WHomework set 7_solns_2011

WHomework set 7_solns_2011 - Homework set # 7 Problems:...

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Homework set # 7 Problems: 18.14, 15, 48 Solutions: 18.14 Going counterclockwise around the upper loop, applying Kirchhoff’s loop rule, gives      1 15.0 V 7.00 5.00 2.00 A 0 I  or 1 15.0 V 10.0 V 0.714 A 7.00 I  From Kirchhoff’s junction rule, 12 2.00 A 0 II  so 21 2.00 A 2.00 A 0.714 A 1.29 A  Going around the lower loop in a clockwise direction gives      2 2.00 5.00 2.00 A 0 I E or       2.00 1.29 A 5.00 2.00 A 12.6 V   E
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18.15 We name the currents 12 3 , , and II I as shown. Using Kirchhoff’s loop rule on the rightmost loop gives    3 2 12.0 V- 1.00+3.00 5.00 1.00 4.00 V 0 I I  or     32 2.00 3.00 4.00 V  (1) Applying the loop rule to the leftmost loop yields     21 4.00 V+ 1.00+5.00 8.00 0  or     4.00 3.00 2.00 V  (2) From Kirchhoff’s junction rule, 123 += III (3) Solving equations (1), (2) and (3) simultaneously gives 3 =0.846 A, =0.462 A, and 1.31 A I All currents are in the directions indicated by the arrows in the circuit diagram.
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18.48
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WHomework set 7_solns_2011 - Homework set # 7 Problems:...

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