qz3sol_8701sm10 - figure. The loop rule applied to the...

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TA: Tomoyuki Nakayama Thursday, June 10, 2010 PHY 2048: Physic 2, Discussion Section 8701 Quiz 3 (Homework Sets #5) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ Three batteries and two resistors are connected as shown below. Battery 1 and 2 are ideal, and they have emfs E 1 = 6.0 V and E 2 = 24.0 V. The third battery of emf E = 12.0 V has internal resistance r = 5.0 . The two external resistors have resistances R 1 = 15.0 and R 2 = 10.0 . a) Choose the directions of currents flowing through E 1 , E , and R 1 . (Just draw arrows in the figure, with labels.) Then write down the loop rules applied to the top loop and bottom loop, and the junction rule applied at point a . We take the positive directions of the currents as shown in the
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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.

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