11/2/2021
Chapter 13
Electric Circuits
General Physics
Fall, 2021
Dr. William Underwood
Chesapeake Energy Chair
of Geosciences
1
Griffith, 2007
2
REVIEW
Assignment
Griffith pp: 274‐275
Exercises: All even exercises.
Griffith pp: 275‐276
Synthesis Problems: All
Due: Thursday, November 11th
Schaum’s: Chapter 26
Problems: 26.22, 26.26, 26.33
Schaum’s: Chapter 27
Problems: 27.14, 27.15, 27.27
Schaum’s: Chapter 28
Problems: 28.30, 28.31, 28.32, 28.34
3
4
Outline
• Electric circuits and electric current
what electric current is and how flows
• Ohm’s Law and resistance
the relationship between current and voltage
• Series and parallel circuits
current flow in series and parallel circuits
• Electric energy and power
energy and power in electric circuits
• Alternating current and household circuits
applying these concepts to our homes
5
TEST
Saturday, 13 November
Comprehensive
Chapters 1 to 8
6
1
11/2/2021
Electric Circuits
Circuits and Current
• Have you wondered how a flashlight works?
• Could you make a light from these 3 things?
• Do you know why the light works (not just the
mechanics of a
circuit)?
• What about
electric current?
Griffith, 2007
• There are 3 ways we can connect the bulb,
battery and wire. Only 1 works. Why?
• Option (c) forms a simple circuit and current
can flow
through
all three
elements
7
Circuits and Current
Circuits and Current
• What’s wrong with the others?
• (a) is not a circuit and current cannot return to
the battery
• (b) leaves
out the
bulb; only
the wire
forms a
circuit
Griffith, 2007
8
Griffith, 2007
• Circuits let current flow. But what is current?
• A battery separates positive and negative
charges
• We saw before that charges want to flow to
equalize charge
• However, they can only flow through an
external conductor (not through the battery)
• This conductor is the wire and bulb
• The flow of charge is electric current
9
Circuits and Current
10
Circuits and Current
• Electric current is the rate of flow of electric
charge
I = q/t
• Where:
I = Current (A)
q = charge (C)
t = time (s)
• Units: 1 Ampere = 1 C/1 s
• Ampere = amp = A
• The direction of the current is defined as the
direction of flow of positive charge (a)
• In fact, electrons
(negative charge)
are what actually
move (b)
• We sometimes
consider the flow
of “holes”
11
Griffith, 2007
12
2
11/2/2021
Circuits and Current
Circuits and Current
• Electricity flows at ~3.0 x 108 m/s in wire. Do
electrons move that fast?
• Think of a straw. If you push a BB in, what
happens?
• You have to fill the straw before any come out
• What if the straw is full of BBs? How long
does it take for one to fall out?
• Pushing a BB in makes one fall out, even
though that one doesn’t
13
Circuits and Current
•
•
•
•
How do we limit current?
(a) is open (broken) so no current flows
(b) has only the wire, so current is unrestricted
(c) includes bulb, which resists flow
Griffith, 2007
Circuits and Current
• An open circuit is not a circuit—it has a gap in
it. This can be caused by a switch or break
• A short circuit has no (little) resistance and
current flow is uninhibited
• The bulb provides resistance, which slows the
flow of current, preserves battery life, and
gives light
• Whenever there is current flow there is
resistance which produces heat
• How does a bulb work?
• Current enters through
the metal contact and
exits through the sheath
• The filament resists the
flow, and heats
• The glow of the heated
filament gives light
15
Circuits and Current
Griffith, 2007
16
Circuits and Current
• We can compare electric flow to water
• Both are closed systems
• Both have a source of energy, flow, restriction,
and can do work
• However, there are differences!
Griffith, 2007
14
17
Water flow
Water
Pump
Pipes
Narrow pipe
Valve
Pressure
Electric circuit
Charge
Battery
Wires
Wire filament
Switch
Voltage
18
3
11/2/2021
Summary
In‐class Questions
Electric current is the rate at which charge
flows through a closed circuit. The unit of
current is the Ampere, which is charge per
time. 1 A = 1 C/s. Current is defined to flow
in the direction of positive charges in spite of
the fact that it is electrons which move.
Griffith p. 272‐273
Q1, Q3, Q4, Q6, Q8; E1
Griffith, 2007
19
Griffith, 2007
Ohm’s Law
20
Ohm’s Law
• How do we determine the amount of current
that flows?
• How does current depend on voltage?
• What about the resistance?
• These are answered by Ohm’s law
• Discovered experimentally this is not a
fundamental law (like Newton’s laws)
• Discovered by Georg Ohm in the early 1800s
• The current flowing in a part of a circuit is
equal to the ratio of the change in voltage to
the resistance
I = V/R
Where
I = the current in amperes
V = the change in voltage in volts
R = the resistance in ohms
21
22
Ohm’s Law
Ohm’s Law
• Resistance is the property that restricts
current flow
• From Ohm’s law
R = V/I
• Unit: 1 ohm () = 1 V/A
• In our circuit the bulb had the most resistance,
but the wire has resistance, too
• Resistance of wire is based on its conductivity
• Conductivity is the ability of a material to
transmit current
• Conductivity is inverse of resistance
• Conductivity depends on material
• Thicker wires have higher conductivity
23
24
4
11/2/2021
Example
Ohm’s Law
• What is the current in this circuit?
I = V/R
I = 1.5V/20
I = 0.075 A
I = 75 mA
• We have ignored
resistance in the
wire and battery
Griffith, 2007
• A battery has a chemical driver (based on an
electrolyte) to separate charges
• We call this increase in potential energy
electromotive force (EMF), 
• However, it is not really a force!
• Units are volts (V), or joules/coulomb
• This is the voltage we measure for a battery
25
26
Ohm’s Law
Example
• A new battery produces full charge
• As a battery ages, it develops internal
resistance, which reduces the effective voltage
• As internal resistance increases, a battery
supplies less EMF to the circuit
• The actual chemical voltage is unchanged, so a
battery without current measures full voltage
• A dead battery may measure full voltage when
removed from a circuit
27
Example
• Battery: 1.5 V; Rinternal = 5
• Bulb: Rbulb = 20
• What is the
current in the
circuit?
• The voltage
difference
across the
bulb?
28
Griffith, 2007
Example
• Battery: 1.5 V; Rinternal = 5
• Bulb: Rbulb = 20
• What is the current in the circuit?
Rtotal = Rinternal + Rbulb
Rtotal = 5 + 20 = 25
I = /R
I = 1.5V/25
I = 0.06 A = 60 mA
• Battery: 1.5 V; Rinternal = 5
• Bulb: Rbulb = 20
• What is the voltage difference across the
bulb?
V = IR
V = (0.06A)(20)
V = 1.2V
29
30
5
11/2/2021
Summary
In‐class Questions
Resistance is the property of a circuit that
opposes current flow. Ohm’s law states that
current is proportional to the voltage across a
particular element, and inversely proportional
to its
resistance.
Griffith, 2007
Griffith p. 273
Questions: Q10, Q11, Q12
Exercises: E3, E5
31
32
Series and Parallel
Series Circuits
• Resistance (light bulbs, etc.) can be connected
in many ways
• Reduced to the simplest form we have series
and parallel circuits
• Series circuits have elements in a single loop
• Parallel circuits have elements connected to
each other at each end
33
Series Circuits
Griffith, 2007
34
Series Circuits
• When elements are connected in series, the
same current flows through all elements
• Each element restricts flow
• Resistances are added
Griffith, 2007
• We’ve seen a battery, switch, and light bulb
circuit before
• A schematic is a diagram representing the
elements of a circuit
• Note the symbol for resistance
• Resistances in a series circuit are added
Rseries = R1 + R2 + R3
• Add more resistances for more elements
• The same current passes through all elements
• Current is reduced as resistance increases
• The voltage changes across each element
V = IR
• If one resistance (bulb) breaks, entire circuit is
open (no current flows)
35
36
6
11/2/2021
Parallel Circuits
Parallel Circuits
• Elements in a parallel circuit are connected at
both ends
• Same voltage
is applied
across all
• More current
flows because
of multiple
paths
• Voltage is the same across all paths
• Current depends on individual resistance
• Effective resistance is not added, but is
inverse of sum of inverses
1/Rparallel = 1/R1 + 1/R2 + 1/R3
• More resistances add more elements
• Effective resistance is less than smallest
resistance
37
Griffith, 2007
Example
Example
• How much current passes through each bulb?
I = V/R
I = 6V / 10
I = 0.6 A
• 0.6 A flows through
each bulb for a total
of 1.2 A, as seen
before
Given this diagram, what is the total current?
1/Rp = 1/R1 + 1/R2
1/Rp = 1/10 + 1/10
Rp = 5
I = /R
I = 6V / 5
I = 1.2 A
Griffith, 2007
38
39
Example
Griffith, 2007
40
Example
• What if the bulbs were in series? What is the
current?
Rs = R1 + R2
Rs = 10 + 10 = 20
I = V/R
I = 6V / 20
I = 0.3 A
• What is the voltage drop across each bulb?
V = IR
V = (0.3A)(10)
V = 3.0 V
• A 3.0V drop across each of 2 bulbs gives a
total of 6.0 V, the charge of the battery
41
42
7
11/2/2021
Series and Parallel
Series and Parallel
• The resistance of a series circuit is the sum of
the individual resistances
• The resistance of a parallel circuit is the
inverse of the sum of the inverses of the
individual resistances
• The current in a series circuit is the same for
all elements
• The current in a parallel circuit is the sum of
the currents through each element
• The voltage across each element of a series
circuit depends on the resistance of each
element
• The voltage across each element of a parallel
circuit is the same
• Circuits can be combined with both parallel
and series elements to make complex circuits
43
Measuring
Measuring
• Voltmeters measure voltage, ohmmeters
measure resistance and ammeters current
• Multimeters are
used to measure
all of them
• Most modern
instruments are
multimeters
• Current must be measured inline (as part of
the circuit)
• Voltage is measured across the element
• Resistance is measured with no power applied
45
Griffith, 2007
Summary
Griffith, 2007
46
In‐class Questions
• Elements in a series circuit are connected
inline so the same current passes through all.
Total resistance is the sum of
the resistances of each.
• Elements in a parallel circuit
are connected such that they
form different paths. Total
resistance is less than the
least resistance.
Griffith, 2007
44
Griffith p. 273
Questions: Q13, Q14, Q15, Q16, Q17
Griffith, 2007
47
48
8
11/2/2021
In‐class Questions
Energy and Power
• Using our water analogy again, we can
understand the energy of an electric circuit
Griffith p. 274
Questions: Q18, Q19, Q20
Exercises: E7, E9, E11
Griffith, 2007
49
Griffith, 2007
50
Energy and Power
Energy and Power
• In the same way the pump supplies potential
energy by filling the tank, the battery supplies
electric potential energy
• The bulb removes
energy as heat
and light
• Power supplied by
the battery is used
for light
• In an electric circuit potential energy from the
battery is converted to kinetic energy which
produces heat (and light)
• Energy source 
potential energy 
kinetic energy 
heat (and light)
• This is why wires and electronics get warm
Griffith, 2007
51
Energy and Power
52
Energy and Power
• Electromotive force (EMF, ) is potential
energy per unit charge supplied by a source
• EMF is the difference in PE per unit charge
 = PE/q
• And
q = PE
• Multiplying by current instead
I = PE/t = P (power)
P = I
• Since  = V and V = IR
P = I2R
• The power dissipated by resistance R is
proportional to the square of the current
• The power supplied by the battery must equal
the power dissipated
I = I2R
• Energy is conserved
53
54
9
11/2/2021
Example
Example
• What is the power dissipated by a 20 light
bulb powered by 2 1.5V batteries in series?
 = 1 + 2 = 3.0V
I = /R
I = 3.0V / 20
I = 0.15A
P = I2R
P = (0.15A)2(20)
P = 0.45W
• Is this the power delivered by the batteries?
P = I
P = (3.0V)(0.15A)
P = 0.45W
• Yes, this is the same as we had before.
55
56
Energy and Power
Energy and Power
• Where does our electric power come from?
• Gravity (hydroelectric), chemical (fossil fuels),
potential (nuclear)
• Heat is converted to electricity by a turbine
• Electricity arrives in wires (a circuit)
• Appliances use the power, converting it to
heat or
mechanical
energy
• How do we measure electric usage?
Energy = Pt
• 1 kilowatt‐hour is the work done by 1000 W of
energy in 1 hour
1 J = 1Ws
• So
1 kWh = (1000W)(3600s)
1 kWh = 3.6 x 106 J
57
Griffith, 2007
Summary
In‐class Questions
Multiplying a voltage difference by charge yields
energy; multiplying a voltage difference by
current yields power, the amount of energy use.
The power supplied
by a source must
equal the power
dissipated in the
resistance.
Griffith, 2007
58
59
Griffith p. 274
Questions: Q21, Q22, Q23, Q24
Exercises: E13, E15
60
10
11/2/2021
AC and Household Circuits
AC and Household Circuits
• Direct current, like that supplied by a battery,
goes in only one direction.
• Alternating current, like that in our homes,
changes polarity regularly (60 Hz in the US).
• A galvanometer (reading + and – current) will
usually flutter around 0, the average
• If the voltage and current change, and average
is 0, how do we define their values?
61
AC and Household Circuits
Griffith, 2007
62
AC and Household Circuits
• Effective current is the square root of the
average of the square of the current
• Called RMS or root mean square value
• Standard method for calculating value of
sinusoidal functions
Ieffective = 0.707 Ipeak
• Note: This is the sine of 45
• Voltage also alternates with a sine curve.
Veffective = 0.707 Vpeak
• If nominal voltage is 115 V, peak voltage is
approximately 160 V (115 V/0.707 = 163 V)
63
AC and Household Circuits
•
•
•
•
•
• An oscilloscope shows voltage vs. time, and
can show an accurate representation of
alternating current, a sine curve
• Power is proportional to I2; which is always +
Griffith, 2007
64
AC and Household Circuits
U.S. uses between 110V and 120V
Europe and other places use 220V to 240V
U.S. uses 60 Hz
Europe and other places use 50 Hz
Some places (e.g. Korea) are converting from
110V to 220V and have both, but not different
frequencies
65
Why are household circuits wired in parallel?
1. Parallel circuits do not require all appliances
to be on for any to work
2. Parallel circuits supply the same voltage to all
appliances
3. Unfortunately, the total current increases as
appliances are added (downside to parallel)
66
11
11/2/2021
AC and Household Circuits
AC and Household Circuits
• Why do we have multiple circuits in our
homes?
• Each added appliance adds current
• Standard wires can only handle 15‐20A
• Lights draw little current; heaters a lot
• If too much current is required, circuit breaker
trips and everything shuts off
• Just resetting the breaker is not the solution—
unplug something!
Appliance
Power (W)
Current (A)
Stove
6000 (220V)
27
Clothes drier
5400 (220V)
25
Water heater
4500 (220V)
20
Dishwasher
1200
10
Microwave
1000
8
Toaster
850
7
Hair drier
650
5
Large fan
240
2
Television
100
0.8
Computer
45
0.4
Clock radio
12
0.1
67
68
Example
Summary
Household circuits use alternating current,
which regularly changes direction. Effective
current (or voltage) is 0.707 times peak
current (or voltage). We connect appliances in
parallel. Because the current in each circuit is
additive, we
must be
careful not
to overload.
• A 60W light bulb operates on 120V AC. What
is the effective current it draws?
I = P/V
I = 60W / 120V
I = 0.5A
• What is the bulb’s resistance?
R = V/I
R = 120V / 0.5A
R = 240
Griffith, 2007
69
In‐class Questions
70
Summary
Electric current is the rate at which charge
flows through a closed circuit. The unit of
current is the Ampere, which is charge per
time. 1 A = 1 C/s. Current is defined to flow
in the direction of positive charges in spite of
the fact that it is electrons which move.
Griffith p. 274
Questions: Q26, Q27, Q28, Q29, Q30
Exercises: E17
71
Griffith, 2007
72
12
11/2/2021
Summary
Summary
Resistance is the property of a circuit that
opposes current flow. Ohm’s law states that
current is proportional to the voltage across a
particular element, and inversely proportional
to its
resistance.
• Elements in a series circuit are connected
inline so the same current passes through all.
Total resistance is the sum of
the resistances of each.
• Elements in a parallel circuit
are connected such that they
form different paths. Total
resistance is less than the
least resistance.
73
Griffith, 2007
Griffith, 2007
Summary
Summary
Multiplying a voltage difference by charge yields
energy; multiplying a voltage difference by
current yields power, the amount of energy use.
The power supplied
by a source must
equal the power
dissipated in the
resistance.
Griffith, 2007
74
75
Schaum’s 26.33
Household circuits use alternating current,
which regularly changes direction. Effective
current (or voltage) is 0.707 times peak
current (or voltage). We connect appliances in
parallel. Because the current in each circuit is
additive, we
must be
careful not
to overload.
Griffith, 2007
76
Schaum’s 27.14
A battery has an emf of 13.2 V and an internal
resistance of 24.0 m. The load current is 20.0
A. What is the terminal voltage.
V=I*R
Vt = Vb – (I * R)
Vt = 13.2 V – (20.0 A * .024 )
Vt = 12.7 V
A conductor connected across a potential
difference of 100 V carries a current of 1.5 A.
What is the total charge transferred in one
minute?
I=q/t
q = It
q = (1.5 A)(60 s)
q = 90 C
77
78
13
11/2/2021
Schaum’s 27.14
Schaum’s 27.14
A conductor connected across a potential
difference of 100 V carries a current of 1.5 A.
What is the work done transferring the charge?
W=q*V
W = (90 C)(100 V)
W = 9000 J
W = 9.0 kJ
A conductor connected across a potential
difference of 100 V carries a current of 1.5 A.
What is the power expended to heat the wire?
P = IV
P = (1.5 A)(100 V)
P = 150 W
79
Schaum’s 27.14
80
Schaum’s 28.31
A conductor connected across a potential
difference of 100 V carries a current of 1.5 A.
What is the power expended to heat the wire?
P=W/t
P = (9000 J)(60 s)
P = 150 W
For the circuit shown in Fig. 28‐14 find its
equivalent resistance.
Rs = R1 + R2 + R3
1/Rp = 1/R1 + 1/R2 + 1/R3
R = Rab + Rcd + Rde
1/Rcd = 1/10 + 1/15 = 1/6
1/Rde = 1/9 + 1/18 + 1/30 = 1/5
R = 4 + 6 + 5 = 15
81
Schaum’s 28.31
82
Schaum’s 28.31
For the circuit shown in Fig. 28‐14 find the
current drawn from the power source.
V = IR
I = V/R
I = 300V / 15
I = 20 A
For the circuit shown in Fig. 28‐14 find the
potential difference across: ab, cd, de.
V = IR
Vab = I Rab = (20A)(4.0) = 80 V
Vcd = I Rcd = (20A)(6.0) = 120 V
Vde = I Rde = (20A)(5.0) = 100 V
Vtot = Vab + Vcd + Vde = 300 V
83
84
14
11/2/2021
Schaum’s 28.31
For the circuit shown in Fig. 28‐14 find the
current in each resistor.
I4 = 20A
I = V/R
I10 = 120/10 = 12A
I15 = 120/15 = 8 A
I9 = 100/9 = 11.1 A
I18 = 100/18 = 5.6 A
I30 = 100/30 = 3.3 A
85
15