203. Ohm’s Law: Resistance and Simple CircuitsLearning Objectives
What drives current? We can think of various devices—such as batteries, generators, wall outlets, and so on—which are necessary to maintain a current. All such devices create a potential difference and are loosely referred to as voltage sources. When a voltage source is connected to a conductor, it applies a potential difference $V$ that creates an electric field. The electric field in turn exerts force on charges, causing current. Ohm’s LawThe current that flows through most substances is directly proportional to the voltage $V$ 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: $$I\propto V\text{.}$$This important relationship is known as Resistance and Simple CircuitsIf voltage drives current, what impedes it? The electric property that impedes current (crudely similar to friction and air resistance) is called Thus, for example, current is cut in half if resistance doubles. Combining the relationships of current to voltage and current to resistance gives $$I=\frac{V}{R}\text{.}$$This relationship is also called Ohm’s law. Ohm’s law in this form really defines resistance for certain materials. Ohm’s law (like Hooke’s law) is not universally valid. The many substances for which Ohm’s law holds are called Figure 1 shows the schematic for a simple circuit. A Example 1: Calculating Resistance: An Automobile HeadlightWhat is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it? Strategy We can rearrange Ohm’s law as stated by $I=\text{V/R}$ and use it to find the resistance. Solution Rearranging $I=\text{V/R}$ and substituting known values gives $$R=\frac{V}{I}=\frac{\text{12}\text{.}\text{0 V}}{2\text{.}\text{50 A}}=\text{4}\text{.}\text{80 \Omega}\text{.}$$Discussion This is a relatively small resistance, but it is larger than the cold resistance of the headlight. As we shall see in Resistance and Resistivity, resistance usually increases with temperature, and so the bulb has a lower resistance when it is first switched on and will draw considerably more current during its brief warmup period. Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of ${\text{10}}^{\text{12}}\phantom{\rule{0.25em}{0ex}}\Omega $ or more. A dry person may have a handtofoot resistance of ${\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\Omega $, whereas the resistance of the human heart is about ${\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\Omega $. A meterlong piece of largediameter copper wire may have a resistance of ${\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\Omega $, and superconductors have no resistance at all (they are nonohmic). Resistance is related to the shape of an object and the material of which it is composed, as will be seen in Resistance and Resistivity. Additional insight is gained by solving $I=\text{V/R}$ for $V,\phantom{\rule{0.25em}{0ex}}$ yielding $$V=\text{IR.}$$This expression for $V$ can be interpreted as the voltage drop across a resistor produced by the flow of current $I$. The phrase $\text{IR}$ drop is often used for this voltage. For instance, the headlight in Example 1 has an $\text{IR}$ drop of 12.0 V. If voltage is measured at various points in a circuit, it will be seen to increase at the voltage source and decrease at the resistor. Voltage is similar to fluid pressure. The voltage source is like a pump, creating a pressure difference, causing current—the flow of charge. The resistor is like a pipe that reduces pressure and limits flow because of its resistance. Conservation of energy has important consequences here. The voltage source supplies energy (causing an electric field and a current), and the resistor converts it to another form (such as thermal energy). In a simple circuit (one with a single simple resistor), the voltage supplied by the source equals the voltage drop across the resistor, since $\text{PE}=q\mathrm{\Delta}V$, and the same $q$ flows through each. Thus the energy supplied by the voltage source and the energy converted by the resistor are equal. (See Figure 2.) Making Connections: Conservation of EnergyIn a simple electrical circuit, the sole resistor converts energy supplied by the source into another form. Conservation of energy is evidenced here by the fact that all of the energy supplied by the source is converted to another form by the resistor alone. We will find that conservation of energy has other important applications in circuits and is a powerful tool in circuit analysis. PhET Explorations: Ohm's LawSee how the equation form of Ohm's law relates to a simple circuit. Adjust the voltage and resistance, and see the current change according to Ohm's law. The sizes of the symbols in the equation change to match the circuit diagram. Section Summary
Conceptual QuestionsExercise 1The $\text{IR}$ drop across a resistor means that there is a change in potential or voltage across the resistor. Is there any change in current as it passes through a resistor? Explain. Exercise 2How is the $\text{IR}$ drop in a resistor similar to the pressure drop in a fluid flowing through a pipe? Problems & ExercisesExercise 1What current flows through the bulb of a 3.00V flashlight when its hot resistance is $3\text{.}\text{60 \Omega}$? Show/Hide Solution Solution0.833 A Exercise 2Calculate the effective resistance of a pocket calculator that has a 1.35V battery and through which 0.200 mA flows. Exercise 3What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 11.0 V to the motor? Show/Hide Solution Solution$7\text{.}\text{33}\times {\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}\Omega $ Exercise 4How many volts are supplied to operate an indicator light on a DVD player that has a resistance of $1\text{40}\phantom{\rule{0.25em}{0ex}}\Omega $, given that 25.0 mA passes through it? Exercise 5(a) Find the voltage drop in an extension cord having a $0\text{.}\text{0600}\Omega $ resistance and through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of $0\text{.}\text{300}\phantom{\rule{0.25em}{0ex}}\Omega $. What is the voltage drop in it when 5.00 A flows? (c) Why is the voltage to whatever appliance is being used reduced by this amount? What is the effect on the appliance? Show/Hide Solution Solution(a) 0.300 V (b) 1.50 V (c) The voltage supplied to whatever appliance is being used is reduced because the total voltage drop from the wall to the final output of the appliance is fixed. Thus, if the voltage drop across the extension cord is large, the voltage drop across the appliance is significantly decreased, so the power output by the appliance can be significantly decreased, reducing the ability of the appliance to work properly. Exercise 6A power transmission line is hung from metal towers with glass insulators having a resistance of $1\text{.}\text{00}\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\Omega .$ What current flows through the insulator if the voltage is 200 kV? (Some highvoltage lines are DC.)
