temperature called the critical temperature.
 Once a current is established in a superconductor it will
continue even if the potential difference is removed
 Electric power is the rate at which electrical energy is
converted to other types of energy
 Power is measured in Watts
V
P  IV  I R 
R
2
2
Circuits and Circuit Elements
Chapter 20
Schematic Diagrams (p. 731)
 A diagram that depicts the construction of an
electrical apparatus is a schematic diagram
Electric Circuits
 An electric circuit is a path through which charges can
be conducted
Necessary Parts of an electrical circuit
 The wire: Current flows through the wire
 The resistor: Can be a light bulb
 The emf source: Provides the potential difference…it’s
usually a battery
Series Circuits
 When resistors are connected in series, all the charges
have to follow a single path
 When one light bulb goes out, they all go out 
Series Circuits
 When resistors are connected in series, the current in
each resistor is the same!!!
 The total current in the circuit depends on how many
resistors are present
 The equivalent resistance is the sum of the circuit’s
resistances
Req  R1  R2  R3 ...
 THE EQUIVALENT RESISTANCE SHOULD ALWAYS
BE GREATER THAN THE LARGEST RESISTANCE IN
THE CIRCUIT
Series Current
 To find the total current in the circuit, first find the
equivalent resistance and then use Ohm’s Law
V
I
Req
 Although the current in each resistor has to be the same,
the potential difference doesn’t have to be the same.
Sample Problem p. 739 #2
 A 4.0 Ω resistor, an 8.0 Ω resistor and a 12.0 Ω
resistor are connected in series with a 24.0 V battery
 A. Calculate the equivalent resistance
Req  R1  R2  R3 ...  4  8  12  24
 B. Calculate the current in the circuit
I
V 24V

1 A
Req 24
 What is the current in each resistor?
 For resistors in series, the current in each resistor is the
same…so the answer is 1.0 A
Parallel Circuits
 A parallel circuit is a wiring arrangement that provides
alternative pathways for the movement of charges
Parallel Circuits
 The total current in a parallel circuit is equal to the sum of the
current in each resistor
I total  I1  I 2  I3 ...
 The equivalent resistance in a parallel circuit is calculated
using the following equation
1
1 1
1
   ...
Req R1 R2 R3
 The potential difference across each resistor is the same
Sample Problem p. 744 # 2
 An 18.0 Ω, 9.00 Ω, and 6.00 Ω resistor are connected in parallel
to an emf source. A current of 4.0 A is in the 9.00 Ω resistor.
 a. Calculate the equivalent resistance of the circuit.
1
1 1
1
1
1
1
  



 3.0
Req R1 R2 R3 18 9 6
 B. What is the potential difference across the source?
V  IReq  (4.0 A)(9)  36 V
 C. Calculate the current in the other resistors
I
V 36V

 2A
Req 18
I
V 36V

 6A
Req 6
Complex Circuits
 Most circuits have both series and parallel components
Complex Circuits (p. 747)
 To determine the equivalent resistance for a complex
circuit, you have to simplify the circuit into groups of
series and parallel resistors
 Sample Problem 20C (p. 747)
Since the 6.0 Ω and 2.0 Ω
resistor are connected in
series, their equivalent
resistance is 8.0 Ω
Req  R1  R2  R3 ...
Sample Problem 20C (p. 747)
 The new 8.0 Ω resistor and
4.0 Ω resistor are connected
in parallel. Their equivalent
resistance can be found using
the following equation:
1
1 1
1
   ...
Req R1 R2 R3
Req= 2.7 Ω
 Finally, the last three resistors are connected in series
so their equivalent resistance= 9.0 Ω + 2.7 Ω + 1.0 Ω=
12.7 Ω
 The circuit can now be redrawn with the equivalent
resistance connected to the original emf source
 To find the current and/or potential difference across a
particular resistor in a complex circuit you must first
find the equivalent resistance for the circuit
 Then you must rebuild the circuit in steps and
calculate the current and potential difference for each
group
 Sample problem 20D is a continuation of sample
problem 20C.
 We already determined the equivalent resistance for
the circuit…12.7 Ω
 Next we need to rebuild the circuit and find the
potential difference and current for each group.
V
9.0V
I

 0.71 A
R 12.7
 Work backward to find the
current and potential
difference for the next
group.
 These three resistors are
connected in series. That
means the current across all
three resistors is the same
(I=0.71 A).
 We only care about the
middle resistor because it’s
the only one that leads to
the 2.0 Ω resistor
V  IR  (0.71A)( 2.7)  1.9V
 Work backward to find the current and
potential difference for the next group.
 The 2.7 Ω resistor is composed of the
8.0 Ω and 4.0 Ω resistors in parallel
 This means they have the same
potential difference. (V=1.9 V)
 We only care about the 8.0 Ω resistor
because it’s the only one that leads to
the 2.0 Ω resistor
V 1.9V
I

 0.24 A
R
8 .0 
 Work backward to find the
current and potential difference
for the next group.
 The 8.0 Ω resistor is composed of
the 6.0 Ω and 2.0 Ω resistors
connected in series.
 This means they share the same
current (I=0.24 A)
V  IR  (0.24 A)( 2.0)  0.48 V
 Solve for the potential difference
and you’re done 
 http://alkalinebatteries.us/images/batteries3.jpg
 http://www.renewablepowernews.com/wp-
content/uploads/2008-04-wind-farm-hawaii22.jpg