Ohm's Law
OBJECTIVES
After
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•
•
•
•
•
•
•
•
•
completing this chapter, the student should be able to:
Identify the three basic parts of a circuit.
Identify three types of circuit configurations.
Describe how current flow can be varied in a circuit.
State Ohm's law with reference to current, voltage, and resistance.
Solve problems using Ohm's law for current, resistance, or voltage in series,
parallel, and series-parallel circuits.
Describe how the total current flow differs between series and parallel circuits.
Describe how the total voltage drop differs between series and parallel circuits.
Describe how the total resistance differs between series and parallel circuits.
State and apply Kirchhoff's current and voltage laws.
Verify answers using Ohm's law with Kirchhoff's law.
See accompanying CD for interactive presentations, tutorials and Ohm's Law
examples in MultiSim, Chapter 5.
Ohm's law defines the relationship among three
fundamental quantities: current, voltage, and resistance. It states that current is directly proportional to voltage and inversely proportional to
resistance.
This chapter examines Ohm's law and how it
is applied to a circuit. Some of the concepts are introduced in previous chapters.
m:I
ELECTRIC CIRCUITS
As stated earlier, current flows from a point with
an excess of electrons to a point with a deficiency
of electrons. The path that the current follows is
called an electric circuit. All electric circuits consist
of a voltage source, a load, and a conductor. The
voltage source establishes a difference of potential
49
SECTION 1 DC CIRCUITS
that forces the current to flow. The source can be
a battery, a generator, or another of the devices
described in Chapter 3, Voltage. The load consists
of some type of resistance to current flow. The resistance may be high or low depending on the
purpose of the circuit. The current in the circuit
flows through a conductor from the source to the
load. The conductor must give up electrons easily. Copper is used for most conductors.
The path the electric current takes to the load
may be through any of three types of circuits: a series circuit, a parallel circuit, or a series-parallel circuit. A series circuit (Figure 5-1) offers a single
continuous path for the current flow, going from
the source to the load. A parallel circuit (Figure 5-2)
offers more than one path for current flow. It allows
the source to apply voltage to more than one load.
It also allows several sources to be connected to a
FIGURE 5-3
A series-parallel circuit is a combination of a series circuit
and a parallel circuit.
R3 •
L-.J
1
FIGURE 5--4
Current flow in an electric circuit flows from the negative
side of the voltage source, through the load, and returns to
the voltage source through the positive terminal.
I
FIGURE 5-1
LOAD
A series circuit offers a single path for current flow.
I
FIGURE 5-5
A closed circuit supports current flow.
-=-
E -T
FIGURE 5-2
A parallel circuit offers more than one path for current flow.
-=iT
L-
~
~
~ R,
J J
~ R2
J
single load. A series-parallel circuit (Figure 5-3) is a
combination of the series and parallel circuit.
Current in an electric circuit flows from the
negative side of the voltage source through the
load to the positive side of the voltage source
(Figure 5--4). As long as the path is not broken, it
is a closed circuit and current flows (Figure 5-5).
However, if the path is broken, it is an open circuit and no current can flow (Figure 5-6).
CHAPTER 5 OHM'S LAW
FIGURE 5-6
FIGURE 5-8
An open circuit does not support current flow.
Current flow in an electric circuit can also be changed by
varying the resistance in the circuit.
tR
lR
I!
It
FIGURE 5-7
Current flow in an electric circuit can be changed by varying
the voltage.
IE
It
!E
I!
The current flow in an electric circuit can be
varied by changing either the voltage applied to the
circuit or the resistance in the circuit. The current
changes in exact proportions to the change in the
voltage or resistance. If the voltage is increased, the
current also increases. If the voltage is decreased,
the current also decreases (Figure 5-7). On the
other hand, if the resistance is increased, the current decreases (Figure 5-8). This relationship of
voltage, current, and resistance is called Ohm's law.
5-1
3. Draw a diagram of a circuit showing how
current would flow through the circuit.
(Use arrows to indicate current flow.)
4. What is the difference between an open
circuit and a closed circuit?
5. What happens to the current in an
electric circuit when the voltage is
increased? When it is decreased? When
the resistance is increased? When it is
decreased?
DB OHM'S
LAW
Ohm's law, or the relationship among current,
voltage, and resistance, was first observed by
George Ohm in 1827. Ohm's law states that the current in an electric circuit is directly proportional
to the voltage and inversely proportional to the
resistance in a circuit. This may be expressed as:
QUESTIONS
current
1. What are the three basic parts of an
electric circuit?
c. Series-parallel circuit
resistance
or
2. Define the following:
a. Series circuit
b. Parallel circuit
voltage
=
E
1=-
R
where:
I
E
R
=
=
=
current in amperes
voltage in volts
resistance in ohms
SECTION 1 DC CIRCUITS
-
FIGURE 5 9
-i
R,-1 "Of
FIGURE 5-10
-
FIGURE 5-11
E
1~O-V
L-
IT=2A
R'-?f
Solution:
=
ET
RT
0.02
=
ET
24 volts
IT
Whenever two of the three quantities are
known, the third quantity can always be determined.
EXAMPLE: How much current flows in the circuit shown in Figure 5-9?
Given:
=
ET
1200
EXAMPLE: What resistance value is needed for
the circuit shown in Figure 5-11 to draw 2 amperes of current?
Given:
2 amps
ET = 120 volts
~ =
~=?
~=?
ET = 12 volts
~ = 1000 ohms
Solution:
ET
IT =RT
120
2=-
RT
IT
=
~ =
60 ohms
12
1000
0.012 amp or 12 milliamps
5-2
=
R,
QUESTIONS
1. State Ohm's law as a formula.
EXAMPLE: In the circuit shown in Figure 5-10,
how much voltage is required to produce 20 milliamps of current flow?
Given:
~ =
20 rnA
ET
=? .
~
=
=
0.02 amp
1.2 kD. = ohms
2. How much current flows in a circuit
with 12 volts applied and a resistance of
2400 ohms?
3. How much resistance is required to limit
current flow to 20 milliamperes with 24
volts applied?
CHAPTER 5 OHM'S LAW
FIGURE
4:. How much voltage is needed to produce
3 amperes of current flow through a
resistance of 100 ohms?
DIll APPLICATION
OHM'S LAW
5-13
In a parallel circuit, the current divides among the
branches of the circuit and recombines on returning to the
voltage source.
-
OF
In a series circuit (Figure 5-12), the same current
flows throughout the circuit.
IT = IR\ = IR2 = IR, ...
= IRn
The total voltage in a series circuit is equal to the
voltage drop across the individual loads (resistance) in the circuit.
ET
ER\
=
+ ER2 + ER,
. . .
+ ERn
=
R]
+ R2 +R3
...
+Rn
In a parallel circuit (Figure 5-13), the same
voltage is applied to each branch in the circuit.
ET
=
ER\
=
ER2
=
ER, . . .
=
ERn
The total current in a parallel circuit is equal to
the sum of the individual branch currents in the
circuit.
FIGURE
-
5-12
In a series circuit, the current flow is the same
throughout the circuit.
.....
--=-
R .••~
ET-
2-4
IT
J
•..••. R3
T'T
1
1
RT
R]
-=-+-+-
The total resistance in a series circuit is equal to
the sum of the individual resistances in the circuit.
Ry
The reciprocal of the total resistance is equal to
the sum of the reciprocals of the individual
branch resistances.
III
R2
R3
... +-
Rn
The total resistance in a parallel circuit will always
be smaller than the smallest branch resistance.
Ohm's law states that the current in a circuit
(series, parallel, or series-parallel) is directly proportional to the voltage and inversely proportional to the resistance.
E
1=R
In determining unknown quantities in a circuit, follow these steps:
1. Draw a schematic of the circuit and label all
known quantities.
2. Solve for equivalent circuits and redraw the
circuit.
3. Solve for the unknown quantities.
NOTE: OHM'S LAW IS TRUE FOR ANY POINT IN A
CIRCUIT AND CAN BE APPLIED AT ANY TIME. THE SAME
CURRENT FLOWS THROUGHOUT A SERIES CIRCUIT AND
THE SAME VOLTAGE IS PRESENT AT ANY BRANCH OF A
PARALLEL CIRCUIT
SECTION
1 DC CIRCUITS
-
FIGURE 5-14
FIGURE 5-16
R1 = 1.2 k.Q
R1 = 560 Q
-=-E =12V
T
-
FIGURE 5-15
RT = 2240
FIGURE 5-17
Q
Given:
~=?
EXAMPLE: What is the total current
circuit shown in Figure 5-14?
ET
flow in the
Given:
~=?
ET
=
12 volts
R]
=
560 ohms
R2
=
680 ohms
=
I kG
=
~ =
=
=
R,
R,
Solution:
of the circuit:
~
=
+ R2 + R3
560 + 680 + 1000
R,
=
2240 ohms
~ = R]
Draw an equivalent
circuit. See Figure
Now solve for the total current flow:
IT
=
1200 ohms
R2
=
3.9 kD
=
3900 ohms
R3
=
5.6 kD
=
5600 ohms
R] + R2 + R3
1200 + 3900 + 5600
10,700 ohms
Draw the equivalent
circuit. See Figure
Solve for the total current in the circuit:
5-17.
RT
5-15.
RT
~=
=
ET
IT =-
ET
=-
~=?
R] = 1.2 kD
First solve for the total circuit resistance:
1000 ohms
First solve for the total resistance
IT
48 volts
Solution:
~=?
R3
=
12
2240
48
I ---T 10,700
~ =
0.0045 amp or 4.5 milliamps
Remember, in a series circuit, the same current
flows throughout
the circuit. Therefore, IR, = IT'
ER,
IR =,
R2
0.0054 amp or 5.4 milliamp
0.0045
EXAMPLE: How much voltage is dropped across
resistor R2 in the circuit shown in Figure 5-16?
ER,
ER
= --'
3900
=
17.55 volts
CHAPTER
FIGURE
-
5-18
gT=
120V
T L-.
.•.• R -?
2-·
R3= 5.6 kQ
IT= 200 mA
EXAMPLE: What is the value of R2 in the circuit
shown in Figure 5-18?
First solve for the current that flows through
R[ and R3. Because the voltage is the same in
each branch of a parallel circuit, each branch
voltage is equal to the source voltage of 120 volts.
Given:
Resistor
Ohm's law.
R2 can now
ER! = 120 volts
120
I =-R!
1000
Rj
IR! = 0.12 amp
1000 ohms
Given:
using
IR2 = 0.059 amp
ER2 = 120 volts
R2
=? .
Solution:
0.059
120
=-
R2
R2
Solution:
ER3
IR3 =~
be determined
Given:
Solution:
ER
IR! =~
=
5 OHM'S LAW
= 2033.9 ohms
ER3= 120 volts
120
I --R3- 5600
EXAMPLE: What is the current through R3 in
the circuit shown in Figure 5-19?
First determine the equivalent resistance (RA)
for resistors R[ and R2.
R3
IR3= 0.021 amp
Given:
=
5600 ohms
In a parallel circuit, the total current is equal
to the sum of the currents in the branch currents.
RA
R[
=
1000 ohms
Given:
R2
=
2000 ohms
Ir
=
0.200 amp
=? .
Solution:
IR! = 0.120 amp
(adding fractions
requires a common
denominator)
IR2=?
IR3= 0.021 amp
1
1
1
RA
1000
2000
-=--+--
Solution:
IT = IR, + IR2 + 1R3
0.200
=
0.120
+ IR2 + 0.021
0.200
=
0.141
+ IR2
0.200 - 0.141
= IR2
RA = 666.67 ohms
Then determine
the equivalent
resistance
(RB) for resistors R4' R5' and R6• First, find the total series resistance (Rs) for resistors R5 and R6.
SECTION 1 DC CIRCUITS
-
FIGURE 5-19
..••..
-...=-
-
R4
R5
= 4.7
..•.
=
1.5 k.Q
..•...
.. ..
'
Given:
Solution:
R5
Rs = x, + R6
Rs = 1500 + 3300
Rs = 4800 ohms
=? .
Rs = 1500 ohms
R6 = 3300 ohms
=? .
R4 = 4700 ohms
Rs = 4800 ohms
R6
..
~
=
R,
=
~
=
IT
Given:
~=?
RB =
=
=
120 volts
8641.97 ohms
120
8641.97
0.0139 amp or 13.9 milliamps
In a series circuit, the same current flows
throughout the circuit. Therefore, the current
flowing through R3 is equal to the total current in
the circuit.
IR, = 4
IR3= 13.9 milliamps
FIGURE 5-20
5- 3
RB = 2375.3 Q
=
4=
666.67 ohms
5600 ohms
2375.30 ohms
RA = 666.67 Q
ET
=-
RT
IT
Redraw the equivalent circuit substituting RA and
RB' and find the total series resistance of the
equivalent circuit. See Figure 5-20.
=
+ 2375.30
Solution:
(In this case, a common
denominator would be too
4800 large. Use decimals!)
=
RA + R3 + RB
666.67 + 5600
8641.97 ohms
4=?
= 2375.30 ohms
RA
R3
k.Q ~
,
- 1 = -- 1 + -- 1
RB
= 5.6
Given:
s,
4700
R3
Solution:
111
RB
= 2 k.Q
= ...•.•.
3.3 kQ
ET
RT
R4
R2
kQ
Solution:
RB
= 1 k.Q
Now solve for the total current through the
equivalent circuit using Ohm's law.
Given:
RB
R1
R3 = 5.6 kQ
QUESTIONS
1. State the formulas necessary for
determining total current in series and
parallel circuits when the current flow
CHAPTER
FIGURE 5-21
..
R1
= 500
Q.
I
-
T
through the individual components is
known.
2. State the formulas necessary for
determining total voltage in series and
parallel circuits when the individual
voltage drops are known.
3. State the formulas for determining total
resistance in series and parallel circuits
when the individual resistances are known.
4. State the formula to solve for total
current, voltage, or resistance in a series
01' parallel circuit when at least two of
the three values (current, voltage, and
resistance) are known.
5. What is the total circuit current in
Figure 5-2l?
4 =?
Er = 12 volts
R] = 500 ohms
R2 = 1200 ohms
R3 = 2200 ohms
11II KIRCHHOFF'S
CURRENT LAW
In 1847 G. R. Kirchhoff extended Ohm's law
with two important statements that are referred
to as Kirchhoff's laws. The first law-known as
Kirchhoff's current laW-states:
-,
FIGURE 5-22
r
-....:-
5 OHM'S LAW
-=--
~
•
L-
B
r :tl :
R1
R2
..2J
~2t
J
11
-
A
• The algebraic sum of all the currents entering
and leaving a junction is equal to zero.
Another way of stating Kirchhoff's current
law is:
• The total current flowing into a junction is
equal to the sum of the current flowing out
of that junction.
A junction is defined as any point of a circuit
at which two or more current paths meet. In a
parallel circuit, the junction is where the parallel
branches of the circuit connect.
In Figure 5-22, point A is one junction and
point B is the second junction. Following the current in the circuit, 4 flows from the voltage source
into the junction at point A. There the current
splits among the three branches as shown. Each of
the three branch currents (II' 12' and 13) flows out
of junction A. According to Kirchhoff's current
law, which states that the total current into a junction is equal to the total current out of the junction, the
current can be stated as:
4 = I] + 12 + 13
Following the current through each of the
three branches finds them coming back together
at point B. Currents 1],12' and 13 flow into junction
B, and 4 flows out. Kirchhoff's current law formula at this junction is the same as at junction A:
I]
+ 12 + I) = 4
SECTION 1 DC CIRCUITS
FIGURE 5-23
14
5-4
QUESTIONS
= 3mA
-
FIGURE 5-24
1. State Kirchhoff's current law.
2. A total of 3 A flows into a junction of
three parallel branches. What is the sum
of the three branch currents?
3. If 1 mA and 5 mA of current flow into a
junction, what is the amount of current
flowing out of the junction?
4. In a parallel circuit with two branches,
one branch has 2 IDA of current flowing
through it. The total current is 5 mA. What
is the current through the other branch?
5. Refer to Figure 5-23. What are the values
of 12 and I3?
In Figure 5-24 there are three voltage drops and
one voltage source (voltage rise) in the circuit. If
the voltages are summed around the circuit as
shown, they equal zero.
ET - Ej
_
KIRCHHOFF'S
VOLTAGE LAW
Kirchhoff's second law is referred to as Kirchhoff's
voltage law, and it states:
• The algebraic sum of all the voltages around
a closed circuit equals zero.
Another way of stating Kirchhoff's voltage
law is:
• The sum of all the voltage drops in a closed
circuit will equal the voltage source.
.
-----------------------
-
E2 - E3
=
a
Notice that the voltage source (ET) has a sign
opposite that of the voltage drops. Therefore the
algebraic sum equals zero.
Looking at this another way, the sum of all
the voltage drops will equal the voltage source.
ET
=
Ej
+ E2 + E3
Both of the formulas shown are stating the
same thing and are equivalent ways of expressing
Kirchhoff's voltage law.
The key to remember is that the voltage
source's polarity in the circuit is opposite to that of
the voltage drops.
CHAPTER 5 OHM'S LAW
FIGURE 5-25
1V
2V
R1
R2
R3
ET =?
5-5
Rs
R4
4V
3V
2V
QUESTIONS
1. State Kirchhoff's voltage law two
different ways.
2. A series resistive circuit is connected to a
12-volt voltage source. What is the total
voltage drop in the circuit?
3. A series circuit has two identical resistors
connected in series with a 9-volt battery.
What is the voltage drop across each
resistor?
4. A series circuit is connected to a 12-volt
voltage source with three resistors. One
resistor drops 3 V and another resistor
drops 5 V.What is the voltage drop across
the third resistor?
5. Refer to Figure 5-25. What is the total
voltage applied to the circuit?
SUMMARY
• An electric circuit consists of a voltage
source, a load, and a conductor.
• The current path in an electric circuit can
be series, parallel, or series-parallel.
• A series circuit offers only one path for
current to flow.
• A parallel circuit offers several paths for
the flow of current.
• A series-parallel circuit provides a
combination of series and parallel paths
for the flow of current.
• Current flows from the negative side of the
voltage source through the load to the
positive side of the voltage source.
• Current flow in an electric circuit can be
varied by changing either the voltage or
the resistance.
• The relationship of current, voltage, and
resistance is given by Ohm's law.
• Ohm's law states that the current in an
electric circuit is directly proportional to
the voltage applied and inversely
proportional to the resistance in the
circuit.
E
1=R
• Ohm's law applies to all series, parallel,
and series-parallel circuits.
• To determine unknown quantities in a
circuit:
Draw a schematic of the circuit and
label all quantities.
Solve for equivalent circuits and
redraw the circuit.
Solve for all unknown quantities.
• Kirchhoff's current law: The algebraic
sum of all the currents entering and
leaving a junction is equal to zero; it
may be restated as the total current
flowing into a junction is equal to the
sum of the current flowing out of that
junction.
• Kirchhoff's voltage law: The algebraic sum
of all the voltages around a closed circuit
equals zero; it may be restated as the sum
of all the voltage drops in a closed circuit
will equal the voltage source.
SECTION 1 DC CIRCUITS
Using Ohm's law, find the unknown value for the following:
1. I = ?
E = 9V
R = 4500 ohms
2. I = 250 IDA
E = ?
R = 470 ohms
3. I = 10 A
E = 240 V
R= ?
4. Find the current and voltage drop through each component for the circuits shown below.
5. Use Kirchhoff's laws to verify answers for question 4.
(A)
(B)
R, =50Q
.•..•..•.
R, = 150 Q
R2
(C)
75Q
75Q
75Q
= 300 Q