P28_068 - i s , i 1 and i ): i s R s i 1 R 1 ir = i 1 R 2 i...

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68. (a) Placing a wire (of resistance r ) with current i running directly from point a to point b in Fig. 28-41 divides the top of the picture into a left and a right triangle. If we label the currents through each resistor with the corresponding subscripts (for instance, i s goes toward the lower right through R s and i x goes toward the upper right through R x ), then the currents must be related as follows: i 0 = i 1 + i s and i 1 = i + i 2 i s + i = i x and i 2 + i x = i 0 where the last relation is not independent of the previous three. The loop equations for the two triangles and also for the bottom loop (containing the battery and point b )leadto i s R s i 1 R 1 ir =0 i 2 R 2 i x R x ir =0 E− i 0 R 0 i s R s i x R x =0 . We incorporate the current relations from above into these loop equations in order to obtain three well-posed “simultaneous” equations, for three unknown currents ( i s ,i
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Unformatted text preview: i s , i 1 and i ): i s R s i 1 R 1 ir = i 1 R 2 i s R x i ( r + R x + R 2 ) = E i s ( R + R s + R x ) i 1 R iR x = The problem statement further species R 1 = R 2 = R and R = 0, which causes our solution for i to simplify signicantly. It becomes i = E ( R s R x ) 2 rR s + 2 R x R s + R s R + 2 rR x + R x R which is equivalent to the result shown in the problem statement. (b) Examining the numerator of our nal result in part (a), we see that the condition for i = 0 is R s = R x . Since R 1 = R 2 = R , this is equivalent to R x = R s R 2 /R 1 , consistent with the result of Problem 43....
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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.

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