<< Chapter < Page Chapter >> Page >

Resistances range over many orders of magnitude. Some ceramic insulators, such as those used to support power lines, have resistances of 10 12 Ω size 12{"10" rSup { size 8{"12"} } ` %OMEGA } {} or more. A dry person may have a hand-to-foot resistance of 10 5 Ω size 12{"10" rSup { size 8{5} } ` %OMEGA } {} , whereas the resistance of the human heart is about 10 3 Ω size 12{"10" rSup { size 8{3} } ` %OMEGA } {} . A meter-long piece of large-diameter copper wire may have a resistance of 10 5 Ω size 12{"10" rSup { size 8{ - 5} } ` %OMEGA } {} , and superconductors have no resistance at all (they are non-ohmic). 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 = V/R size 12{I = ital "V/R"} {} for V , size 12{V} {} yielding

V = IR. size 12{V = ital "IR."} {}

This expression for V size 12{V} {} can be interpreted as the voltage drop across a resistor produced by the flow of current I size 12{I} {} . The phrase IR size 12{ ital "IR"} {} drop is often used for this voltage. For instance, the headlight in [link] has an IR size 12{ ital "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 PE = q Δ V size 12{"PE"=qΔV} {} , and the same q size 12{q} {} flows through each. Thus the energy supplied by the voltage source and the energy converted by the resistor are equal. (See [link] .)

The figure shows a simple electric circuit. A battery is connected to a resistor with resistance R, and a voltmeter is connected across the resistor. The direction of current is shown to emerge from the positive terminal of the battery of voltage V, pass through the resistor, and enter the negative terminal of the battery, in a clockwise direction. The voltage V in the circuit equals I R, which equals 18 volts.
The voltage drop across a resistor in a simple circuit equals the voltage output of the battery.

Making connections: conservation of energy

In 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 law

See 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.

Ohm's Law

Section summary

  • A simple circuit is one in which there is a single voltage source and a single resistance.
  • One statement of Ohm’s law gives the relationship between current I , voltage V , and resistance R in a simple circuit to be I = V R . size 12{I = { {V} over {R} } } {}
  • Resistance has units of ohms ( Ω ), related to volts and amperes by 1 Ω = 1 V/A size 12{1 %OMEGA =" 1 V/A"} {} .
  • There is a voltage or IR size 12{ ital "IR"} {} drop across a resistor, caused by the current flowing through it, given by V = IR size 12{V = ital "IR" } {} .

Conceptual questions

The IR size 12{ ital "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.

How is the IR size 12{ ital "IR"} {} drop in a resistor similar to the pressure drop in a fluid flowing through a pipe?

Problems&Exercises

What current flows through the bulb of a 3.00-V flashlight when its hot resistance is 3 . 60 Ω size 12{3 "." "60" %OMEGA } {} ?

0.833 A

Calculate the effective resistance of a pocket calculator that has a 1.35-V battery and through which 0.200 mA flows.

What 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?

7 . 33 × 10 2 Ω size 12{7 "." "33"´"10" rSup { size 8{-2} } %OMEGA } {}

How many volts are supplied to operate an indicator light on a DVD player that has a resistance of 1 40 Ω size 12{1"40 " %OMEGA } {} , given that 25.0 mA passes through it?

(a) Find the voltage drop in an extension cord having a 0 . 0600- Ω size 12{0 "." "0600-" %OMEGA } {} resistance and through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a resistance of 0 . 300 Ω size 12{0 "." "300" %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?

(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.

A power transmission line is hung from metal towers with glass insulators having a resistance of 1 . 00 × 10 9 Ω . size 12{1 "." "00"´"10" rSup { size 8{9} } %OMEGA } {} What current flows through the insulator if the voltage is 200 kV? (Some high-voltage lines are DC.)

Practice Key Terms 5

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics ii. OpenStax CNX. Nov 29, 2012 Download for free at http://legacy.cnx.org/content/col11458/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics ii' conversation and receive update notifications?

Ask