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74. (a) From symmetry we see that the current through the top set of batteries (
i
)i
sthesamea
sthe
current through the second set. This implies that the current through the
R
=4
.
0 Ω resistor at
the bottom is
i
R
=2
i
.T
h
u
s
±w
i
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Unformatted text preview: counterclockwise: 3 ( E − ir ) − (2 i ) R = 0 . This yields i = 3 . 0 A. Consequently, i R = 6 . 0 A. (b) The terminal voltage of each battery is E − ir = 8 . 0 V. (c) Using Eq. 2814, we obtain P = i E = (3)(20) = 60 W. (d) Using Eq. 2722, we have P = i 2 r = 36 W....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Current

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