Ch27-p084 - must be i = i'’ i 1 = 11.25 A upward(which is the"forward" direction for that battery Thus battery 3 is supplying energy(h

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(c) Since the current through the ε 1 = 20.0 V battery is “forward”, battery 1 is supplying energy. (d) The rate is P 1 = (5.00 A)(20.0 V) = 100 W. (e) Reducing the parallel pair (which are in parallel to the 2 = 10.0 V battery) to a single R' = 1.00 resistor (and thus with current i ' = (10.0 V)/(1.00 ) = 10.0 A downward through it), we see that the current through the battery (by the junction rule) must be i = i ' i 1 = 5.00 A upward (which is the "forward" direction for that battery). Thus, battery 2 is supplying energy. (f) Using Eq. 27-17, we obtain P 2 = 50.0 W. (g) The set of resistors that are in parallel with the 3 = 5 V battery is reduced to R '’' = 0.800 (accounting for the fact that two of those resistors are actually reduced in series, first, before the parallel reduction is made), which has current i ''’ = (5.00 V)/(0.800 ) = 6.25 A downward through it. Thus, the current through the battery (by the junction rule)
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Unformatted text preview: must be i = i ''’ + i 1 = 11.25 A upward (which is the "forward" direction for that battery). Thus, battery 3 is supplying energy. (h) Eq. 27-17 leads to P 3 = 56.3 W. 84. (a) We reduce the parallel pair of resistors (at the bottom of the figure) to a single R’ =1.00 Ω resistor and then reduce it with its series ‘partner’ (at the lower left of the figure) to obtain an equivalence of R” = 2.00 Ω +1.00 Ω =3.00 Ω . It is clear that the current through R” is the i 1 we are solving for. Now, we employ the loop rule, choose a path that includes R” and all the batteries (proceeding clockwise). Thus, assuming i 1 goes leftward through R” , we have 5.00 V + 20.0 V − 10.0 V − i 1 R” = 0 which yields i 1 = 5.00 A. (b) Since i 1 is positive, our assumption regarding its direction (leftward) was correct....
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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.

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