EET 114 Unit 5. Ohm's Law

Unit 5. Ohm's Law

The main topic in this unit is the most fundamental equation in electronics: Ohm's Law. This is an equation that relates voltage, current, and resistance. We'll start to discuss a topic that's not covered in the book, but that's important for a technician: how to estimate answers to mathematical problems without using a calculator. Finally, we'll look beyond series circuits and parallel circuits to more complicated circuits called series-parallel circuits.

First, you should read the following chapter of Thomas Floyd's Principles of Electric Circuits (8th edition):

Ohm's Law (Chapter 3)

Then work through the e-Lesson and self-test questions below.

After completing the e-Lesson, take Quiz #5, perform Lab #5, and do Homework #5.

Unit 4 Review

This unit will build on material that you studied in Unit 4. So let's begin by taking this self-test to review what you learned in that unit.

A Simple Circuit

Here is a very simple circuit that contains just a voltage source (V_S) and a resistor (R):
Note: In drawings like this, sometimes our textbook uses the label V_S for the source voltage (as we have done here), and sometimes the textbook uses the label V for the source voltage. The two labels mean the same thing, so don't think that there's an important difference between using V_S and using V.

Some textbooks use the label E for a source voltage, so you may see this label from time to time.

How will the current in this circuit change if we change either the source voltage or the resistance? (Read on for the answer.)

Current Depends on Voltage (Floyd, p. 73)

Voltage is a measure of a source's ability to produce current. The greater the voltage, the greater the current that the source can produce in a fixed resistance.
In other words, current produced in a resistance is directly proportional to the voltage of the source.
So, for instance, in the circuit shown above, if we double the source voltage V_S while we keep the resistance R constant, then the current through the circuit will double.
On the other hand, if we reduce the source voltage V_S by half while we keep the resistance R constant, then the current through the circuit will be reduced by half.

Current Depends on Resistance (Floyd, p. 73)

Resistance reduces the flow of current. The greater the resistance, the smaller the current produced through it by a fixed source voltage.
In other words, current is inversely proportional to resistance.
So, for instance, in the circuit shown above, if we double the resistance R while we keep the source voltage V_S constant, then the current through the circuit will be reduced by half.
On the other hand, if we reduce the resistance R by half while we keep the source voltage V_S constant, then the current through the circuit will double.

Ohm's Law (Floyd, p. 74)

The points made above can be combined in a single mathematical expression called Ohm's law:
I = V ÷ R (Equation 3-1)
In words, the current in a circuit is equal to the source voltage divided by the circuit's resistance.

Other Forms of Ohm's Law (Floyd, p. 74)

Using basic algebra, we can rewrite Ohm's law in two different forms. One is:

V = I × R (Equation 3-2)
In words, this says that voltage equals current times resistance.
The other form of Ohm's law is:

R = V ÷ I (Equation 3-3)
In words, resistance equals voltage divided by current.

Ohm's Law Triangle

Many people use the following memory aid to help them remember the three different forms of Ohm's law:
1. Draw the triangle shown below.
2. Use your finger to cover up the letter that you're solving for.
3. The two remaining letters will then be in the correct position:
  - If the two letters are side-by-side, then you have to multiply them.
  - If one letter is positioned over the other one, then you have to divde.
For example, if you cover up the I, then you're left with a V over an R, which tells you that I = V ÷ R.
Of course, when you draw the triangle you have to put the letters in the correct positions. To help you remember that the V goes on top and the I and R go below, remember this little saying: "The Vulture flies over the Indian, who walks by the River."
Try it out on the following questions!

Voltage Drops

So far we've been discussing a circuit with just one resistor connected across a voltage source. In such a case, we say that the entire source voltage is dropped across the resistor. This means that if we use a voltmeter to measure the voltage across the resistor, we'll get the same value that we would get if we measured across the voltage source itself.
But most circuits contain more than one resistor. In this case, each resistor will have what's called a voltage drop, but usually this is not equal to the source voltage.
Example: The circuit shown below has two resistors, R₁ and R₂, connected to a 12-volt source. In this circuit, R₁ will have a voltage drop of 4 volts, and R₂ will have a voltage drop of 8 volts. (In a later course, EET 150, you'll learn the equations that let you find these values.)
We express these facts by writing V₁ = 4 V, and V₂ = 8 V. As you can probably guess, V₁means the voltage drop across R₁, and V₂means the voltage drop across R₂.

Ohm's Law Again

Ohm's law applies to all circuits containing resistors, but you have to be careful how you use it.
- When you apply it to the entire circuit, Ohm's law says that the circuit's current is equal to the source voltage divided by the circuit's total resistance.
- When you apply it to a single resistor, Ohm's law says that a resistor's current is equal to the resistor's voltage drop divided by the resistor's resistance.
In symbols, when we're talking about the entire circuit, we'll write

I_T = V_S ÷ R_T
where I_T is the circuit's total current (that's the current coming out of the battery), and V_S is the source voltage, and R_T is the circuit's total resistance.
But when we're talking about a single resistor, we'll write

I₁ = V₁ ÷ R₁
where I₁ is the current through resistor R1, and V₁ is the voltage drop across resistor R1, and R₁ is the resistance of resistor R1.
This relation holds for every resistor. So we can also write I₂ = V₂ ÷ R₂, and so on. (And we can also rearrange these equations to solve for voltage or resistance.)

Linearity (Floyd, p. 74)

An electrical device is said to be linear if a graph of the voltage across it versus the current through it is a straight line.
Some circuit analysis techniques that we'll study can only be applied to circuits composed of linear devices.
A resistor is an example of a linear device. Other examples are capacitors and inductors, which you will study later. If you drew a graph of voltage versus current for a 3.3 kΩ resistor, the graph would be a straight line, as shown below.
A diode is an example of a device that is not linear. If you drew a graph of voltage versus current for a diode, the graph would look something like the one shown below. We will not study diodes or any other non-linear devices in this course, but you will study them in later courses.

Estimation

When multiplying or dividing two quantities, you should be able to come up with a ball-park estimate without using a calculator.
Example: Suppose you know that a 2.06 kΩ resistor has a current of 5.75 mA, and you want to estimate the resistor's voltage drop. You know by Ohm's law that
V = I × R = 5.75 mA × 2.06 kΩ
I'm going to re-write this using colors to draw attention to three different parts of the problem:
V = I × R = 5.75 mA × 2.06 kΩ
Now let's use a three-step approach to this problem:
1. First, deal with the number parts of the two quantities. In this case, the number parts are 5.75 and 2.06. Since we're just looking for a ball-park estimate, we'll round those numbers off to 6 and 2. Multiplying 6 times 2 is easy; the answer is 12.
2. Next, deal with the metric prefixes of the two quantities. In this case, the metric prefixes are milli- and kilo-. One trick for doing this is to remember that milli- and kilo- are “opposites,” in the sense explained below. In particular, when you multiply milli- times kilo-, they cancel each other, leaving you with no metric prefix.
3. Finally, deal with the units of the two quantities. In this case, the units are amps and ohms. You know (from your knowledge of Ohm's law) that when you multiply amps times ohms, you get volts.
Putting it all together, our answer has an approximate number part of 12, and it has no engineering prefix, and it has a unit of volts. So our estimate is 12 V. If you use your calculator to get the exact answer, you'll get 11.8 V (rounded to three digits). So our estimate is pretty close.

k- and m- Are “Opposites”

When I say that kilo- and milli- are opposites, what I mean mathematically is that kilo- stands for 10³ and milli- stands for 10^-3. Note that the number part of the exponents are the same, but one exponent is positive while the other exponent is negative.
Here are a few situations in which it's useful to know that kilo- and milli- are the "opposite" of each other:
- When you multiply kilo- times milli-, they cancel each other, leaving you with no metric prefix:
  - Example: 6 mA × 2 kΩ = 12 V.
- When you're doing a division, if you have kilo- in the denominator and no metric prefix in the numerator, your answer's metric prefix will be milli-:
  - Example: 10 V ÷ 5 kΩ = 2 mA.
- When you're doing a division, if you have milli- in the denominator with no metric prefix in the numerator, your answer's metric prefix will be kilo- :
  - Example: 15 V ÷ 3 mA = 5 kΩ.

M- and μ- Are “Opposites”

In the same way that kilo- and milli- are the opposite of each other, mega- and micro- are also the opposite of each other, since mega- stands for 10⁶ and micro- stands for 10^-6.
Here are a few situations in which it's useful to know that mega- and micro- are the "opposite" of each other:
- When you multiply mega- times micro-, they cancel each other, leaving you with no metric prefix:
  - Example: 5 µA × 3 MΩ = 15 V.
- Also, when you're doing a division, if you have mega- in the denominator and no metric prefix in the numerator, your answer's metric prefix will be micro- :
  - Example: 12 V ÷ 4 MΩ = 3 µA.
- And if you have micro- in the denominator with no metric prefix in the numerator, your answer's metric prefix will be mega- :
  - Example: 16 V ÷ 8 µA = 2 MΩ.

Series-Parallel Circuit

A series-parallel circuit is one that contains combinations of series- and parallel-connected components, which are in turn connected in series and/or parallel with other such combinations.
Example: The circuit shown below, which appeared in some of the self-test questions for Unit 4, is a series-parallel circuit.

Seeing Series Connections and Parallel Connections

In working with series-parallel circuits, it's very important that you be able to see which components are connected in series with each other, and which are connected in parallel with each other.
Example: In the circuit above, R2 is connected in parallel with R3. Also, V_S is connected in series with R1, and V_S is connected in series with R4. (If you don't understand why, go back and review the meanings of "connected in series" and "connected in parallel" in Unit 4.)

Breadboarding Series-Parallel Circuits

Once you've spent some time studying how the components in a particular circuit are connected with each other, you should be ready to build the circuit on a breadboard.
Example: Repeated below is the series-parallel circuit from above. Beneath the schematic diagram is a breadboard diagram showing one way of building this circuit. (Of course, there are many other correct ways of building it.)
Study this example until you are satisfied that the schematic diagram and the breadboard diagram represent the same circuit.
Now here's a photograph of the same circuit built on an actual breadboard.
Be sure to try these Self-Test questions!

Measurements in Series-Parallel Circuits

Once you've built a series-parallel circuit, you'll want to be able to make measure voltage drops and currents in the circuit.
Measuring voltage drops is easy. To measure the voltage drop across a resistor, you set the meter to measure voltage, and then you touch the meter's two leads to the two legs of the resistor.
- Another way of saying this is that the voltmeter must be connected in parallel with the resistor whose voltage you wish to measure.
Measuring current is usually trickier. To measure the current through a resistor, you must first lift one leg of that resistor out of the breadboard. (It does not matter which of the resistor's legs you lift.) Then set the meter to measure current, and connect one meter lead to the detached leg of the resistor, and connect the other meter lead to whatever the resistor leg was originally attached to.
- A simpler way of saying this is that the ammeter must be connected in series with the resistor whose current you wish to measure.
Example: In the circuit that we've been discussing, suppose you wish to measure R2's current. Shown below is one way of doing this. Notice that one leg of R2 has been detached from what it was originally attached to (namely, R1 and R3), and then the meter was inserted between R2 and those things to which R2 was originally attached (R1 and R3).

Unit 5 Review

This e-Lesson has covered several important topics, including:
- Ohm's law
- estimation
- series-parallel circuits.
To finish the e-Lesson, take this self-test to check your understanding of these topics.

Congratulations! You've completed the e-Lesson for this unit. What's next?

Take Quiz #5.
Perform Lab #5.
Do Homework #5.
For more practice with the material from this Unit, visit the textbook's Chapter 3 web page and take the multiple-choice, true/false, and fill-in-the-blank quizzes provided there.
Keep practicing your skills by playing the games on the Games page.

Then you'll be ready to go on to Unit 6's e-Lesson .

Nick Reeder | Electronics Engineering Technology | Sinclair Community College

Send comments to nick.reeder@sinclair.edu