17_Chapter 18 HomeworkCH18 Direct-Current Circuits

17_Chapter 18 HomeworkCH18 Direct-Current Circuits - 115...

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Direct-Current Circuits 115 18.21 First simplify the circuit by combining the series resistors. Then, apply Kirchhoff’s junction rule at point a to find Next, we apply Kirchhoff’s loop rule to the rightmost loop to obtain 12 2.00 A II += ( ) ( ) 12.0 0 () 8.00 V 6.00 −+ = or ( ) 1 12.0 2.00 I 1 I = 8.00 V 6.00 ( A 0 This yields Finally, apply Kirchhoff’s loop rule to the leftmost loop to obtain 1 1.78 A I = )( ) ( ) 11 0 A 6.00 I 4.00 2.0 8.00 V 0 ε +− ( ) ( + = ) or 1 4.00 2.00 A 6.00 1.78 =+ A 8.00 V 10.7 V = 18.22 From Kirchhoff’s point rule, note that 1 I 2 = in the circuit shown at the right. Going counterclockwise around the upper pane of the circuit, Kirchhoff’s loop rule gives ( )( ) 2 0 I I −= 22 50 30 90 −Ω−Ω+Ω or 2 170 90  =   (1) Now, going counterclockwise around the outer perimeter
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