ch27-p041 - 41. (a) The batteries are identical and,...

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20 . 2 ir iR i rR ε −− = ⇒ = + The power dissipated in R is Pi R R == + () . 2 4 2 2 2 2 We find the maximum by setting the derivative with respect to R equal to zero. The derivative is dP dR r R R = + + = + 4 2 16 2 42 2 2 3 2 3 2 3 εε . The derivative vanishes (and P is a maximum) if R = r /2. With r = 0.300 , we have 0.150 R =Ω . (b) We substitute R = r /2 into P = 4 ε 2 R /( r + 2 R ) 2 to obtain 22 2 max 2 4 ( / 2) (12.0 V) 240 W. [ 2( / 2)] 2 2(0.300 ) r P rr r = = +Ω 41. (a) The batteries are identical and, because they are connected in parallel, the
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