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Unformatted text preview: figure. The loop rule applied to the upper loop yields +E + E 1 + I 1 R 2 E 2 rI 2 = 0 10I 1 5I 2 = 6 (1) The loop rule applied to the lower loop yields E + I 3 R 1  I 2 r = 0 5I 2 15I 3 = 12 (2) The junction rule at point a is I 1 + I 2 + I 3 = 0 (3) b) What is the current through internal resistor r ? We solve the three equations in part a) simultaneously for I 2 . First we remove I 3 from our equations. Solving (3) for I 3 and plugging into (2), we have 5I 2 15(I 1 I 2 ) = 6 15I 1 + 20I 2 = 6 (4) We multiply (1) by 3 and subtract it from 2 times (4) to obtain 55I 2 = 6 I 2 = +0.109 A c) What is the potential difference from point a to point b , namely V b V a (The sign is important, so do not flip them!) We start from point a and count up the change in potential across each element all the way to point b: V = E + I 2 r + E 2 = +12.55 V...
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This note was uploaded on 12/13/2011 for the course PHY 2049 taught by Professor Any during the Summer '08 term at University of Florida.
 Summer '08
 Any
 Physics, Work

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