NIDA SERIES 130E
Block 3
BASIC DC CIRCUITS
BASIC ELECTRICITY
UNIT I - DC CIRCUITS
LESSON 2
SERIES CIRCUITS
OBJECTIVES
OVERVIEW
Upon completion of this lesson, the
student should be able to:
In this lesson, series circuits are
discussed in terms of current, voltage,
resistance, and power.
1. Recognize a series circuit and define
the relationship of current, resistance,
and power in a series circuit.
2. Determine the current, total resistance
RT, and total voltage ET in series
circuits.
3. Determine voltage drops across
resistors in a series circuit.
4. Determine power dissipated in the
resistors in a series circuit.
5. Solve for unknown circuit values in a
series circuit.
6. Measure current and voltage drops in
series circuits.
PREREQUISITES
None
EQUIPMENT REQUIRED
Nida Model 130E Test Console
Nida Series 130 Experiment Card
PC130-6A
Nida Model 480/488 Multimeter, or
equivalent
Copyright © 2002 by Nida Corporation
Students learn to determine total
resistance RT and total voltage ET in
series circuits and to calculate voltage
drops across resistors, using Ohm's Law.
The relationship between voltage drops
and applied voltage is then established.
Students solve for unknown circuit
values and learn to calculate power
dissipation in resistors.
In the experiment, students measure
current and voltage drops in series
circuits and then solve problems to
demonstrate the relationship of current,
voltage, resistance, and power in series
circuits.
INTRODUCTION
So far in your study of electronics, you
have learned about a basic, simple
electronic circuit.
3-2-1
LESSON 2
SERIES CIRCUITS
UNIT I
Block 3
Basic DC Circuits
To refresh your memory, take a look at the
circuit in Figure 1.
The circuit consists of a power source which
provides the voltage to force the current to
flow through the circuit across one resistor (the
load).
The load in a complex electronic circuit, as you
will discover, consists of many components,
not just one resistor. You will also learn that
these components can be connected in a
variety of ways.
Figure 1. Simple Circuit
In this lesson you will study the most basic of the complex electronic circuits: the series
circuit.
WHAT IS A SERIES CIRCUIT?
You learned in the last lesson that in an electronic circuit, voltage forces current to flow
from the negative terminal of the power source, through a load, and back to the positive
terminal of the power source.
That statement covers all electronic circuits, so the logical question is: How can you
recognize whether or not an electronic circuit is a series circuit? Just what is a series
circuit? Let's start with the definition.
DEFINITION
SERIES CIRCUIT: An electronic circuit in which the power source
provides voltage to force current to flow in a single path through all
the components in the circuit.
In other words, series circuit current flows in just one path through a load consisting of
two or more components.
Figure 2 shows a basic series circuit in which
the voltage forces the current to flow through a
load consisting of two resistors, R1 and R2.
Notice how R1 and R2 are connected one right
after the other in series, just like two beads on
a string. The current, like the string on which
the beads are strung, goes through R1 and R2,
in a single path.
Figure 2. A Series Circuit
3-2-2
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 2
SERIES CIRCUITS
The current path is easy to see in the rectangular series circuit diagram in Figure 2. Series
circuits, however, do not always look like nice, neat, easy-to-read rectangles, as you can
see by looking at the series circuit diagrams in Figures 3 and 4.
With a pencil or with your finger, trace the path
of the current in the circuits of Figure 3.
Follow each circuit all the way from the
negative terminal of the power source, through
each resistor, and back to the positive terminal
without lifting your pencil or finger off the
paper.
Figure 3. Series Circuits
Now look at Figure 4. Here, too,
is a series circuit. This circuit,
however, needs to have the line
drawn to represent the current
flowing through the components.
Take your pencil and, without
breaking the line or lifting your
pencil off the paper, draw the
current path to connect R1, R2,
R3, and R4 in series, in
numerical order.
Figure 4. Draw the Current Path
You can see from Figure 3 and Figure 4 that the current in a series circuit goes through all
the resistors in the circuit. Since the current must go in one unbroken path through all the
series resistances, the total of all the series resistors is the total resistance across the
voltage source. This fact can be stated in the following rule.
RULE
TOTAL RESISTANCE RT OF THE RESISTORS IN A SERIES CIRCUIT
EQUALS THE SUM OF THE INDIVIDUAL RESISTORS.
Look at the circuit diagrams in Figure 5A and Figure 5B. These circuits, which are
equivalent circuits, illustrate the total resistance rule.
5A. Series Circuit, 3 Resistors
5B. Equivalent Circuit to 5A
Figure 5. Equivalent Circuits Show Total Resistance Rule
Copyright © 2002 by Nida Corporation
3-2-3
LESSON 2
SERIES CIRCUITS
UNIT I
Block 3
Basic DC Circuits
The series circuit in Figure 5A contains 3 resistors: R1, which equals 12 ohms; R2, which
equals 6 ohms; and R3, which equals 6 ohms. The equivalent circuit in Figure 5B
contains 1 resistor: R1, which equals 24 ohms.
Since total resistance RT of the resistors in a series circuit equals the sum of the individual
resistors, we can find the value of RT for the circuit in Figure 5A as follows:
RT = R1 + R2 + R3 = 12 Ω + 6 Ω + 6 Ω = 24 Ω
RT for the circuit in Figure 5A equals 24 Ω. R1 in the circuit of Figure 5B also equals
24 Ω. Thus the circuit of Figure 5B is equivalent to the circuit of Figure 5A.
Now that you can recognize a series circuit and can see the relationship of current,
voltage, and total resistance in a series circuit, you should be able to solve for unknown
circuit values in a series circuit.
All you do is use Ohm's Law, just like you did in the last lesson. But first, let's have a
brief review of how to solve for unknown circuit values in a simple electronic circuit using
Ohm's law.
Example:
How to Solve for Current in a Simple Electronic Circuit when E = 8 volts
and R = 8 ohms.
Refer back to the simple circuit diagram in Figure 1. Remember the formula you learned in
the last lesson to solve for current?
I=E÷R
Right. The formula is I = E/R, where E is the power source, R is the load, and I is the
current through the load.
To find the current, simply insert the known circuit values into the formula I = E/R.
I=E÷R=8V÷8Ω=1A
The current is 1 amp.
Now look back at the circuit diagram in Figure 2. Note that resistors R1 and R2 are in
series. How would you solve for current in this circuit?
Example:
How to Solve for Current in a Series Circuit when E = 8 volts,
R1 = 4 ohms, and R2 = 4 ohms.
The first step is to find total resistance RT.
RT = R1 + R2 = 4 Ω + 4 Ω = 8 Ω
Now insert the values for E and RT into the formula I = E/R.
I = E ÷ RT = 8 V ÷ 8 Ω = 1 A
3-2-4
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 2
SERIES CIRCUITS
You can also solve for current in this circuit if you substitute the values for R1 and R2 for
RT directly into the formula I = E/R.
I = E ÷ (R1 + R2) = 8 V ÷ (4 Ω + 4 Ω) = 8 V ÷ 8 Ω = 1 A
Whichever way you work the problem, the current equals 1 amp.
Here's another example showing how to solve for current in a series circuit. Refer to the
circuit diagram in Figure 5A for this problem.
Example:
How to Solve for Current in the Series Circuit of Figure 5A.
First, find total resistance RT.
RT = R1 + R2 + R3 = 12 Ω + 6 Ω + 6 Ω = 24 Ω
Then insert the values for E and RT into the formula for current.
I = E ÷ RT = 12 V ÷ 24 Ω = 0.5 A
The current in the series circuit of Figure 5A equals 0.5 amp.
As you can see, the current in a series circuit depends upon the total resistance in the
circuit. The current, therefore, is the same at any point in a series circuit. Whether the
circuit contains one resistor or any number of resistors does not matter. What does
matter is that series circuit current follows only one path through all components in the
circuit to get from the negative terminal to the positive terminal of the power supply.
Copyright © 2002 by Nida Corporation
3-2-5
LESSON 2
SERIES CIRCUITS
Exercise 1:
UNIT I
Block 3
Basic DC Circuits
Solve for the Unknown Quantities in the Circuits in Figure 6.
Problem 1.
RT =
I=
6A: Series Circuit
Problem 2.
R3 =
6B: Equivalent Circuit
I=
6C: Series Circuit
Problem 3.
RT =
6E: Series Circuit
6D: Equivalent Circuit
E=
6F: Equivalent Circuit
Figure 6. Series Circuits and Equivalent Circuits for Exercise 1
3-2-6
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 2
SERIES CIRCUITS
VOLTAGE DROPS
When current flows through a resistance in a circuit, the voltage across that resistor
equals the current times the resistance, or E = IR, according to Ohm's Law. Since
current in a series circuit flows from the negative terminal of the power source, through all
resistors in its path, and then into the positive terminal of the power source, we know
that E = IRT. The formula can also be written as:
E = I(R1 + R2 + ...) or E = IR1 + IR2 + ....
The voltage across each resistor (ER) is called the IR voltage or voltage drop. Thus,
ER1=IR1, ER2=IR2, and so on.
Series circuit voltage drops are defined as follows:
DEFINITION
VOLTAGE DROP: The voltage across each passive component, such
as a resistor, in a series circuit.
An IR voltage drop must have polarity, meaning one end is more positive or negative than
the other end. Without polarity, no current can flow through the resistor to produce the
voltage drop. Because current flows from the negative terminal to the positive terminal,
negative polarity develops at the point where current enters the resistor and positive
polarity develops at the point where current leaves the resistor.
Since E = IRT, the sum of the voltage drops of the resistors should equal the total applied
voltage, ET. To demonstrate this, take a look at the example below and Figure 7.
Example:
How to Determine Voltage Drops in Series Circuits, Using the Circuit of
Figure 7.
First you need to determine RT so that you
can calculate the current in the circuit.
RT = R1 + R2 + R3 + R4
RT = 1 kΩ
Ω + 3 kΩ
Ω + 3 kΩ
Ω + 5 kΩ
Ω
Ω
RT = 12 kΩ
Now you can solve for the current, using
the formula I = ET ÷ RT.
I = ET ÷ RT = 24 V ÷ 12 kΩ
Ω = 2 mA
Figure 7. Series Circuit with Four Resistors
To find the voltage drops, insert the known values for the current and the four resistors
into the formula E = IR.
ER1
ER2
ER3
ER4
=
=
=
=
IR1
IR2
IR3
IR4
=
=
=
=
2
2
2
2
mA
mA
mA
mA
x
x
x
x
1
3
3
5
kΩ
Ω
kΩ
Ω
kΩ
Ω
kΩ
Ω
=
=
=
=
0.002
0.002
0.002
0.002
Copyright © 2002 by Nida Corporation
A
A
A
A
x
x
x
x
1000
3000
3000
5000
Ω
Ω
Ω
Ω
=
=
=
=
2V
6V
6V
10 V
3-2-7
LESSON 2
SERIES CIRCUITS
UNIT I
Block 3
Basic DC Circuits
Now add the voltage drop values to show that the ET of 24 volts equals the sum of the
voltage drops.
ET = ER1 + ER2 + ER3 + ER4 = 2 V + 6 V + 6 V + 10 V = 24 V
This fact can be stated in the following rule.
RULE
THE TOTAL APPLIED VOLTAGE ET OF A SERIES CIRCUIT EQUALS
THE SUM OF THE VOLTAGE DROPS ACROSS THE INDIVIDUAL
RESISTORS IN THE CIRCUIT.
Now that you can determine both current and voltage drops in series circuits, you can
calculate the power dissipated by each resistor (PR) and the total power (PT) dissipated in
the circuit. The basic formula to use, as you remember, is P = EI.
Example:
How to Determine Power Dissipated by Resistors and Total Power
Dissipated in Series Circuits, Using the Circuit of Figure 7.
First, calculate the total power dissipated in the circuit by inserting the known values for
current and total voltage into the formula.
PT = ETI = 24 V x 0.002 A = 0.048 W = 48 mW
Next, calculate the power dissipated by each resistor in the circuit.
PR1
PR2
PR3
PR4
=
=
=
=
ER1I
ER2I
ER3I
ER4I
=
=
=
=
2 V x 0.002 A = 0.004 W = 4 mW
6 V x 0.002 A = 0.012 W = 12 mW
6 V x 0.002 A = 0.012 W = 12 mW
10 V x 0.002 A = 0.020 W = 20 mW
Now add the values of power dissipated in each resistor to show that PT of 48 mW is the
same as the total power dissipated by all four resistors.
PT = PR1 + PR2 + PR3 + PR4 = 4 mW + 12 mW + 12 mW + 20 mW = 48 mW
This fact can be stated in the following rule:
RULE
THE TOTAL POWER DISSIPATED PT IN A SERIES CIRCUIT EQUALS
THE SUM OF THE POWER DISSIPATED BY THE INDIVIDUAL
RESISTORS IN THE CIRCUIT.
As you can see, the sum of the voltage drops in a series circuit equals the total applied
voltage to the circuit. Also, the sum of the power dissipated by each resistor in a series
circuit equals the total power dissipated by the circuit.
3-2-8
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
Exercise 2:
UNIT I
LESSON 2
SERIES CIRCUITS
Solve for the Unknown Circuit Values in Table 1.
Based on the series circuit in Figure 8, use
the formulas given to solve for all the
unknowns in the problems in Table 1 below.
ET = ER1 + ER2 + ER3 + ER4 + ER5
RT = R1 + R2 + R3 + R4 + R5
I = ET ÷ RT
PT = ETI
PT = PR1 + PR2 + PR3 + PR4 + PR5
Figure 8. Series Circuit with Five Resistors
Table 1. Solving for Unknown Circuit Values
NO.
1.
2.
3.
TOTAL VALUES
VALUES
OF RESISTORS
VOLTAGE DROPS
ACROSS RESISTORS
POWER DISSIPATED
BY EACH RESISTOR
ET = 30 V
R1 = 100 Ω
ER1 =
PR1 =
RT =
R2 = 400 Ω
ER2 =
PR2 =
I =
R3 = 250 Ω
ER3 =
PR3 =
PT =
R4 = 300 Ω
ER4 =
PR4 =
R5 = 450 Ω
ER5 =
PR5 =
ET =
R1 =
ER1 = 1.2 V
PR1 =
RT =
R2 =
ER2 = 1.8 V
PR2 =
I = 1.2 mA
R3 =
ER3 = 3.0 V
PR3 =
PT =
R4 =
ER4 = 3.6 V
PR4 =
R5 =
ER5 = 2.4 V
PR5 =
ET = 40 V
R1 = 12 Ω
ER1 =
PR1 =
RT =
R2 = 20 Ω
ER2 =
PR2 =
I =
R3 = 8 Ω
ER3 =
PR3 =
PT = 20 W
R4 = 25 Ω
ER4 =
PR4 =
R5 =
ER5 =
PR5 =
Copyright © 2002 by Nida Corporation
3-2-9
LESSON 2
SERIES CIRCUITS
UNIT I
Block 3
Basic DC Circuits
Now compare the values in Table 4 with the values in Table 3. Are the values the same?
Any small differences you might have could be from your multimeter readings.
SUMMARY
With this lesson on series circuits, you have started to explore the behavior of current and
voltage in circuits that contain more than one component. So far, you have learned the
following:
 Components in a series circuit are connected like beads on a string.
 A series circuit is one in which the current leaves the negative terminal of the power
source, flows in a single path through all components in the circuit, and enters the
positive terminal of the power source.
 Current is the same at any point in a series circuit.
 To determine total resistance RT, add the resistance of each resistor in the circuit.
 To solve for current in a series circuit, divide voltage by the total resistance RT.
 Voltage drops across resistors in series circuits, caused by current flowing through
the resistors, are easily measured with a voltmeter.
 The sum of the voltage drops across all resistors in series is equal to the voltage
applied across the resistors.
 To calculate the voltage drop across an individual resistor, multiply the resistor value
by the current flowing through the resistor.
3-2-18
Copyright © 2002 by Nida Corporation