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Unformatted text preview: Simplifying the circuit one more level gives us the current flowing through the battery and R 1 . i 1 = E R 1 + P Stepping back one level gives us the potential difference across P . V P = E  i 1 R 1 = E 1 R 1 R 1 + P = E P R 1 + P In the end we will need to express this in terms of R 3 , but we will keep it in this form for the time being. Since R 2 and R 3 are in parallel, they must both have a potential V P across them. Hence, we can express the current traveling through R 3 as i 3 = V P R 3 = E R 3 P R 1 + P Finally, putting all of this together and expressing P in terms of R 1 and R 2 , we see that the power dissipated by resistor 3 is P 3 = i 3 V 3 = 1 R 3 E P R 1 + P 2 = E 2 R 3 R 2 R 3 R 2 + R 3 R 1 + R 2 R 3 R 2 + R 3 ! 2 = E 2 R 3 R 2 R 3 R 1 R 2 + R 2 R 3 + R 3 R 1 2 = R 2 2 E R 3 ( R 1 R 2 + R 2 R 3 + R 3 R 1 ) 2 Now that we know the power dissipated we simply need to maximize this with respect to R 3 ....
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 Spring '08
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 Physics, Current

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