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solutions6_2011

solutions6_2011 - Math 152 Spring 2011 Solutions...

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Math 152, Spring 2011 Solutions Assignment #6 Notes: Each question is worth 5 marks. Due in class: Monday, February 21 for MWF sections; Tuesday, Feb- ruary 22 for TTh sections. Solutions will be posted Tuesday, February 22 in the afternoon. No late assignments will be accepted. (1) Consider the resistor network given in the following picture: t V V\ - /\0\ Use the method of loop currents to find the current though each of the resistors and the change in voltage E across the current source Solution: We are going to solve this problem by applying Kirchhoff’s second law to each fundamental loop. For the loop on the left we get i 1 + 2( i 1 - i 2 ) + 1( i 1 - i 3 ) - 10 = 0; that is, 4 i 1 - 2 i 2 - i 3 = 10 . For the right upper loop we get 3 i 2 + 2( i 2 - i 3 ) + 2( i 2 - i 1 ) = 0; that is, - 2 i 1 + 7 i 2 - 2 i 3 = 0 . For the right lower loop we get ( i 3 - i 1 ) + 2( i 3 - i 2 ) + E = 0; 1

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2 that is, - i 1 - 2 i 2 + 3 i 3 + E = 0 . Finally, the current source makes the current through the right lower loop to be 2 A in the direction shown in the picture. Thus i 3 = - 2 . We can use Gaussian elimination to solve this system. The associated augmented matrix is 4 - 2 - 1 0 | 10 - 2 7 - 2 0 | 0 - 1 - 2 3 1 | 0 0 0 1 0 | - 2 We can permute the rows and change the signs and we obtain the matrix 1 2 - 3 - 1 | 0 4 - 2 - 1 0 | 10 0 0 1 0 | - 2 - 2 7 -
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