Ohm's Law
The interdependence among current, voltage, and resistance in dc circuits is called Ohm’s law , named after the scientist who supposedly first expressed it. Three formulas denote this law:
E = IR
I = E/R
R = E/I
You need only remember the first of these formulas to be able to derive the others. The easiest way to remember it is to learn the abbreviations E for emf, I for current, and R for resistance; then remember that they appear in alphabetical order with the equals sign after the E . Thus E = IR .
It is important to remember that you must use units of volts, amperes, and ohms in order for Ohm’s law to work right. If you use volts, milliamperes (mA), and ohms or kilovolts (kV), microamperes (μA), and megohms (MΩ), you cannot expect to get the right answers. If the initial quantities are given in units other than volts, amperes, and ohms, you must convert to these units and then calculate. After that, you can convert the units back again to whatever you like. For example, if you get 13.5 million ohms as a calculated resistance, you might prefer to say that it is 13.5 megohms. However, in the calculation, you should use the number 13.5 million (or 1.35 × 10 ^{7} ) and stick to ohms for the units.
Current Calculations
The first way to use Ohm’s law is to find current values in dc circuits. In order to find the current, you must know the voltage and the resistance or be able to deduce them.
Refer to the schematic diagram of Fig. 12-8. It consists of a variable dc generator, a voltmeter, some wire, an ammeter, and a calibrated wide-range potentiometer. Actual component values are not shown here, but they can be assigned for the purpose of creating sample Ohm’s law problems. While calculating the current in the following problems, it is necessary to mentally “cover up” the meter.
Fig. 12-8 . Circuit for working Ohm’s law problems.
Current Practice Problems
Problem 1
Suppose that the dc generator (see Fig. 12-8 ) produces 10 V and that the potentiometer is set to a value of 10 Ω. What is the current?
Solution 1
This is solved easily by the formula I = E/R . Plug in the values for E and R ; they are both 10, because the units are given in volts and ohms. Then I = 10/10 = 1.0 A.
Problem 2
The dc generator (see Fig. 12-8 ) produces 100 V, and the potentiometer is set to 10.0 kΩ. What is the current?
Solution 2
First, convert the resistance to ohms: 10.0 kΩ = 10,000 Ω. Then plug the values in: I = 100/10,000 = 0.0100 A.
Voltage Calculations
The second use of Ohm’s law is to find unknown voltages when the current and the resistance are known. For the following problems, uncover the ammeter and cover the voltmeter scale in your mind.
Voltage Practice Problem
Fig. 12-8 . Circuit for working Ohm’s law problems.
Problem
Suppose that the potentiometer (see Fig. 12-8 ) is set to 100 Ω, and the measured current is 10.0 mA. What is the dc voltage?
Solution
Use the formula E = IR . First, convert the current to amperes: 10.0 mA = 0.0100 A. Then multiply: E = 0.0100 × 100 = 1.00 V. This is a low, safe voltage, a little less than what is produced by a flashlight cell.
Resistance Calculations
Ohm’s law can be used to find a resistance between two points in a dc circuit when the voltage and the current are known. For the following problems, imagine that both the voltmeter and ammeter scales in Fig. 12-8 are visible but that the potentiometer is uncalibrated.
Resistance Practice Problem
Problem
If the voltmeter reads 24 V and the ammeter shows 3.0 A, what is the value of the potentiometer?
Solution
Use the formula R = E/I , and plug in the values directly because they are expressed in volts and amperes: R = 24/3.0 = 8.0 Ω.
Power Calculations
You can calculate the power P (in watts, symbolized W) in a dc circuit such as that shown in Fig. 12-8 using the following formula:
P = EI
where E is the voltage in volts and I is the current in amperes. You may not be given the voltage directly, but you can calculate it if you know the current and the resistance.
Remember the Ohm’s law formula for obtaining voltage: E = IR . If you know I and R but don’t know E , you can get the power P by means of this formula:
P = (IR) I = I ^{2} R
That is, take the current in amperes, multiply this figure by itself, and then multiply the result by the resistance in ohms.
You also can get the power if you aren’t given the current directly. Suppose that you’re given only the voltage and the resistance. Remember the Ohm’s law formula for obtaining current: I = E/R . Therefore, you can calculate power using this formula:
P = E (E/R) = E ^{2} / R
That is, take the voltage, multiply it by itself, and divide by the resistance.
Stated all together, these power formulas are
P = El = I ^{2} R = E ^{2} /R
Now we are all ready to do power calculations. Refer once again to Fig. 12-8.
Fig. 12-8 . Circuit for working Ohm’s law problems.
Power Practice Problem
Problem
Suppose that the voltmeter reads 12 V and the ammeter shows 50 mA. What is the power dissipated by the potentiometer?
Solution
Use the formula P = EI . First, convert the current to amperes, getting I = 0.050 A. Then P = EI = 12 × 0.050 = 0.60 W.
Practice problems of these concepts can be found at: Direct Current Practice Test
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From Physics Demystified: A Self-Teaching Guide. Copyright © 2002 by The McGraw-Hill Companies, Inc. All Rights Reserved.