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Unformatted text preview: resistor. On the way to V 2 we pass through two resistors and E 1 . V 2 = 0iRiR E 1 =E 12 E 2 E 1 P 2 R =E 14 7 ( E 2 E 1 ) =4 7 E 23 7 E 1 =9 V Finding V 1 is now a simple matter of going one resistor further. V 1 = V 2iR =4 7 E 23 7 E 12 7 ( E 2 E 1 ) =6 7 E 21 7 E 1 =11 V Problem 27.35 In ﬁgure 27, R 1 = 2Ω, R 2 = 5Ω, and the battery is ideal. What value of R 3 maximizes the dissipation rate in resistance 3? Figure 27: The circuit for Problem 27.35 The power dissipated by resistor three is given by: P 3 = i 3 V 3 . Hence, our ﬁrst job is to ﬁnd the current ﬂowing through and the potential diﬀerence across resistor three. Lets start with the potential diﬀerence. If we simplify R 2 and R 3 down to an equivalent resistor we ﬁnd that their equivalent resistance is P = R 2 R 3 R 2 + R 3 32...
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This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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