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In this animation, you can close and open switches to determine the resistance of each lightbulb. The current and voltage readings show the current through the battery and the voltage across the battery (voltage is given in volts and current is given in amperes). In order to solve such problems, you can always use Kirchhoff loop equations. If it is possible, a faster way to work problems is in terms of the effective resistance of the network of resistors. You consider the resistors that are in parallel and resistors that are in series. It is, however, worth looking at the circuit in a bit more detail first, before diving into equations, to see if there is a way to understand the problem conceptually and to solve it faster. Restart.
Notice that when both switches are closed, bulbs A and B are dimmer than bulb C. This should not be surprising because the current through bulb C is the sum of the currents through A and B. Open one of the switches and leave the other one closed. Now bulb C is in series with one of the bulbs (which one?). Notice that the total current from the battery is less, but that either bulb A or bulb B is brighter than it was before.
Go back to the case in which both switches are closed and notice that bulbs A and B look to be the same brightness. If the brightness were exactly the same, they would have the same resistance.
Now for some math.
Notice that in this problem, trying to understand conceptually what was happening helped to guide the problem-solving process. Although the Kirchhoff loop rules will work, they are not necessarily the easiest way to solve a problem.
Exploration authored by Anne J. Cox.
Script authored by Wolfgang Christian.
Instructor's Resource CD Edition: Do not post or distribute.
The complete version of Physlet Physics is available as a text with CD; Physlet Physics can be bundled with this Prentice Hall textbook.
© 2004 by Prentice-Hall, Inc. A Pearson Company