Review of the Fundamentals of DC Circuits
Overview
This module covers fundamental DC circuit principles to help post-secondary electrical engineering technology students and engineering bound high school students, understand basic concepts and perform limited circuit analysis.
Introduction
This module covers some of the fundamentals of DC circuits and circuit analysis, as a supplement to a post-secondary electrical engineering technology in-class course with 128 hours of lectures and labs. Basic concepts are covered with practice questions related to each topic.
In this module you will learn some basic concepts about electric circuits and how to analyze them from a Direct Current (DC) perspective. The following topics will be covered:
Electron Current Flow
Circuit Terminology
Series and parallel Circuits
Ohm’s Law
Power
Kirchhoff’s Laws
Total Time Required: Approximately 1 hour 45 minutes minutes.
Electron Current Flow
Learning Outcomes:
Be able to describe how electrons flow through a conductor as current.
Discussion
Before discussing electric circuits and DC Circuit analysis, it is important to understand the concept of how electrons can flow through a conductor to create current. A constant current or Direct Current (DC) is created when a constant voltage or DC voltage source is applied to a conductor.
The following video explains how an electrical conductor works to carry current.
Terms of Use: This video, How Conductors Work, by Engineering Technology is licensed under a Creative Commons Attribution 4.0 International License. |
Circuit Terminology
Learning Outcomes:
Be able to describe the symbols for a constant voltage source and a resistor.
Be able to describe resistance values with different ways to express values.
Be able to describe various circuit terminology.
Be able to describe an electric circuit using symbols.
Discussion
In order to do analysis of DC circuits you will need to understand some circuit terminology, as well as how circuit elements are represented graphically. Elements in a circuit are described using symbols to represent their physical nature and have units of measure associated with their characteristics.
Some basic circuit elements are necessary to discuss DC circuits: a constant voltage source (which includes batteries) and resistors. Current is usually represented by an arrow (for the direction of the current along a path in the circuit) and the designation I for direct current. (DC)
A DC voltage source is represented by the following symbols:
A resistor is represented by the following symbol:
Note that the term for resistance is called an Ohm and it is represented by the symbol Ω. Resistance can be expressed in a few ways:
100 Ω or 100 ohms
1kΩ or 1 kilohm or 1,000 Ω or 1,000 ohms
1MΩ or 1 megohm or 1,000,000 Ω or 1,000,000 ohms
Here is a reference site for electrical symbols:
Follow the tutorial below, focusing on the sections called Circuit and Schematic.
Circuit Terminology, Khan Academy
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Series and Parallel Circuits
Answers to Practice Calculating Resistance Questions:
Question 1: Circuits A, C, and F are series circuits.
Instructor Notes:
The purpose of this question is to get students to identify what distinguishing characteristic uniquely identifies a circuit as being “series.” Once this has been identified, there are several conclusions which may be deduced (regarding voltage drops, currents, resistances, etc.).
Circuit F is thrown in the mix just to show students that the non-battery components don’t have to all be the same (resistors) in order for a circuit to qualify as “series.”
Learning Outcomes:
Be able to describe and calculate the resistance in a series configuration circuit.
Be able to describe and calculate the resistance in a parallel configuration circuit.
Be able to describe and calculate the resistance in a series-parallel configuration circuit.
Discussion
There are two basic configurations of electrical circuits that electrical components can be organized in: series and parallel circuits.
Series Configuration Circuit
First, an example of a series circuit:
Here, we have three resistors (labeled R1, R2, and R3) connected in a long chain from one terminal of the battery to the other. (It should be noted that the subscript labeling—those little numbers to the lower-right of the letter “R”—are unrelated to the resistor values in ohms. They serve only to identify one resistor from another.)
The defining characteristic of a series circuit is that there is only one path for current to flow. In this circuit, the current flows in a clockwise direction, from point 1 to point 2 to point 3 to point 4 and back around to 1.
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The total resistance in a series configuration circuit is the sum of all resistance values in the circuit. In the example above RTOTAL = R1 + R2 + R3.
Parallel Circuit Configuration
Now, let’s look at the other type of circuit, a parallel configuration:
Again, we have three resistors, but this time they form more than one continuous path for current to flow. There’s one path from 1 to 2 to 7 to 8 and back to 1 again. There’s another from 1 to 2 to 3 to 6 to 7 to 8 and back to 1 again. And then there’s a third path from 1 to 2 to 3 to 4 to 5 to 6 to 7 to 8 and back to 1 again. Each individual path (through R1, R2, and R3) is called a branch.
The defining characteristic of a parallel circuit is that all components are connected between the same set of electrically common points. Looking at the schematic diagram, we see that points 1, 2, 3, and 4 are all electrically common. So are points 8, 7, 6, and 5. Note that all resistors, as well as the battery, are connected between these two sets of points.
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The total resistance in a parallel configuration circuit is the inverse of the sum of the inverse values of each resistance values in the circuit.
Terms of Use: This material, The Equation for Resistance in Parallel Circuits, by All About Circuits is available through the terms of their User Agreement, which stipulates the use of their materials for educational purposes only and that the user of this information will not hold All About Circuits liable for its content. |
Series-Parallel Configuration Circuit
In this circuit, we have two loops for the current to flow through: one from 1 to 2 to 5 to 6 and back to 1 again, and another from 1 to 2 to 3 to 4 to 5 to 6 and back to 1 again. Notice how both current paths pass through R1 (from point 1 to point 2). In this configuration, we’d say that R2 and R3 are in parallel with each other, while R1 is in series with the parallel combination of R2 and R3.
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Practice Calculating Resistance
Question 1: Identify which of these circuits is a series circuit (there may be more than one shown!):
Question 2: Calculate the resistance between points A and B (RAB) for the following resistor networks using the mathematical relationships presented in the sections for series and parallel configuration circuits.
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This video provides further understanding of series and parallel circuits.
Terms of Use: This video, Series and Parallel Circuits, by Engineering Technology is licensed under a Creative Commons Attribution 4.0 International License |
Ohm’s Law
Answers to Practice Using Ohm’s Law
Question 1:
IR1 = 2.22 mA ; VR1 = 3.33 V
IR2 = 2.22 mA ; VR2 = 22.2 V
IR3 = 2.22 mA ; VR3 = 10.4 V
Instructor Notes:
Students often just want to memorize a procedure for determining answers to questions like these. Challenge your students to not only understand the procedure, but to also explain why it must be followed.
Learning Outcomes:
Be able to describe Ohm’s Law.
Be able to use Ohm’s Law to calculate current and voltage in a circuit with a voltage source and resistive load.
Description of Ohm’s Law
The current that flows through most substances is directly proportional to the voltage applied to it. The German physicist Georg Simon Ohm (1787–1854) was the first to demonstrate experimentally that the current in a metal wire is directly proportional to the voltage applied:
I α V
This important relationship is the basis for Ohm’s law. It can be viewed as a cause-and-effect relationship, with voltage the cause and current the effect. This is an empirical law, which is to say that it is an experimentally observed phenomenon, like friction. Such a linear relationship doesn’t always occur. Any material, component, or device that obeys Ohm’s law, where the current through the device is proportional to the voltage applied, is known as an ohmic material or ohmic component. Any material or component that does not obey Ohm’s law is known as a nonohmic material or nonohmic component.
Ohm’s Experiment
In a paper published in 1827, Georg Ohm described an experiment in which he measured voltage across and current through various simple electrical circuits containing various lengths of wire. A similar experiment is shown in Figure 1. This experiment is used to observe the current through a resistor that results from an applied voltage. In this simple circuit, a resistor is connected in series with a battery. The voltage is measured with a voltmeter, which must be placed across the resistor (in parallel with the resistor). The current is measured with an ammeter, which must be in line with the resistor (in series with the resistor).
Figure 1 The experimental set-up used to determine if a resistor is an ohmic or nonohmic device. (a) When the battery is attached, the current flows in the clockwise direction and the voltmeter and ammeter have positive readings. (b) When the leads of the battery are switched, the current flows in the counterclockwise direction and the voltmeter and ammeter have negative readings.
In this updated version of Ohm’s original experiment, several measurements of the current were made for several different voltages. When the battery was hooked up as in Figure 1(a), the current flowed in the clockwise direction and the readings of the voltmeter and ammeter were positive. Does the behavior of the current change if the current flowed in the opposite direction? To get the current to flow in the opposite direction, the leads of the battery can be switched. When the leads of the battery were switched, the readings of the voltmeter and ammeter readings were negative because the current flowed in the opposite direction, in this case, counter-clockwise. Results of a similar experiment are shown in Figure 2.
Figure 2 A resistor is placed in a circuit with a battery. The voltage applied varies from -10.00 V to +10.00 V, increased by 1.00 V increments. A plot shows values of the voltage versus the current typical of what a casual experimenter might find.
In this experiment, the voltage applied across the resistor varies from -10.00 V to +10.00 V by increments of 1.00 V. The current through the resistor and the voltage across the resistor are measured. A plot is made of the voltage versus the current, and the result is approximately linear. The slope of the line is the resistance, or the voltage divided by the current. This result is known as Ohm’s law:
V = I R
Where V is the voltage measured in volts across the object in question, I is the current measured through the object in amps, and R is the resistance in units of ohms. As stated previously, any device that shows a linear relationship between the voltage and the current is known as an ohmic device. A resistor is therefore an ohmic device.
Terms of Use: This material is adapted by Joe Carey from Introduction to Electricity, Magnetism, and Circuits by Daryl Janzen and is used under a Creative Commons Attribution 4.0 International license. |
The following video provides further understanding of Ohm’s Law.
Terms of Use: This video, Ohm’s Law, by Engineering Technology is licensed under a Creative Commons Attribution 4.0 International License. |
Practice Using Ohm’s Law
Explain, step by step, how to calculate the amount of current (I) that will go through each resistor in this series circuit, and also the voltage (V) dropped by each resistor:
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Here is a link to a series of calculators for electric circuits.
Examples for Electric Circuits, WolframAlpha
Terms of Use: This material, Examples for Electric Circuits, by WolframAlpha is available through their General Terms of Use agreement, which stipulates the use of their materials for non-commercial purposes only and that the user must attribute WolframAlpha for results generated from its tools through a link to the specific web page where results to the query were generated. |
Power
Answers to Practice Calculating Power in Electric Circuits Questions
Question 1:
6.48 watts
Here, the battery is the source and the light bulb is the load.
Instructor Notes:
The solution to this problem simply requires the application of Ohm’s Law. As for the question of source versus load, ask your students to define each of these words in terms of energy transfer.
Question 2:
4,840 watts at 110 volts; 19,360 watts at 220 volts.
Instructor Notes:
Many students will mistakenly calculate 9,680 watts as the power dissipation at 220 volts. However, power dissipation does not increase linearly with increases in voltage!
Question 3:
E1 Ω = 4 volts
E2 Ω = 8 volts
E3 Ω = 12 volts
P1 Ω = 16 watts
P2 Ω = 32 watts
P3 Ω = 48 watts
Follow-up question: Compare the direction of current through all components in this circuit with the polarities of their respective voltage drops. What do you notice about the relationship between current direction and voltage polarity for the battery, versus for all the resistors? How does this relate to the identification of these components as either sources or loads?
Instructor Notes:
The answers to this question should not create any surprises, especially when students understand electrical resistance in terms of friction: resistors with greater resistance (more friction to electron motion) require greater voltage (push) to get the same amount of current through them. Resistors with greater resistance (friction) will also dissipate more power in the form of heat, given the same amount of current.
Another purpose of this question is to instill in students’ minds the concept of components in a simple series circuit all sharing the same amount of current.
Challenge your students to recognize any mathematical patterns in the respective voltage drops and power dissipations. What can be said, mathematically, about the voltage drop across the 2 Ω resistor versus the 1 Ω resistor, for example?
Learning Outcomes:
Be able to describe what power is in an electrical circuit.
Be able to calculate the power dissipated by a load in an electric circuit.
Power as a Function of Voltage and Current
Power is a measure of how much work can be performed in a given amount of time. In electric circuits, power is a function of both voltage and current:
P = I V
In this case, however, power (P) is exactly equal to current (I) multiplied by voltage (V), rather than merely being proportional to IV. When using this formula, the unit of measurement for power is the watt, abbreviated with the letter “W.”
It must be understood that neither voltage nor current by themselves constitute power. Rather, power is the combination of both voltage and current in a circuit. Remember that voltage is the specific work (or potential energy) per unit charge, while current is the rate at which electric charges move through a conductor. Voltage (specific work) is analogous to the work done in lifting a weight against the pull of gravity. Current (rate) is analogous to the speed at which that weight is lifted. Together as a product (multiplication), voltage (work) and current (rate) constitute power.
Just as in the case of the diesel tractor engine and the motorcycle engine, a circuit with high voltage and low current may be dissipating the same amount of power as a circuit with low voltage and high current. Neither the amount of voltage alone nor the amount of current alone indicates the amount of power in an electric circuit.
Practice Calculating Power in Electric Circuits
Question 1: How much electrical power is being dissipated by the light bulb in this circuit?
In this circuit, identify which component is the source and which is the load.
Question 2: Calculate the amount of power dissipated by this electric heating element, if the generator’s output voltage is 110 volts and the heater’s resistance is 2.5 ohms:
Now, calculate the power dissipated by the same heater if the generator’s output voltage is doubled.
Question 3: n this circuit, three resistors receive the same amount of current (4 amps) from a single source. Calculate the amount of voltage “dropped” by each resistor, as well as the amount of power dissipated by each resistor:
Terms of Use: This material, Energy, Work, and Power and Series DC Circuits Practice Worksheet with Answers, by All About Circuits is available through the terms of their User Agreement, which stipulates the use of their materials for educational purposes only and that the user of this information will not hold All About Circuits liable for its content. |
Here is a link to a series of calculators for electric circuits.
Examples for Electric Circuits, WolframAlpha
Terms of Use: This material, Examples for Electric Circuits, by WolframAlpha is available through their General Terms of Use agreement, which stipulates the use of their materials for non-commercial purposes only and that the user must attribute WolframAlpha for results generated from its tools through a link to the specific web page where results to the query were generated. |
Kirchhoff's Circuit Laws
Answers to Practicing Kirchhoff’s Voltage Law Questions
Question 1:
Red lead on “A”, black lead on ground (Digital voltmeter reads 15 volts)
Red lead on “B”, black lead on ground (Digital voltmeter reads -15 volts)
Red lead on “A”, black lead on “B” (Digital voltmeter reads 30 volts)
Red lead on “B”, black lead on “A” (Digital voltmeter reads -30 volts)
Instructor Notes:
This question may be easily answered with only a voltmeter, two batteries, and a single “jumper” wire to connect the two batteries in series. It does not matter if the batteries are 15 volts each! The fundamental principle may still be investigated with batteries of any voltage, so this is a very easy demonstration to set up during discussion time.
Question 2:
Instructor Notes:
The answer to this question is fairly simple, but the real point of it is to get students thinking about how and why it is the way it is. One thing I’ve noticed as an instructor of electronics is that most students tend to follow the rule of proximity: the resistor’s voltage drop polarity is determined by proximity to poles of the battery. The resistor terminal closest to the battery’s negative terminal must be negative as well, or so the thinking goes.
In this particular circuit, though, the rule of proximity does not hold very well, and a different rule is necessary.
Answers to Practicing Kirchhoff’s Current Law Questions
Question 1:
The total current in this circuit is 4 mA, and the load voltage is 18.8 volts.
Follow-up question: indicate the polarity of the voltage across the load resistor with “+” and “-” symbols.
Instructor Notes:
Have your students collectively agree on a procedure they may use to accurate discern series voltage sums and polarities. Guide their discussion, helping them identify principles that are true and valid for all series circuits.
Question 2:
Instructor Notes:
It is not necessary to know anything about series-parallel or even parallel circuits in order to solve the R4‘s current - all one needs to know is how to use Kirchhoff’s Current Law.
Question 3:
Instructor Notes:
Discuss with your students techniques for calculating the node currents. What Laws did your students apply, and more importantly, in what order did they apply them?
Learning Outcomes:
Be able to describe Kirchhoff’s Voltage Law.
Be able to calculate the voltage and polarity across a load in an electric circuit.
Be able to describe Kirchhoff’s Current Law.
Be able to calculate the currents and their directions into and out of a node in an electric circuit.
What is Kirchhoff’s Voltage Law (KVL)
Gustav R. Kirchhoff, a German physicist, discovered in 1847 that the algebraic sum of all voltages in a loop must equal zero. The following video explains the concept of Kirchhoff’s Voltage Law (KVL).
Kirchhoff’s Voltage Law, Kahn Academy
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Demonstrating Kirchhoff’s Voltage Law in a Series Circuit
Let’s take another look at our example series circuit, this time numbering the points in the circuit for voltage reference:
If we were to connect a voltmeter between points 2 and 1, red test lead to point 2 and the black test lead to point 1, the meter would register +45 volts. Typically, the “+” sign is not shown but rather implied, for positive readings in digital meter displays. However, for this lesson, the polarity of the voltage reading is very important and so I will show positive numbers explicitly:
E2-1 = +45 V
When a voltage is specified with a double subscript (the characters “2-1” in the notation “E2-1”), it means the voltage at the first point (2) as measured in reference to the second point (1).
If we were to take a voltmeter and measure the voltage drop across each resistor, stepping around the circuit in a clockwise direction with the red test lead of our meter on the point ahead and the black test lead on the point behind, we would obtain the following readings:
We should already be familiar with the general principle for series circuits stating that individual voltage drops add up to the total applied voltage, but measuring voltage drops in this manner and paying attention to the polarity (mathematical sign) of the readings reveals another facet of this principle: that the voltages measured as such all add up to zero:
In the above example, the loop was formed by the following points in this order: 1-2-3-4-1. It doesn’t matter which point we start at or which direction we proceed in tracing the loop; the voltage sum will still equal zero. To demonstrate, we can tally up the voltages in loop 3-2-1-4-3 of the same circuit:
This may make more sense if we re-draw our example series circuit so that all components are represented in a straight line:
It’s still the same series circuit, just with the components arranged in a different form. Notice the polarities of the resistor voltage drops with respect to the battery: the battery’s voltage is negative on the left and positive on the right, whereas all the resistor voltage drops are oriented the other way: positive on the left and negative on the right. This is because the resistors are resisting the flow of electric charge being pushed by the battery. In other words, the “push” exerted by the resistors against the flow of electric charge must be in a direction opposite the source of electromotive force.
Here we see what a digital voltmeter would indicate across each component in this circuit, black lead on the left and red lead on the right, as laid out in horizontal fashion:
If we were to take that same voltmeter and read voltage across combinations of components, starting with the only R1 on the left and progressing across the whole string of components, we will see how the voltages add algebraically (to zero):
The fact that series voltages add up should be no mystery, but we notice that the polarity of these voltages makes a lot of difference in how the figures add. While reading voltage across R1—R2, and R1—R2—R3 (I’m using a “double-dash” symbol “—” to represent the series connection between resistors R1, R2, and R3), we see how the voltages measure successively larger (albeit negative) magnitudes, because the polarities of the individual voltage drops are in the same orientation (positive left, negative right). The sum of the voltage drops across R1, R2, and R3 equals 45 volts, which is the same as the battery’s output, except that the battery’s polarity is opposite that of the resistor voltage drops (negative left, positive right), so we end up with 0 volts measured across the whole string of components.
That we should end up with exactly 0 volts across the whole string should be no mystery, either. Looking at the circuit, we can see that the far left of the string (left side of R1: point number 2) is directly connected to the far right of the string (right side of battery: point number 2), as necessary to complete the circuit. Since these two points are directly connected, they are electrically common to each other. And, as such, the voltage between those two electrically common points must be zero.
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Practicing Kirchhoff’s Voltage Law
Question 1: Many electronic circuits use what is called a split or a dual power supply:
Determine what a digital voltmeter would indicate if connected between the following points:
Red lead on “A”, black lead on ground
Red lead on “B”, black lead on ground
Red lead on “A”, black lead on “B”
Red lead on “B”, black lead on “A”
NOTE: in electronic systems, “ground” is often not associated with an actual earth-soil contact. It usually only refers to a common point of reference somewhere in the circuit used to take voltage measurements. This allows us to specify voltages at single points in the circuit, with the implication that “ground” is the other point for the voltmeter to connect to.
Question 2: Determine both the polarity of voltage across the resistor in this circuit, and how much voltage will be dropped across the resistor:
Explain the procedure(s) you used to answer both these questions.
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What is Kirchhoff’s Current Law (KCL)
Kirchhoff’s Current Law (KCL) states that the sum of the currents flowing into a node equals the sum of the currents flowing out of the node. The following video explains the concept of KCL.
Kirchhoff’s Current Law, Kahn Academy
Terms of Use: This video, Kirchhoff’s Current Law, by Khan Academy is licensed under CC BY-NC-SA 4.0. |
Also, look at this article on Kirchhoff’s Laws, paying attention the sections called Currents into a Node, Kirchhoff’s Current Law and Kirchhoff's Current Law - concept checks.
Kirchhoff’s Laws, Kahn Academy
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Practicing Kirchhoff’s Current Law
Question 1: Calculate the total current output to the load resistor by this set of parallel-connected current sources:
Also, calculate the voltage dropped across Rload.
Question 2: Use Kirchhoff’s Current Law to calculate the magnitude and direction of the current through resistor R4 in this resistor network:
Question 3: Calculate and label the currents at each node (junction point) in this circuit:
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