algebra 5 - Chapter One Linear Systems 76(c This is a...

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76 Chapter One. Linear Systems (c) This is a reasonably complicated network. 9 volt 3 ohm 3 ohm 2 ohm 2 ohm 3 ohm 4 ohm 2 ohm 2 In the first network that we analyzed, with the three resistors in series, we just added to get that they acted together like a single resistor of 10 ohms. We can do a similar thing for parallel circuits. In the second circuit analyzed, 20 volt 12 ohm 8 ohm the electric current through the battery is 25/6 amperes. Thus, the parallel portion is equivalent to a single resistor of 20/ ( 25/6 ) = 4.8 ohms. (a) What is the equivalent resistance if we change the 12 ohm resistor to 5 ohms? (b) What is the equivalent resistance if the two are each 8 ohms? (c) Find the formula for the equivalent resistance if the two resistors in parallel are r 1 ohms and r 2 ohms. 3 A Wheatstone bridge is used to measure resistance. r 1 r 3 r g r 2 r 4 Show that in this circuit if the current flowing through r g is zero then r 4 = r 2 r 3 /r 1 .
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  • Fall '12
  • Durant
  • Algebra, Resistor, ohm, Electrical resistance, Series and parallel circuits, Willow Jay Shelburne

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