20.2

Ohm’s Law

The current that a battery can push through a wire is analogous to the water flow that a pump can push through a pipe. Greater pump pressures lead to larger water flow rates, and, similarly, greater battery voltages lead to larger electric currents. In the simplest case, the current I is directly proportional to the voltage V; that is, . Thus, a 12-V battery leads to twice as much current as a 6-V battery, when each is connected to the same circuit.

CONCEPTS AT A GLANCE In a water pipe, the flow rate is not only determined by the pump pressure but is also affected by the length and diameter of the pipe. Longer and narrower pipes offer higher resistance to the moving water and lead to smaller flow rates for a given pump pressure. A similar situation exists in electric circuits, and to deal with it we introduce the concept of electrical resistance. As the Concepts-at-a-Glance chart in Figure 20-5 indicates, electrical resistance is defined in terms of two ideas that have already been discussed—the electric potential difference, or voltage (see Section 19.2), and the electric current (see Section 20.1).

The resistance (R) is defined as the ratio of the voltage V applied across a piece of material to the current I through the material, or R = V/I. When only a small current results from a large voltage, there is a high resistance to the moving charge. For many materials (e.g., metals), the ratio V/I is the same for a given piece of material over a wide range of voltages and currents. In such a case, the resistance is a constant. Then, as Figure 20-5 indicates, the relation R = V/I is referred to as Ohm’s law, after the German physicist Georg Simon Ohm (1789–1854), who discovered it.

Figure 20-5 CONCEPTS AT A GLANCE The concepts of electric potential difference, or voltage V, and the electric current I are brought together to define the electrical resistance R. Ohm’s law specifies that the resistance remains constant as the voltage and current change. A current exists in the gap of this spark plug when the voltage between the two ends of the gap is high enough. (Fotopic/Index Stock)

| Ohm’s Law
| The ratio V/I is a constant, where V is the voltage applied across a piece of material (such as a wire) and I is the current through the material:

| (20.2)

R is the resistance of the piece of material.
SI Unit of Resistance: volt/ampere (V/A) = ohm ()

The SI unit for resistance is a volt per ampere, which is called an ohm and is represented by the Greek capital letter omega (). Ohm’s law is not a fundamental law of nature like Newton’s laws of motion. It is only a statement of the way certain materials behave in electric circuits.
To the extent that a wire or an electrical device offers resistance to the flow of charges, it is called a resistor. The resistance can have a wide range of values. The copper wires in a television set, for instance, have a very small resistance. On the other hand, commercial resistors can have resistances up to many kilohms or megohms . Such resistors play an important role in electric circuits, where they are used to limit the amount of current and establish proper voltage levels.
In drawing electric circuits we follow the usual conventions: (1) a zigzag line represents a resistor and (2) a straight line (——) represents an ideal conducting wire, or one with a negligible resistance. Example 2 illustrates an application of Ohm’s law to the circuit in a flashlight.

| Example 2 | | | A Flashlight

The filament in a light bulb is a resistor in the form of a thin piece of wire. The wire becomes hot enough to emit light because of the current in it. Figure 20-6 shows a flashlight that uses two 1.5-V batteries (effectively a single 3.0-V battery) to provide a current of 0.40 A in the filament. Determine the resistance of the glowing filament.

Figure 20-6 The circuit in this flashlight consists of a resistor (the filament of the light bulb) connected to a 3.0-V battery (two 1.5-V batteries).

Reasoning The filament resistance is assumed to be the only resistance in the circuit. The potential difference applied across the filament is that of the 3.0-V battery. The resistance, given by Equation 20.2, is equal to this potential difference divided by the current.
Solution The resistance of the filament is

| (20.2)

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.