NIDA SERIES 130E
Block 3
BASIC DC CIRCUITS
BASIC ELECTRICITY
UNIT I - DC CIRCUITS
LESSON 4
PARALLEL CIRCUITS
OBJECTIVES
OVERVIEW
On completion of this lesson, the student
will be able to:
This lesson introduces students to the
relationship of current, voltage
resistance, and power in parallel circuits.
1. Identify a parallel circuit and
determine that applied voltage Ea is
the same across each parallel branch
in a parallel circuit.
2. Calculate current in a parallel circuit
using Ohm's Law and prove that total
current IT equals the sum of the
branch currents.
3. Calculate the total resistance RT of a
parallel circuit.
4. Analyze a parallel circuit and solve for
unknown circuit values.
5. Calculate power dissipated in any or
all of the resistors in a parallel circuit.
6. Measure voltage, current, and
resistance in parallel circuits.
PREREQUISITES
None
EQUIPMENT REQUIRED
Students first study the behavior of
current in parallel circuits. They then
learn to calculate branch currents and
total current in parallel circuits, using
Ohm's Law.
Next, students study the relationship
which determines total resistance in
parallel circuits. They also learn to
perform calculations to prove this
relationship.
Using Ohm's Law, students then learn to
solve for unknown circuit values in
parallel circuits. Using the power
formula, they also calculate power
dissipation in resistors.
The experiment provides students with
hands-on practice in measuring voltage,
current, and resistance in parallel circuits.
From their analysis of experiment results,
they see the proof of the relationship of
current, voltage, resistance, and power in
parallel circuits.
Nida Model 130E Test Console
Nida Series 130 Experiment Card
PC130-8A
Nida Model 480/488 Multimeter, or
equivalent
Copyright © 2002 by Nida Corporation
3-4-1
LESSON 4
PARALLEL CIRCUITS
UNIT I
Block 3
Basic DC Circuits
INTRODUCTION
In the last two lessons, you learned how series circuits operate. Thus, you know that a
series circuit consists of a power supply and several resistors which are connected in a
string-like circuit. The same current flows through all components in the string. If one
resistor opens or an open occurs in the circuit, the current flow through all the resistors
stops.
The old-fashioned strings of Christmas tree lights were typical series circuits. When one
bulb burned out, all the lights in the string went out. Trying to find the burned-out light
bulb was often quite a job.
Today's strings of Christmas tree lights are typical parallel circuits. When one bulb burns
out, all the rest of the bulbs stay lit. Most of the electrical wiring in your house also
consists of parallel circuits. What happens when you turn on all the lights in your house
and one bulb burns out? Do any of the other lights go out? Of course not.
This lesson teaches you about parallel circuits. In parallel circuits, the loads connect
across each other, like rungs on a ladder, with the voltage source across each load. If one
load (resistor) opens, the current stops flowing through that resistor only. Current
through the other resistors does not change.
To calculate current, voltage, and resistance in parallel circuits, you will again use the
Ohm's Law formulas. You already know that calculating current in a series circuit is
easier if you calculate the total resistance in the circuit first. You'll learn how to calculate
the total resistance in parallel circuits, too. The calculations, however, are not as simple
as those for series circuits. You'll also learn to use the power formulas to solve for power
dissipated by the resistors.
The experiment provides plenty of hands-on experience in measuring parallel circuit values.
By measuring current, you'll prove that the total current developed by the voltage source
is the sum of the branch currents in the circuit. You'll also measure resistance and prove
that you can calculate total resistance in a parallel circuit by using Ohm's Law. All you do
is divide the voltage across the parallel resistances by the total current of all branches.
Sounds complicated, doesn't it? Don't worry. You'll find that parallel circuits are not all
that difficult when you take things step by step. Let's get started.
WHAT IS A PARALLEL CIRCUIT?
How can you tell when a circuit is a parallel circuit? Let's start with the definition.
DEFINITION
PARALLEL CIRCUIT: An electronic circuit that has two or
more paths (or branches) for current flow.
3-4-2
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 4
PARALLEL CIRCUITS
What's the difference between a parallel circuit and a series circuit? That is an easy
question to answer. The difference is in the current path. You already know that current
in a series circuit flows in a single path through all components in the circuit. In a parallel
circuit, current flows in more than one complete path through the circuit.
Look at the diagram of a typical series
circuit in Figure 1. The circuit consists of
power source E and two resistors, R1 and
R2, hooked up in series.
Current I leaves the negative terminal of
the power source. It flows around the
circuit through points 4 and 3, R2,
point 2, R1, and point 1. It then flows
back into the positive terminal of the
power source. As you can see, the
current has only one possible path it
can flow through.
Figure 1. Series Circuit
Now look at the schematic diagram of
a typical parallel circuit in Figure 2.
Here, too, the circuit consists of
power source E and two resistors, R1
and R2. This time, however, they are
hooked up in parallel.
Figure 2. Parallel Circuit
In other words, R1 is connected
across E and R2 is also connected
across E.
Leaving the negative terminal of E, total current IT flows through point 6 to point 5. At
point 5, IT finds that it can take two paths: it can flow to R1 and it can flow to point 4.
Thus, the current splits up. Part of the current, IR1, flows through R1 to point 2. The
other part, IR2, flows through point 4, R2, point 3, and to point 2. IR1 and IR2 come
together at point 2 and again become IT. IT then flows through point 1 and into the
positive terminal of E.
You can connect any number of resistors in a parallel circuit just like you can in a series
circuit. There is no limit to the number, as long as the current flows through every one of
the resistors. As you add resistors in parallel to the resistors and voltage source in an
existing circuit, the total current increases to provide for them.
Copyright © 2002 by Nida Corporation
3-4-3
LESSON 4
PARALLEL CIRCUITS
UNIT I
Block 3
Basic DC Circuits
TOTAL CURRENT IN A PARALLEL CIRCUIT
In the last lesson you learned to solve for current in a series circuit. That was easy, for
the current flowed through one path only. How do you solve for current in a parallel
circuit, when you have more than one current path? It's not really that hard, if you just
take it step by step.
Look at the parallel circuit in the schematic diagram of Figure 3. Notice the three
branches or paths which the current can flow through after it comes from the power
supply.
Figure 3. Current Flow in a Parallel Circuit
Since this circuit has three branches, IT actually splits up into three parts: IR1, IR2, and IR3.
IR1 flows through R1, IR2 flows through R2, and IR3 flows through R3. IR1, IR2, and IR3 then
come back together again and, as IT, flow into the positive terminal of E.
You can easily see from this description that IT = IR1 + IR2 + IR3. In other words, the
total current is equal to the sum of all branch currents in a parallel circuit. Here's an
example to illustrate this statement.
Example:
Solving for IT in a Parallel Circuit.
Let's solve for the total current being drawn from the power supply in the parallel circuit
of Figure 3. The first step is to solve for the current in each of the three branches of the
parallel circuit. We do this by using Ohm's Law.
We have these circuit values:
E = 12 V
R1 = 24 Ω
R2 = 12 Ω
R3 = 6 Ω
For Branch 1:
E = IR1 x R1
Therefore:
IR1 = E ÷ R1
For Branch 2:
E = IR2 x R2
Therefore:
IR2 = E ÷ R2
3-4-4
= 12 V ÷ 24 Ω = 0.5 A
= 12 V ÷ 12 Ω = 1.0 A
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 4
PARALLEL CIRCUITS
For Branch 3:
E = IR3 x R3
Therefore:
IR3 = E ÷ R3
We know that:
IT = IR1 + IR2 + IR3
Therefore:
IT = 0.5 A + 1 A + 2 A = 3.5 A
= 12 V ÷ 6 Ω
= 2.0 A
As you can see, total current for this parallel circuit is 3.5 amperes.
TOTAL RESISTANCE IN A PARALLEL CIRCUIT
You discovered that resistance in series circuits was easier to calculate if you combined all
the resistors in the circuit into one resistance, RT. You can do the same thing with parallel
circuits. Simply combine all the resistors into total resistance, RT. It's easy.
Look again at the schematic diagram in
Figure 3, which shows a parallel circuit with
three branches.
We want to combine the individual resistors
in the three branches so we have total
resistance in one branch.
Now look at the circuit in Figure 4. It looks
familiar, doesn't it? You have worked with
circuits like this before.
Figure 4.
Parallel Circuit, with Resistors
Combined
Let's use these circuits in Figures 3 and 4 to show you how to calculate resistance in
parallel circuits.
Example:
How to Solve for RT in a Parallel Circuit with Three Branches.
First, we need to find a formula to solve for RT in a parallel circuit. Using the circuit in
Figure 3, we'll show you how to get the formula.
You know that:
IT = IR1 + IR2 + IR3
From Figure 4, you know that:
IT = E ÷ RT
Therefore:
IR1 =
When you substitute, you get:
E
E
E
E
=
+
+
RT
R1 R2 R3
Condense the equation, and you get:
E
1
1
 1
=E
+
+

 R1 R2 R3 
RT
Copyright © 2002 by Nida Corporation
E
R1
IR2 =
E
R2
IR3 =
E
R3
3-4-5
LESSON 4
PARALLEL CIRCUITS
UNIT I
Block 3
Basic DC Circuits
Now divide both sides by E:
1
1
E
E  1
=
+
+


ERT E  R1 R2 R3 
This gives us Formula 1:
1
1
1
1
=
+
+
RT
R1 R2 R3
Thus, to find total resistance of a parallel circuit, take the reciprocal of RT (divide 1 by RT).
Then make the reciprocal of RT equal to the sum of the reciprocals of all the resistors in
parallel. The circuit we used in the example above has three resistors. You can, however,
have any number of resistors.
Now let's solve for RT in the circuit of Figure 3. We'll use the same circuit values we used
to calculate IT.
Example:
Solving for RT in the Circuit of Figure 3, Using Formula 1.
We know that:
E = 12 V
R1 = 24 Ω
R2 = 12 Ω
R3 = 6 Ω
Thus:
1
1
1
1
=
+
+
RT 24 Ω 12 Ω 6 Ω
Thus:
1
7
=
RT 24 Ω
Therefore:
RT =
=
1
2
4
+
+
24 Ω 24 Ω 24 Ω
=
7
24 Ω
and cross-multiplying: 7 RT = 24 Ω
24 Ω
= 3.428 Ω
7
Now let's check this answer with the current value we calculated in the first example.
This calculation showed that IT = 3.5 A. We know that E=12 V, so we use Ohm's Law
to calculate the total resistance.
Using Ohm's Law, we get:
RT =
E
IT
=
12 V
3.5 A
= 3.428 Ω
These calculations do prove that our formula for determining total resistance RT is valid.
Let's summarize the formulas we've used so far.
 Total Resistance--Formula 1:
3-4-6
1
1
1
1
=
+
+
+ etc.
RT
R1 R2 R3
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 4
PARALLEL CIRCUITS
 Total Current:
IT = IR1 + IR2 + IR3 + etc.
 Ohm's Law:
E = IT (RT )
or
RT =
E
IT
or
IT =
E
RT
You'll often have to determine the total
resistance in parallel circuits with two
branches, like the one shown in
Figure 5.
How would you calculate the total
resistance in this circuit? You learn
another formula. We do have one that
makes the calculations easier.
Here's an example that will show you
how to use this formula.
Figure 5. Parallel Circuit with Two Branches
Example:
Solving for RT in a Parallel Circuit with Two Branches.
We know that:
1
1
1
=
+
RT R1 R2
This gives us Formula 2:
RT =
If circuit values are:
R1=24 Ω,
Then:
RT =
R1 × R2
R1 + R2
R2=12 Ω, and
24 Ω × 12 Ω
24 Ω + 12 Ω
=
288 Ω
36
E=12 V
= 8Ω
Now let's check the answer by calculating the current values.
For Branch 1:
IR1 =
E
12 V
=
= 0.5 A
R1 24 Ω
For Branch 2:
IR2 =
E
12 V
=
= 1.0 A
R2 12 Ω
For circuit:
IT = IR1 + IR2 = 0.5 A + 1.0 A = 1.5 A
Therefore:
RT =
E
12 V
=
= 8Ω
IT 1.5 A
Our answers check--total resistance equals 8 ohms.
Copyright © 2002 by Nida Corporation
3-4-7
LESSON 4
PARALLEL CIRCUITS
UNIT I
Block 3
Basic DC Circuits
Now let's add a third branch with a resistor of 6 ohms to the parallel circuit of Figure 5.
Take a look at the diagram of this three-branch parallel circuit in Figure 6.
As you can see, this circuit
is the same as our original
parallel circuit in Figure 3.
Can you use Formula 2 to
calculate the total resistance
in this circuit?
Figure 6.
Of course you can. Here's how
you do it.
Example:
Parallel Circuit with Three
Branches
Solving for RT in a Three-branch Parallel Circuit, Using Formula 2.
Look again at the parallel circuit in Figure 6. In the last example, we used Formula 2 to
calculate total resistance in the R1 and R2 branches of this circuit. RT of these branches
was 8 ohms.
Knowing this value, we can redraw the three-branch parallel circuit, making it a
two-branch parallel circuit. Figure 7 illustrates this two-branch circuit.
One branch of the circuit has a total resistance
of 8 ohms. This is the total resistance of
resistors R1 and R2.
The other branch of the circuit has a resistance
of 6 ohms. This is the value of resistor R3.
Figure 7. Parallel Circuit with
Two Branches
Now we can use Formula 2 to find the total
resistance just like we did in the circuit of
Figure 5. This gives us:
RT =
8 Ω × 6 Ω 48 Ω2
=
= 3.428 Ω
8 Ω + 6 Ω 14 Ω
Notice that we got the same results with Formula 2 as we did when we used Formula 1.
You can solve for total resistance, therefore, in two different ways when you have a
parallel circuit with several branches or resistors.
 With Formula 1, you use all the resistors in parallel at once.
1
1
1
1
=
+
+
+ etc.
RT
R1 R2 R3
 With Formula 2, you use only two resistors in parallel at a time.
RTA =
3-4-8
R1 × R2
;
R1 + R2
then RTB =
RTA × R3
; etc.
RTA + R3
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 4
PARALLEL CIRCUITS
As we worked through the examples, did you see anything unusual about the total
resistance in parallel circuits? What about the value of the total resistance compared with
the value of the resistors in the circuit?
Take a look again at the examples. Notice that in each one, the value of RT is lower than
the value of any of the resistors in the circuit. This is always true in parallel circuits. The
reverse, however, is always true in series circuits.
Remember:
In series circuits, total resistance is always higher than the value of any of
the resistors in the string of the circuit.
In parallel circuits, total resistance is always lower than the value of any of
the resistors in the branches of the circuit.
Practice Exercise:
Solve for the Unknown Circuit Values.
Let's practice solving for unknown circuit values in parallel circuits, given the circuit in
Figure 8. This exercise consists of two problems which we have worked through for you.
We've even provided the answers.
Follow the problems through, step by step. You'll get to use the new formulas from this
lesson plus the Ohm's Law and power formulas from the previous lessons.
E
= 20 V
R1 = 40 Ω
R2 = 40 Ω
R3 = 20 Ω
Figure 8. Parallel Circuit for Practice Exercise
1. Imagine that resistors R1, R2, and R3 represent light bulbs in the circuit. How
much power does each bulb dissipate? How much total power must the power
source be able to supply? You can solve this problem two different ways.
 Use Ohm's Law to find current through each bulb, use the power formula to
find power in each bulb, and then solve for total power required.
 Substitute known circuit values for the unknown values in the power formula,
use the power formula to find power in each bulb, and then solve for total
power required.
Copyright © 2002 by Nida Corporation
3-4-9
LESSON 4
PARALLEL CIRCUITS
UNIT I
Block 3
Basic DC Circuits
a. First, use Ohm's Law to find the current through each bulb.
IR1 =
E
R1
=
20 V
40 Ω
= 0.5 A
IR2 =
E
R2
=
20 V
40 Ω
= 0.5 A
IR3 =
E
R3
=
20 V
20 Ω
= 1.0 A
Next use power formula (P = EI) to find power in each bulb.
PR1 = E(IR1) = 20 V x 0.5 A = 10 W
PR2 = E(IR2) = 20 V x 0.5 A = 10 W
PR3 = E(IR3) = 20 V x 1.0 A = 20 W
Then solve for total power in the circuit.
PT = PR1 + PR2 + PR3 = 10 W + 10 W + 20 W = 40 W
b. Substitute known voltage and resistance values for the unknown current
value in the power formula.
We know that:
P = EI and I =
E
R
so: P = E ×
E
R
=
E×E
R
Next use the power formula to find the power in each bulb.
PR1 =
E×E
R1
=
20 V × 20 V
40 Ω
= 10 W
PR2 =
E×E
R2
=
20 V × 20 V
40 Ω
= 10 W
PR3 =
E×E
R3
=
20 V × 20 V
20 Ω
= 20 W
Then solve for total power in the circuit.
PT = PR1 + PR2 + PR3 = 10 W + 10 W + 20 W = 40 W
As you can see, PT = 40 W, no matter which way you work the problem.
3-4-10
Copyright © 2002 by Nida Corporation
Block 3
Basic DC Circuits
UNIT I
LESSON 4
PARALLEL CIRCUITS
2. What is the total resistance of the circuit? You can solve this problem in any one
of three ways.
 Use Formula 1 to calculate the total resistance.
 Use Formula 2 to calculate the total resistance.
 Solve for total current and then use Ohm's Law to calculate the total
resistance.
a. Use Formula 1 to calculate the total resistance.
1
1
1
1
=
+
+
RT
R1 R2 R3
Thus:
1
4
=
RT
40 Ω
=
1
1
1
+
+
40 Ω 40 Ω 20 Ω
so:
=
4RT = 40 Ω and
1
1
2
+
+
40 Ω 40 Ω 40 Ω
RT = 10 Ω
b. Use Formula 2 to calculate the total resistance.
RTA =
R1 × R2
R1 + R2
Thus, RT =
=
40 Ω × 40 Ω
40 Ω + 40 Ω
RTA × R3
RTA + R3
=
= 20 Ω
20 Ω × 20 Ω
20 Ω + 20 Ω
= 10 Ω
c. Solve for total current in the circuit.
IT =
20 V 20 V 20 V
E
E
E
+
+
=
+
+
40 Ω 40 Ω 20 Ω
R1 R2 R3
= 0.5A+0.5A+1.0A =2 A
Use Ohm's Law to calculate the total resistance.
Thus, RT =
E
I
=
20 V
2A
= 10 Ω
Thus, RT = 10 Ω, no matter which way you work the problem.
These problems were fairly simple because the resistor values were easy to work with.
Now that you've had some practice, you should be ready for the next exercise. Circuit
values in these problems are more realistic.
Copyright © 2002 by Nida Corporation
3-4-11
LESSON 4
PARALLEL CIRCUITS
Exercise 1:
UNIT I
Block 3
Basic DC Circuits
Solve for the Unknown Circuit Values in Table 1.
Fill in the blanks in Table 1 for the unknown values, given the circuits in Figure 9. Be sure
to include the appropriate symbols with all your answers.
9A. Parallel Circuit
9B. Equivalent Circuit
Figure 9. Parallel Circuits for Exercise Problems
Table 1. Exercise Problems
VALUE
PROBLEM NUMBER
1
2
3
4
5
6
Ea
12 V
24 V
15 V
18 V
10 V
R1
12 kΩ
1 kΩ
2 kΩ
360 Ω
R2
6 kΩ
2 kΩ
360 Ω
R3
4 kΩ
360 Ω
RT
IR1
1 mA
IR2
3 mA
IR3
2 mA
9 mA
IT
PR1
45 mW
PR2
45 mW
PR3
135 mW
PT
3-4-12
54 mW
200 mW
120 W
Copyright © 2002 by Nida Corporation
LESSON 4
PARALLEL CIRCUITS
UNIT I
Block 3
Basic DC Circuits
SUMMARY
This lesson has introduced you to parallel circuits and the relationship between current,
voltage, resistance, and power in parallel circuits. Here are the important points about
parallel circuits for you to remember.
 A parallel circuit is an electronic circuit that has two or more branches through
which the current flows.
 Applied voltage is the same across each resistor in the circuit.
 All resistors in the circuit are across each other as well as across the power source.
 Total current is equal to the sum of all the branch currents in the circuit.
 The reciprocal of the total resistance is equal to the sum of the reciprocals of all
the resistors in that circuit.
 Total resistance is always lower than the value of any resistor in any of the
branches of the circuit.
 Formulas you use with parallel circuits:
Total current:
IT = IR1 + IR2 + IR3 + etc.
Total resistance:
Formula 1:
1
1
1
1
=
+
+
+ etc.
RT
R1 R2 R3
Formula 2:
RT =
R1 × R2
R1 + R2
Ohm's law:
E = I T (RT )
3-4-18
RT =
E
IT
IT =
E
RT
Copyright © 2002 by Nida Corporation