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: Δ V = E + I 2 r + E 2 = +12.55 V...
View
Full Document
 Summer '08
 Any
 Physics, Work, Resistor, Electrical resistance, Tomoyuki Nakayama, loop yields, upper loop yields

Click to edit the document details