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Lesson#23
Topic: Simple Circuits
11/13/06
Objectives: (After this class I will be able to)
1.
2.
3.
4.
5.
Explain the difference between wiring light bulbs
in series and in parallel
Describe EMF and terminal voltage of batteries.
Describe how the difference between EMF and
terminal voltage can be treated as internal
resistance of the battery.
Find voltage, current, and resistance for a simple
circuit
Explain the loop rule for a single loop.
Assignment:
Warm Up
►
1.
2.
3.
4.
5.
A simple series circuit consists of three identical lamps
powered by a battery. When a wire is connected
between points a and b.
What happens to the brightness of lamp 3?
Does current in the circuit increase, decrease, or
remain the same?
What happens to the brightness of lamps 1 and 2?
Does the voltage drop across lamps 1 and 2 increase,
decrease, or remain the same?
Is the power dissipated by the circuit increased,
decreased, or does it remain the same?
a
b
Series vs. Parallel Circuits
► Light
bulbs wired in series are lined up one
right after the other.
► The current throughout the circuit remains the
same.
► The circuit makes a single loop from the
positive terminal of the battery to the negative
terminal.
► Light bulbs wired in parallel are each
individually connected to the battery terminals.
► The current is then split and divided amongst
each light bulb.
Series vs. Parallel Circuits
► The
voltage drop across identical light bulbs in
series is divided evenly among each bulb.
► The voltage drop across identical light bulbs in
parallel is equal to the original voltage of the
battery.
► Batteries wired in series add voltage (or
subtract if are reversed)
► Batteries wired in parallel do not add.
► Two batteries in parallel would light the bulb
about the same as one battery by itself.
► Possibly only slightly brighter due to internal
resistance.
EMF
Batteries are rated by the voltage difference between
terminals, but when current begins to flow between
terminals, the voltage drops.
► This is similar to what happens in water pipes when
you turn on a faucet.
► The water pressure is larger when the faucet is closed
than when the water is running.
► The plumbing cannot maintain the same pressure as
the water is removed.
► The more water that flows (current) the lower the
pressure becomes.
► The voltage across the battery terminals goes down as
current is drawn.
► The more current drawn, the more the drop in voltage.
►
EMF
►
Technically, the chemical reactions in the battery
responsible for producing the electric potential energy
cannot proceed quickly enough to maintain the
potential difference.
►
The initial voltage that the battery is rated at for when
there is no current is its EMF.
►
The actual voltage between terminals when the circuit
is closed is the terminal voltage.
►
The more current drawn from the battery, the smaller
the terminal voltage.
Internal Resistance
What is typically done when quantifying the properties
of circuits is to measure the internal resistance of the
battery.
► The difference in voltage between the EMF and the
terminal voltage can be treated as a tiny resistor
located within the battery.
► We can then say that the EMF is the actual voltage of
the battery.
► And that there is an internal resistance that causes a
voltage drop as soon as the circuit is closed.
► This explains why two batteries in parallel will light a
bulb brighter than a single battery in series.
► This is because when used in parallel, less current
needs to be drawn from each battery.
► With less current, the internal resistance of each
battery has less of an effect.
►
Internal Resistance
►
A typical fresh AA dry cell has an EMF of 1.5V
and an internal resistance of .31ohms
What is the terminal voltage of the battery if
the battery is hooked up in series to a resistor
such that a current of 58mA is measured
through the resistor?
2. What is the resistance of the resistor?
3. Suppose the resistor was replaced by a wire of
negligible resistance. What current will flow
through the wire?
1.
Loop Rule
►
If a circuit contains a single battery and a single resistor
it is simple to find the current through the circuit.
►
For circuits with multiple elements a loop rule needs to
be applied.
►
This rule extends from the idea that the voltage between
a point and itself is zero.
►
As we go around a complete loop in a circuit (starting
and ending at the same point), we find that each
element is associated with a voltage across it.
►
Loop Rule: The sum of all of the voltages across each
element in the loop as you go around the loop should
equal zero.
Loop Rule Example
►
Find the current in the following circuit.
8Ω
6V
2Ω
Loop Rule Example
►
Find the current in the following circuit.
8Ω
16V
6V
2Ω
11/9/06
Lesson #24
Topic: Lab: Simple Circuits
Objectives: (After this class I will be able to)
1.
Create a simple series circuit with light bulbs
2.
Measure the voltage drop and current through a light
bulb using a multimeter.
3.
Create a simple parallel circuit with light bulbs
4.
Compare the changes in voltage and current in a series
circuit to a parallel circuit.
5.
Mechanically create the power needed to generate the
electricity needed to light four bulbs and make note of
the difference in torque when bulbs are removed.
Lab Task: Complete the Hewitt lab #91. Make sure all tables are
completed and questions are answered.
Assignment: Lab due at the end of the period.
Lesson #25
Topic: Multi-loop Circuits
Objectives: (After this class I will be able to)
1.
2.
3.
4.
11/16/06
Describe and draw a schematic of a multi-loop
circuit
Pick out single loops from a multiple loop
schematic
Write voltage equations for each individual loop
Solve for multiple unknown currents in different
sections of a circuit
Warm Up: Draw a circuit with a 12V battery and 3 resistors wired in parallel
with one another. If the resistance of each resistor is 2ohms, 4 ohms, and 6ohms,
can we say that the current through each section is the same? How can we
predict the current in each section?
Assignment: Holt Physics p757 #43 b, c, d, e
Multi-loop Circuits
► Most
real circuits have multiple loops and are
much more complicated than the light bulb
circuits we have been dealing with.
► The complication arises when there are different
size resistors located at within different loops.
► This means that the current will be different for
different sections of the circuit.
► To solve for the current in each section, we will
have to create multiple loop equations to solve
for the multiple unknown variables.
► To create these equations we will follow the loop
rule, or specifically:
► Kirchoff’s loop rules
Multiple Loops
►
Consider the following circuit.
Itop = ?
4Ω
16V
8Ω
2Ω
►
►
►
►
6V
Imid = ?
Ibot = ?
Pick out two loops seen within this circuit.
Then use the loop rule to write an equation for each
individual loop.
For this loop there are three unknowns, so you should
have at least three equations.
Notice that there are three independent loops in this
circuit. The top loop, the bottom loop, and the outer
loop.
Junction Rule
► The
loop rule can only be used to get two of the
equations.
► In order to get the third independent equation we
need to examine a point in the circuit where the
current is split or comes back together.
► Because current is conserved, we can add up the
current from two divided sections to obtain the
current from the original source.
4Ω
Itop = ?
Imid = ?
16V
8Ω
2Ω
6V
Ibot = ?
Itop + Imid = Ibot
Solving for multiple unknowns
►
Use the following schematic and direction of
current to create two more equations besides
the junction rule, and solve for each of the
unknowns using algebraic substitution.
4Ω
Itop = ?
Imid = ?
16V
8Ω
2Ω
6V
Ibot = ?
Equations:
Solution
Practice
A simple circuit is set up as shown. If the ammeter reads 1A, what is the
resistance R?
1Ω
7V
R
6V
2Ω
4Ω
A
Lesson #26
Topic: Lab: Predicting Resistance
Objectives: (After this class I will be able to)
1.
2.
3.
4.
11/29/06
Build a circuit from a schematic
Correctly wire a multimeter to measure the
current in a section of wire.
Write equations and solve for multiple unknown
variables
Compare the calculated prediction of resistance
to the measured resistance of a light bulb.
Lab Task: In addition to the usual lab report, complete the calculations and
questions found on the lab sheet.
Assignment: Lab Report due tomorrow.
DC Exam Review due Friday.
Lesson #27
Topic: Simplifying Circuits
Objectives: (After this class I will be able to)
1.
2.
3.
11/30/06
Add resistance of resistors that are wired in
series
Add resistance of resistors that are wired in
parallel.
Simplify a complex circuit to a simple
circuit by calculating total resistance.
Warm Up: If you add resistors to a circuit wired in series, will the overall
resistance go up or down? What if the circuit and the added resistors were wired
in parallel with each other?
Assignment: Holt Physics p755,756 #23, 24, 25, 26, 33
Adding Resistance in Series
►
►
►
►
►
►
When you add resistors to a circuit in series, the overall
resistance of the circuit will increase
This is because the same current is forced through each
resistor
The voltage drop across each resistor is the total voltage
divided by each resistor.
If the resistance of each resistor is constant, and the
voltage drop per resistor goes down with each resistor
added, then the current must drop as well.
Hence, as you add bulbs to a series circuit, bulbs get
dimmer.
The total resistance of a series circuit is the simple
addition of the resistance of each individual resistor.
Rtotal  R1  R2  R3  ....  Rn
Practice
► Find
the total resistance in the circuit.
► Find the total current drawn from the
battery.
1Ω
2Ω
3Ω
4Ω
7V
5Ω
6Ω
7Ω
Adding Resistance in Parallel
►
►
►
►
►
►
When you add resistors to a circuit in parallel, the overall
resistance of the circuit will decrease.
This is because the current is split through each resistor.
The voltage drop across each resistor is the same as the
total voltage of the circuit.
If the resistance of each resistor is constant, and the
voltage drop per resistor remains the same with each
resistor added, then the current drawn from the battery
must increase.
If the voltage is held constant and the overall current
increases, then the resistance must drop as bulbs are
added.
The total resistance of a parallel circuit is the inverse of
the addition of the inverses of each individual resistor.
1
Rtotal
1 1
1
1
 

 .... 
R1 R 2 R 3
Rn
Practice
► Find
the total resistance in the circuit.
► Find the total current drawn from the
battery.
3Ω
2Ω
1Ω
6Ω
7V
Simplifying Circuits
► With
these addition rules, we can simplify
circuits to a single battery and resistor.
► This is useful for finding the total voltage,
resistance, or current drawn from a single
battery in a circuit.
► This cannot be used to solve for the current
within sections of a parallel circuit.
Practice
Find the equivalent resistance of circuit and the current drawn from the battery.
8Ω
1Ω
8Ω
6V
2Ω
4Ω
4Ω