CLASS 21: SIMPLE CIRCUITS
21.1. INTRODUCTION
We have an energy source (the battery) that produces a current, and we know that different objects
have different resistances. In this section, we put together all of these elements to make simple
circuits.
21.2. GOALS
ƒ Understand that current must be able to travel in a complete, unbroken path around the circuit
from one electrode of the battery to the other;
ƒ Explain the difference between open, closed and short circuits;
ƒ Understand how energy is distributed throughout a circuit, including the transformation of
electrical energy into heat, light or sound;
ƒ Be able to draw a schematic diagram of a circuit;
ƒ Explain the difference between conventional and electron current;
ƒ Use Ohm’s law to calculate the voltage drops across loads and the current carried in a simple
circuit; and
ƒ Understand that energy and charge must be conserved in a circuit.
21.3. THE BASIC PRINCIPLES OF SIMPLE CIRCUITS
21.3.1. Circuits. A circuit is a set of electric elements that are connected to each other in such a
way that there is a complete path for current to flow through. A simple circuit is a light bulb, one
wire and a battery. In more technical terms, a circuit must contain a source of voltage (emf) (the
battery), a load (the bulb) and wires to form a complete path.
21.3.2. Rules for Circuits
1. Circuits must contain a source of emf (usually a battery)
2. Current must be able to travel in a complete, unbroken path around the circuit from one terminal
of the battery to the other.
3. The current carriers must “get rid” of all the emf they acquired from the battery by the time they
reach the negative terminal. They do this by converting the energy from the battery to light, heat,
sound or other forms of energy.
21.3.3. Open and Closed Circuits. A closed circuit is an arrangement of electric elements that
provides a complete path from the battery through the wires, through the load(s) and back to the
battery. An open circuit is an arrangement that does not form a complete path and so current does
not flow. Current does flow in a short circuit, but not through the intended path. For example, if
there is a short in a lamp cord, the current could flow directly between the two wires and never reach
the lamp itself.
21.3.1. Conventional Current vs. Electron Current. We mentioned earlier that Ben Franklin
named the two charges such that the electron ended up being negative. People had talked for so long
about positive charges that this was little confusing. The conventional current is current viewed as
being positively charged and traveling in the direction from the positive terminal to the negative
terminal. The electron current which is the net direction of electron motion (i.e. the direction of the
drift velocity). The electron current flows from the negative terminal of the battery around to the
positive terminal. Different texts use different currents. We will use the electron current.
21.4. DRAWING CIRCUITS
It doesn’t make sense to draw a toaster if you are trying to figure out how current is distributed
through a circuit involving a toaster. A set of shorthand symbols has developed and these symbols
are used to draw a schematic, which is a shorthand way of communicating how circuit elements are
connected. Table 21.1 shows the various symbols we will use.
Object
Symbol
Battery: In a battery, the longer line represents
the positive terminal and the shorter line
represents the negative terminal.
Load: Loads also are known as resistances
because they resist the flow of current. The
general symbol is shown at top right; however,
some elementary-level books use a symbol that
looks more like a bulb (bottom right).
Switches are used to start and stop the flow of
electricity by completing or breaking the path
the electrons must travel. The top figure at
right shows an open switch and the bottom
shows a closed switch.
A fuse.
Table 21.1: Symbols for schematics
21.5. WHAT HAPPENS IN A SIMPLE CIRCUIT.
The rules that govern how a circuit behaves (called Kirchoff’s Laws) are based on two fundamental
ideas.
• Energy must be conserved in the circuit.
• Charge must be conserved in the circuit.
Conserved means that none of the quantity (energy or charge) is lost in the circuit.
21.5.1. Conservation of Energy. The first rule says that there is a fixed amount of energy available
to the circuit. If we think about the current as consisting of electrons (but remember that this is a
construct we use!), each electron leaves the battery with voltage V. The electrons travel around the
circuit and, when they encounter the load, they give up the same voltage V to pass through the load –
the load is like a toll bridge. We call the loss of energy as current traverses a load a voltage drop.
When the electrons return to the battery, they have no energy, but they gain voltage V from the
battery. (Sort of like passing ‘go’ in monopoly.) This is repeated for as long as there is a complete
circuit, or until the battery dies. This means that if you add up all the voltage sources (batteries) and
subtract all the voltage drops (loads), you have to end up with zero.
21.5.2. Conservation of Charge. The second rule – conservation of charge – says that the same
amount of current entering a junction (a junction is where one wire splits into two) must equal the
amount of current leaving the junction. In our model of electrons as current carriers, we can say that
all of the electrons that come into a junction must leave the junction.
21.6. OHM’S LAW
The current has a certain potential V that is determined by the battery. The resistance of the load is
determined by the load itself (for example, a dryer vs. a light bulb). The relationship between the
potential V, the current I, and the resistance of the load, R, is given by Ohm’s Law.
V = IR
(21.6.1)
The ohm is related to the volt and the ampere by:
Ω=
V
A
Ohm’s Law is valid only for metallic conductors. More complex expressions are required for other
materials. In many cases, Ohm’s Law provides a good approximation for the actual behavior.
EXAMPLE 21.1: A 110 V potential difference is applied to a blender with a resistance of 45.0 Ω. a) What
is the current going through the blender? b) How much charge will flow through a given point in 2.00
minutes? c) How many electrons is this?
110 V
Draw a picture
45 W
known:
V = potential difference = 110 V
R = resistance = 45.0 Ω
Δt = time elapsed = 2.00 min = 120 s
I = current = 55.8 m
ΔQ = charge in 2.00 min
Ne = number of electrons in 2.00 min
need to find:
V =I R
Equation to use (part a)
I=
Solve for unknown
Insert numbers
Answer:
Equation to use (part b)
V
R
110 V
45.0 Ω
= 2.444444 A
I=
I = 2.44 A
I=
ΔQ
Δt
ΔQ
Δt
Δq = I Δt
I=
Solve for unknown
ΔQ = (2.444444 A)(120 s)
= 293.33333 A s
⎛C⎞
= 293⎜⎜ ⎟⎟⎟ s
⎜⎝ s ⎠
Insert numbers
= 293 C
ΔQ = 293 C
Answer:
Comments:
This is a very large charge – imagine how fast the electrons must be traveling if
this many pass through a point in 2 minutes!
ΔQ = N e e
Equation to use (part c)
Ne =
Solve for unknown
Insert numbers
Ne =
ΔQ
e
293.3333 C
C
1.60×10-19 electron
=1.83333×1021 electrons
N e = 1.83×1021 electrons
Answer:
21.7. POWER
Power P is defined as the energy E used divided by the time over which it is used, t.
P=
Energy
time
P=
E
Δt
(21.7.1)
Be careful! We’re running out of variables, so you have to use the context of the equation to
determine whether E stands for energy or electric field. Power will always involve energy and not
electric field.
When an amount of charge ΔQ, passes through a potential difference, V, it acquires an energy E.
E = ΔQ V
(21.7.2)
If the charge takes time Δt to move across the potential difference, the power is
P=
ΔQ V
Δt
⎛ ΔQ ⎞
P=⎜
⎟V
⎝ Δt ⎠
(21.7.3)a
The quantity in parenthesis is the current, so
P = IV
(21.7.3)b
The unit of power, the watt (W), is equivalent to an ampere times a volt.
W=AV
We can also write the power in terms of the resistance, by using Ohm’s law. There are two
additional forms for the power.
P = IV
= I ( IR)
P = I 2R
(21.7.3)c
and…
P = IV
⎛V ⎞
= ⎜ ⎟V
⎝R⎠
P=
V2
R
(21.7.3)d
Energy is expressed in joules (J), or in kW-h. Since power is energy per time, multiplying power
times time gives you energy. The typical power used by various appliances is shown in Table 21.2
Appliance
Charger
for
toothbrush
an
Power (W)
electric
1
Clothes dryer
5000
Coffeemaker
700
Dishwasher
1600
Fan
Fax machine
150
65
Appliance
Iron
Personal computer
Photocopier
Power (W)
1000
150
1400
Portable sander
400
Refrigerator
400
Stove
12000
Hair dryer
1500
TV Receiver
120
Heater
2000
Vacuum Cleaner
750
Table 21.2: The power used by various common appliances.
EXAMPLE 21.2: What is the resistance of a 100-W light bulb operating at 120 V? (Assume both
values given have three significant figures.)
Draw a picture
known:
V = potential difference = 120 V
P = power = 100 W
No picture is necessary
need to find:
R = resistance
P=
Equation to use:
V2
R
Note: We choose which of the three forms for the power to use depending on which quantities are known. In
this case, we know V and R, so there is only one form that works.
R=
Solve for unknown
Insert numbers
V2
P
(120 V) 2
100 W
=144 Ω
R=
R = 144 Ω
Answer: (3 s.f.)
EXAMPLE 21.3: What is the current passing through a 1250 W hair dryer operating at 120 V? (Assume 3
significant figures for each quantity)
Draw a picture
known:
V = potential difference = 120 V
P = power = 1250 W
No picture is necessary
need to find:
I = current
Equation to use (part a)
P = IV
Solve for unknown
I=
Insert numbers
I=
P
V
1250 W
120 V
=10.4266666 A
Answer:
I =10.4 A
EXAMPLE 21.4: If electricity is $0.054 per kW-hr, how much does it cost to run a 75 W light bulb for 24
hours?
No picture is necessary
Draw a picture
c = cost rate = 0.054
known:
P = power = 75 W
t = time = 24 h s.
$
kW h
need to find:
C = total cost
Equation to use: If one kWh costs$0.054, then two kW-h would cost twice this and so on. The first step is
thus to find out how much energy is used by a 75-W light bulb in 24 hours.
How much energy does
The bulb has a power of 75 W and runs for a time of 24 h.
the bulb need?
Equation to use
Solve for unknown
P=
E
Δt
E = P Δt
E = (75 W )( 24 h )
Insert numbers
= 1800 W h
= 1.800 kW h
The cost rate, c is how much we pay per kW-h, so the total cost is the number of kW-h we use times the
charge per kW-h
Equation to use (already
solved for unknown)
Insert numbers
C=c E
⎛
$ ⎞⎟ ⎛⎜1.800 kW h ⎞⎟
⎟⎜
C = ⎜⎜0.054
⎟
⎜⎝
⎠⎟
kW h ⎠⎟⎟ ⎝⎜
= 9.720×10-2 $
Answer:
Comments
C = 9.7×10-2 $
That’s just a little less than ten cents (10 cents = 0.1 dollar) to run a 75-W light
bulb 24 hours per day. That doesn’t seem a lot, but multiply that number by,
say, the number of students living in the dorms and leaving the bulb on day
after day. It adds up. .
21.8. ANALYZING A SIMPLE CIRCUIT
V
+
-
D
A
The simplest circuit consists of a battery with
potential V and a load of some type, which
might be a lamp, a radio, etc. The most
important characteristic of the load is its
resistance R1.
Letters indicate important points in the circuit.
This may seem silly for a simple circuit;
however, they will familiarize you with the
analysis technique, which is very helpful when
dealing with more complicated circuits.
R1
C
B
21.8.1. Calculate the Current in the Circuit
EXAMPLE 21.5 If V = 9.0 V and R1=3.6 Ω, calculate the current in the circuit.
V
+
-
D
A
Draw a picture
R1
C
known:
V = potential due to battery = 9.0 V
R1 = resistance of load = 3.6 Ω
Equation to use (solved for
unknown)
Insert numbers
B
I = current
need to find:
I=
I=
V
R
9.0 V
3.6 Ω
= 2.5A
Answer:
I = 2.5A
Current is the number of electrons that pass a given point over some time period. All of the current
that starts at point A must eventually reach point D. Current is not “used up” along the way: it just
loses energy. The current we calculated above is the current in the entire circuit, at all points. So if I
asked you, “What is the current in the circuit at point C?”, the answer would be 2.5 A. The answer
to “What is the current in the circuit at point D?” is 2.5 A.
21.8.2. Finding potentials
What is the potential at point B? The current starts at point A and travels to point B. In our ideal
circuits, we will assume always that wires have no resistance, so there is no potential drop in the
wires. This is reasonable because wires are conductors and have very low resistance compared to
typical loads. The current doesn’t lose any energy between point A and point B, so the potential at
point B is the same as the starting potential: 9.0 V.
What is the potential at point C? When the current crosses the load, it transfers energy to the load.
There is only one place to deposit energy in the circuit, and since we know that the current cannot
have any energy when it returns to the battery, the potential drop across the load must be 9.0 V. So
if the potential is 9.0 V at point B and we drop 9.0 V across the load, the potential at point C is 0 V.
This is also the potential at point D. The potential immediately next to the positive battery terminal
always will be 0 V.
21.9. SUMMARIZE
21.9.1. Definitions: Define the following in your own words. Write the symbol used to represent
the quantity where appropriate.
1. Schematic
2.
Voltage drop
3.
Kirchoff’s Laws
21.9.2. Equations: For each question: a) Write the equation that relates to the quantity b) Define
each variable by stating what the variable stands for and the units in which it should be expressed,
and c) State whether there are any limitations on using the equation.
1. The relationship between the current, voltage and resistance of a simple circuit.
2.
The relationship between power, energy and time.
3.
The relationship between energy, charge and potential difference.
4.
The three different relationships between power in terms of any two of the following: current,
resistance and voltage.
21.9.3. Concepts: Answer the following briefly in your own words.
1. Explain the differences between open, closed and short circuits.
2.
What are the necessary elements required to have a closed circuit?
3.
Explain the difference between conventional current and electron current.
4.
Why can we ignore the resistance of the wires in problems we work?
5.
What happens to the energy carried by the current when the current crosses the load?
6.
Explain the ideas of conservation of charge and conservation of energy in circuit using your
own words.
21.9.4. Your Understanding
1. What are the three most important points in this chapter?
2.
Write three questions you have about the material in this chapter.
21.9.5. Questions to Think About
1. What is the advantage of using 240-V power lines (instead of the more common 120 V power
lines) for plugging in appliances like clothes dryers and stoves?
21.9.6. Problems
1. Each string of super-bright Christmas tree lights requires 100 W of power. If the voltage
available is the standard 120V (assuming 3 significant figures for both numbers provided):
a) How much current does each string of lights draw?
b) How many strings of lights can we safely plug into the outlet if the extension cord is rated for
5.00 A?
2. How much does it cost you to keep your cable TV box plugged in for one year continuously?
A typical cable-TV box uses 15.0 Watts of power, even when the TV is not on. (You have to
unplug the box to keep it from drawing current.) Assume that power costs you $0.05 per kWhr. Note: The amount of money you will calculate doesn’t seem like a lot, but when you figure
multiple cable boxes, plus microwave ovens, cordless phones, TVs, VCRs, answering
machines, etc., it ads up to total 30% of the energy usage every year. By 2015, this type of
energy usage is anticipated to be 50% of the total energy usage. Appliances that draw energy
even when not in use are called "energy vampires".
3. A lightning stroke might have a potential difference of 2 million volts, a current of 5000 A and
last for 1 ms.
a) How much power is contained in the lightning strike?
b) How much energy is contained in the lightening strike?
4. How many electrons flow through a 75.0-W, 120-V light bulb each second? (Assume 3
significant figures in the 120 V number.)
5. A 120.0 V coffee pot draws a current of 0.6 A. What is the resistance of the coffee pot?
6. What potential difference is needed across a 1.500×103 Ω resistance at a current of 1 mA?
7. A 4.00×102 W power drill runs off a 240.0 V line. What is the current the drill draws?
The potential difference between a cloud and the ground is 4.00×106 V. 80.0 C of charge is
transferred in 1.00 ms.
a) How much power is contained in the lightning strike?
b) How much energy is contained in the lightening strike?
9. What voltage is necessary to produce a current of 5.00 A in a resistance of 40.0 Ω?
10. What is the resistance of a light that draws a current of 2.00 A and is run from a 12.0 V battery?
8.
PHYS 261 Spring 2007
HW 22
HW Covers Class 21 and is due February 28th, 2007
1.
2.
3.
The potential difference between a cloud and the ground is 4.00×106 V. 80.0C of charge is
transferred in 1.00 ms.
a) How much power is contained in the lightning strike?
b) How much energy is contained in the lightening strike?
What happens to the energy carried by the current when the current crosses the load?
Each string of super-bright Christmas tree lights requires 100 W of power. If the voltage
available is the standard 120V (assuming 3 significant figures for both numbers provided):
a) How much current does each string of lights draw?
b) How many strings of lights can we safely plug into the outlet if the extension cord is rated
for 5.00 A?