RATIO AND PROPORTION

Properties of Circles
Wheels in Motion
Graphing Ratios
Scale Drawings
Putting it Together
Proportions as Ratio
Inverse Proportion
Ohm's Law
Credits
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Ohm's Law
(Teacher Copy)


Activity  14:  An Electrifying Inverse Proportion

Activity Summary

Students build a simple electrical circuit and measure voltage, resistance, and current to discover Ohm's Law.  Students also set up and apply Ohm's Law.

Objective

To understand inverse proportions in terms of Ohm's law.

Introduction:

This lesson continues the discussion of inverse ratios by providing the opportunity to observe an inverse ratio in an electric circuit.  You will work with Ohm's Law and the relationships between voltage, current and resistance.  Although the experiments in this lesson use low voltages which are quite safe, be careful not to try these activities using more powerful power sources.  In other words,  DON'T  TRY THIS AT HOME (unless you have the proper equipment).

Answer Key

Part A

In the first table of Part A, students should list resistances and currents that multiply to get the same voltage.

Students should recognize that, for the same voltage, current and resistance are inversely proportional.

Voltage
Resistance
Current
100
100
1
100
50
2
12
4
3
12
6
2
12
3
4
      Voltage  (volts) = Resistance (ohms) X Current (amps)
 

 Part B

     1.  2.875 amps
     2.  13 amps
     3.  25 2/3 ohms must be added
     4.  $105,000
     5.  5.75 ohms
     6.  22.5 amps

Closing Discussion

A "short circuit" occurs when there is no resistor in a circuit and thus the only resistance is the minimal amount caused by the wires themselves.  Assuming a household circuit is 115 volts, what happens to the current when there is very little resistance in a circuit, say 2 ohms or less?  The current is too great for the wires to handle.

Problems 2 and 6 in activity two could have been solved by finding the product of the amperage and resistance or by using the invert and cross-multiply method for inverse proportions.  Show how the problems could be done using both methods.

For #2, 30 amps X 6.5 ohms = 195 volts.  Then, 195 volts / 15 ohms  is the current desired (13 amps).  Or, 30 amps X 6.5 ohms = I amps X 15 ohms.  Then, cross multiply to get 30 X 6.5 / 15 = 13 amps.

For #6, 15 amps X 90  ohms = 1350 volts.  Then, 1350 volts / 60  ohms is the desired current (22.5 amps).  Or, 15 amps X 90 ohms = I amps X 60 ohms.  Then, cross multiply to get 15 X 90 / 60 = 22.5 amps.



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Area 10 Mathematics and Technology Professional Development Center
Permission is granted to duplicate these materials for classroom use.

Last updated on 1/30/1999
Comments: egalindo@indiana.edu
http://www.indiana.edu/~atmat/units/ratio/ratio_t8.htm