If there were n+1 parallel resistors, then batterybattery1eq12nVVniRnR++==+. For the relative increase to be 0.0125 ( = 1/80 ), we require in+ 1– inin= in+ 1in– 1 = (1)/(2)1/(1)nn++−+= 180. This leads to the second-degree equation n2+ 2n– 80 = (n+ 10)(n–8) = 0. Clearly the only physically interesting solution to this is n= 8. Thus, there are eight resistors in parallel (as well as that resistor in series shown towards the bottom) in Fig.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.