20.2iriRirRε−− = ⇒ =+The power dissipated in Ris PiRR==+().242222We find the maximum by setting the derivative with respect to Requal to zero. The derivative is dPdRrRR=+−+=−+42162422232323εε.The derivative vanishes (and Pis a maximum) if R = r/2. With r= 0.300 Ω, we have 0.150 R=Ω. (b) We substitute R = r/2 into P= 4ε 2R/(r+ 2R)2to obtain 222max24( / 2)(12.0 V)240 W.[2( / 2)]22(0.300 )rPrrr==+Ω41. (a) The batteries are identical and, because they are connected in parallel, the
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