40_Ch 18 College Physics ProblemCH18 Direct-Current Circuits

40_Ch 18 College Physics ProblemCH18 Direct-Current...

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138 CHAPTER 18 Apply Kirchhoff’s loop rule around loop abcfa to find ( ) ( ) 3 6.00 6.00 0 II −Ω +Ω = or I 3 I = (1) Apply Kirchhoff’s loop rule around loop abcda to find ( ) ( ) 2 0 = 6.00 6.00 V 10.0 + − or 2 0.600 A 0.600 = (2) Apply Kirchhoff’s loop rule around loop abcea to find ( ) ( ) 1 0 = 6.00 4.50 V 5.00 + or 1 0.900 A 1.20 = (3) Finally, apply Kirchhoff’s junction rule at either point a or point c to obtain (4) Substitute equations (1), (2), and (3) into equation (4) to obtain the current through the ammeter. This gives or and 312 II I I +=+ 0.900 A 1.20 0.600 I += + 3.80 1.50 A I = A 0 .600 I 1.50 A 3.80 0.395 A I == 18.64 In the figure given below, note that all bulbs have the same resistance, . R B ± ±²³´µ¶ B ±²³´µ· ± ± ± ± ± (a) In the series situation, Case 1, the same current
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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