Download Most useful circuits are more complicated than the simple circuits we

CLASS 22: SERIES AND PARALLEL CIRCUITS
22.1. INTRODUCTION
Most useful circuits are more complicated than the simple circuits we studied in the last chapter;
however, the same basic principles – conservation of energy and conservation of charge – still
determine the behavior of the current.
22.2. GOALS
ƒ Understand that current always prefers the path with the least resistance;
ƒ Batteries connected in series are connected with the positive terminal of one battery to the
negative terminal of the next battery. The voltage of the combined batteries is higher than the
voltage of the individual batteries. If batteries are connected with the positive and negative ends
together, the voltage is less than the voltage or either battery;
ƒ Batteries connected in parallel are connected with the positive terminal of one battery to the
positive terminal of the next battery. The voltage output is not higher, but they will last for a
longer time;
ƒ Loads connected in series are connected such that the current must go through each of the
individual loads. The brightness of a bulb connected in series is the same as the brightness if the
bulb were the only one connected to the battery;
ƒ Loads connected in parallel are connected such that the current does not travel through each
individual load. The brightness of a bulb connected in parallel is less than it would be if it were
the only bulb connected to the battery; and
ƒ Use Ohm’s law, the law of conservation of charge and the law of conservation of voltage to
numerically explain the current and voltage drop in all parts of series and parallel circuits.
22.3. BATTERIES IN SERIES
If there is more than one battery in the circuit, the total voltage supplied to the circuit depends on
how the batteries are hooked up. For example, if you have two ‘D’ batteries (each with 1.5 V), you
can hook them up two different ways, as shown in Figure 22.1.
E
B
C
R
V2
-
-
A
D
E
F
+ +
C
R
V2
B
A
F
+
Figure 22.1: Batteries connected in series.
-
+
-
3V
V1
0V
-
D
+
+
-
-
V1
+
+
The left drawing of Figure 22.1 shows two batteries connected with the positive terminal of one
leading to the negative terminal of the other. The current will have potential V1 + V2 after passing
through both batteries. If V1 and V2 are both have 1.5V emf, the total voltage available from the
combination is 3.0 V. Electronics that require large voltages use multiple batteries connected in
series this way.
In the right-hand-side drawing of Figure 22.1, the battery supplying V1 has been turned around. The
batteries act in opposition to each other when connected this way. If the two batteries have the same
emf, the total voltage from the combination will be 0 V. This is why electronics don’t work if you
don’t put the batteries in the equipment in the right orientation.
You should never open a battery, as most batteries have potentially dangerous chemicals, including
acids. On their website, the howstuffworks.com folks show what’s on the inside of a 9-V battery –
six 1.5V cells.
Figure 22.2: A nine-volt battery is six 1.5 volt batteries. Don’t
open a battery yourself at home – they contain acids that can
burn you and ruin clothing.
22.4. BATTERIES IN PARALLEL.
The voltage of any number of batteries in parallel is the same as the voltage of one battery, so why
bother? The answer is that a parallel arrangement of batteries lasts longer than a single battery – it’s
like using a larger battery. Figure 22.3 shows a schematic for batteries in parallel. Both batteries
supply a potential difference. Current can come from V1 or V2, the same current doesn’t come from
both. Each electron thus has the same energy it would have if only one battery were in the circuit.
The difference is that the two batteries produce more total energy.
V1
+ V2
+ -
R
Figure 22.3: Batteries connected in parallel.
+
+
+
+
-
-
-
-
1.5 V
22.5. RESISTORS IN SERIES
Figure 22.4a shows
a circuit with two
+
+
resistors, R1 and R2
and a battery with
E
A
emf V.
Charge
conservation tells us
that but no current is
Req= R1+R2
R2
R1
lost in the circuit.
D
C
B
All of the current
drawn by the circuit
Figure 22.4: a) Two resistors in b) The equivalent resistance to the must pass through
each of the resistors
series
two resistors in series shown in a.
and return to the
battery. There is no other place for the current to go except through both resistors. The same
current thus passes through all resistors in series with each other.
Passing through two resistors R1 and R2 in series is equivalent to passing through a single resistor
with a resistance that is the sum of the two resistors. We could draw this circuit as a simple circuit,
replacing R1 and R2 by one resistor with resistance Req = R1 + R2, as shown in Figure 22.4b. We call
the single resistor the equivalent resistance. This principle can be extended to any number of
resistors. If Req is the equivalent resistance of resistors in series of resistances R1, R2, etc. is given
by:
V
V
Req = R1 + R2 + R3 + ...
(22.5.1)
The voltage drops across the resistors will be different because the voltage will be given by Ohm’s
Law and different resistances with the same current flowing through them will have different
voltages.
EXAMPLE 22.1: A 9.0 V battery is hooked up to a resistance of 1.0 Ω in series with a resistance of 2.0 Ω.
a) What is the current running through the circuit? b) What is the potential drop across each resistor?
V = 9.0 V
+ E
A
Draw a picture
R2= 2.0 Ω
known:
V = emf = 9.0 V
R1 = resistance 1 = 2.0 Ω
R2 = resistance 2 = 1.0 Ω
R1 = 1.0 Ω
D
C
need to find:
I = current
We first need to find an equivalent resistance.
Equation to use (solved for
unknown)
Req = R1 + R2
B
Insert numbers
Req = 1.0 Ω+2.0 Ω
= 3.0 Ω
V = 9.0 V
+ -
We can draw an equivalent
circuit for this situation
Req= 3.0 Ω
Equation to use (solved for
unknown)
Insert numbers
I=
I=
V
Req
9.0 V
3.0 Ω
= 3.0A
Answer to part a:
I = 3.0 A
Part B. We know the current now, and we know that the same current goes through R1 and R2. The potential
drop across either resistor is given by Ohm’s Law.
Equation to use (solved for
unknown)
Insert numbers
V1 = IR1
V1 = 3.0 A ( 2.0 Ω )
=6.0 V
V2 = IR2
Similarly for R2
= 3.0 A (1.0 Ω )
= 3.0 V
Answer to part b:
Comments
V1 = 6.0 V
V2 = 3.0 V
Note that if you add up all the potential drops around the circuit, you must get
the same value as the total voltage supplied by the battery, which is required by
conservation of energy.
22.6. RESISTORS IN PARALLEL
When resistors are arranged to give the electrons more
than one possible path, we say the resistors are
+
connected in parallel. Let the battery again have
voltage V and the two resistors R1 and R2.
H
A
When the current reaches point B in the circuit, it can
go via one of two paths: through R1 or through R2.
Since current always takes the path with the least
D
C
resistance, the current would prefer to go through the
R1
smaller resistor; however, not all of the current can go
through one resistor, so a larger part of the current
B
G
goes through the path with the lowest resistance. The
two currents separate at point B and re-join each other
E
R2
F
at point G. The current behaves like cars going
through toll booths – when one line gets too long, cars
Figure 22.5: A simple parallel circuit
will go to a booth with a shorter line.
V
The current will split when it reaches point B: some of the
+
current goes toward C and some of the current goes toward point
E; however, when the currents meet up at point G, the total
H
A
current has to have the same value it did at point B according to
I
charge conservation. The sum of the current going into a
D
C
I1
junction must equal the current leaving the junction. If we
I
R1
assign the different currents names as we approach junction B, as
G
shown in Figure 22.6, Kirchoff’s second rule requires that
B
V
F
R2
E
I2
Figure 22.6: A simple parallel
circuit
parallel is:
I = I1 + I 2
(22.6.1)
Resistors add differently in parallel than they do in series. The
equivalent resistance Req of any number of resistors connected in
1
1
1
1
= + + + ...
Req R1 R2 R3
(22.6.2)
The equivalent resistance for the parallel arrangement always is less than any of the individual
resistances. There are more paths for the current to go through, so an arrangement of parallel
resistors has lower overall resistance than if any one of the resistors were used.
The voltage drops are distributed differently because the currents through the two resistors are
unequal. The voltage drop across the two resistors is the same for parallel arrangements. The
potential drop V1 across R1 is
V1 = I1 R1
and the potential drop V2 across R2 is
V2 = I 2 R2
Compare the potential difference across points B and G. The potential at point B is V, and the
potential at point G is zero – regardless of which path the current takes. The potential drop across
R1 must be the same as the potential drop across R2. For this circuit,
V1 = V2 = V
If two bulbs in parallel have the same resistance, half of the current will go through one bulb and
half will go through the other. This means that the current going through each bulb is less than the
current that would go through the bulb if it were the only bulb in the circuit. The result is that both
bulbs are dimmer than if all the current went through one bulb.
A parallel circuit has the advantage that one bulb can be removed and the other will remain lit
because the electrons still have a complete path to travel. In a series circuit, if one bulb is removed,
the complete circuit is broken and current cannot flow.
EXAMPLE 22.2: What is the equivalent resistance of a circuit in which V = 9.0 V, R1 = 2.0 Ω, R2 = 1.0 Ω
where the resistors are arranged in parallel? b) What is the total current in the circuit? c) What is the current
that goes through each resistor?
V = 9.0 V
+ -
Draw a picture
R2= 2.0 Ω
R1 = 1.0 Ω
known:
V = emf = 9.0 V
R1 = resistance 1 = 2.0 Ω
R2 = resistance 2 = 1.0 Ω
Req = equivalent resistance
need to find:
We first need to find an equivalent resistance.
Equation to use (solved for
unknown)
Insert numbers
1
1
1
= +
Req R1 R2
1
1
1
=
+
Req 1.0 Ω 2.0 Ω
=
31
2Ω
Be careful here because we have solved for one over the resistance we want.
1
31
=
R eq 2 Ω
2
Req = Ω
3
Answer (part a)
2
Req = Ω
3
V = 9.0 V
+ -
For part b, we need to
draw an equivalent circuit
I
Req =
Equation to use (solved for
unknown)
I=
I=
2
Ω
3
V
Req
9.0 V
2
3 Ω
9.0 V ⎛ 3 ⎞
⎜ Ω⎟
⎝2 ⎠
27
=
A
2
Insert numbers
=
Leave all the answers as fractions – otherwise, you will end up with errors due to the truncation.
I=
Answer to part b:
27
A
2
This is the current that travels through the entire circuit. Now we have to calculate the current going through
each resistor. We know that the voltage drop across each resistor is the same, and we know that it has to be
9.0 V because all the voltage has to be dropped across the combination of resistors.
Part B. We know the current now, and we know that the same current goes through R1 and R2. The potential
drop across either resistor is the same and, looking at the equivalent circuit, we know that:.
Known:
Equation to use
Solve for the unknown
V1 = voltage across R1 = 9.0 V
V2 = voltage across R2 = 9.0 V
V1 = I1 R1
I1 =
V1
R1
Insert numbers
9.0 V
2.0 Ω
9
= A
2
Equation to use
V2 = I 2 R2
I1 =
Solve for the unknown
I2 =
V2
R2
I2 =
Insert numbers
9.0 V
1.0 Ω
= 9.0 A
Answer
I1 =
9
A I 2 = 9.0 A
2
Check your answer. We know that the sum of the currents has to equal the total current
9
A + 9.0 A
2
9
18
= A+ A
2
2
27
= A
2
I1 + I 2 =
If you have batteries and resistors together, it is easiest to first replace the multiple batteries with a
single battery of the appropriate potential, and then to analyze the resistors as we’ve done above.
SUMMARIZE
22.6.1. Definitions: Define the following in your own words. Write the symbol used to represent
the quantity where appropriate.
1. Equivalent resistance
22.6.2. 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 equivalent resistance and the individual resistance of resistors in
series.
2.
The relationship between the equivalent resistance and the individual resistance of resistors in
parallel.
22.6.3. Concepts: Answer the following briefly in your own words.
1. Which of the following statements are true? a) When resistors are in series, the effective
resistance is greater than any of the individual resistors; b) When resistors are in parallel, the
effective resistance is greater than any of the individual resistances; c) When resistors are in
series, the effective resistance is less than any of the individual resistors; d) When resistors are
in parallel, the effective resistance is less than any of the individual resistances. Explain your
choices.
2.
Which of following statements are true? a) The same current passes through each resistor
when the resistors are connected in series; b) More current passes through the largest resistor
when resistors are connected in series; c) The voltage drop is the same across each resistor
when the resistors are connected in series; d) The largest voltage drop for resistors in series will
occur in the largest resistance. Explain your choices.
3.
Which of following statements are true? a) The same voltage passes through each resistor
when the resistors are connected in parallel; b) More voltage passes through the smallest
resistor when resistors are connected in parallel; c) The current is the same across each resistor
when the resistors are connected in parallel; d) The largest current through resistors in series
will occur in the largest resistance. Explain your choices.
4.
Explain in your own words the difference between batteries connected in series positive to
negative and batteries connected in series positive to positive.
5.
Explain in your own words the difference between batteries connected in series (positive to
negative) and batteries connected in parallel.
22.6.4. 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.
22.6.5. Questions to Think About
1. Make a table to show whether the current or voltage is the same over resistors in parallel and
resistors in series. Indicate how the other quantity (the one that isn’t constant) changes with the
size of the resistor.
2. Explain in your own words what conservation of charge and conservation of energy mean in a
circuit with resistors in parallel and in a circuit with resistors in series.
22.6.6. Problems
1. What is the equivalent resistance of
V
the circuit at right if V = 20 V,
+
R1 = 8 Ω, R2 = 6 Ω and R3 = 3 Ω?
R1
R3
R2
2.
3.
In the circuit shown on the right, a
current of 3A flows through the
circuit. a) What would a voltmeter
read when connected between points
A and B? b) What would it read if
connected between points B and C?
The same current flows through
each resistor.
c) what is the
potential at each point A, B and C?
A
F
30 V
3Ω
7Ω
B
C
D
E
What is the equivalent resistance of the circuit
at right if R1 = 6Ω, R2 = 6Ω and R3 = 3 Ω?
V
H
A
C
R1
R2
B
E
F
R3
D
+
B
R1
C
A
H
F
4. For the circuit shown at left, find:
the current passing through each resistor
the voltage drop across each resistor
-
R3
I
R4
G
R2
D
V
L
R5
E
J
G
K
Take:
R1 = 2/3 Ω
R2 = 2 Ω
R3 = 3 Ω
R4 = 4 Ω
R5 = 12 Ω
V = 12 V
5. For the circuit shown to the right, find:
the current passing through each resistor
the voltage drop across each resistor
Take:
R1 = 2/3 Ω
R2 = 2 Ω
R3 = 3 Ω
R4 = 4 Ω
R5 = 12 Ω
V = 12 V
C
R1
D
B
G
E
A
V
R2
F
I
R3 J
K
R4 L
M
R5 N
H
P
O
PHYS 261 Spring 2007
HW 23
HW Covers Class 22 and is due March 2nd, 2007
In the circuit at right: V = 20 V, R1 = 8 Ω,
R2 = 6 Ω and R3 = 3 Ω?
1. What is the equivalent resistance of the
circuit?
2. What are the voltage drops across each
resistor?
3. What are the currents running through
each resistor?
V
+
-
R1
R3
R2