09exam2 - Husain, Zeena – Exam 2 – Due: Mar 9 2004,...

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Unformatted text preview: Husain, Zeena – Exam 2 – Due: Mar 9 2004, 11:00 pm – Inst: Sonia Paban 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points Given: Circuit shown in the figure below. Assume: The batteries have zero internal resistance. Assume: The currents are flowing in the direction indicated by the arrows. A negative current denotes flow opposite to the direction of the arrow. 7 . 9 Ω 23 . 7 Ω 6 . 4 V 16 . 2 V I I I Find the current through the 7 . 9 Ω resistor and the 6 . 4 V battery at the top of the circuit. Correct answer: 2 . 86076 A. Explanation: Gven : R 1 = 7 . 9 Ω , R 2 = 23 . 7 Ω , E 1 = 6 . 4 V , and E 2 = 16 . 2 V . R 1 R 2 E 1 E 2 I 1 I 2 I 3 At nodes, we have I 1- I 2- I 3 = 0 . (1) Pay attention to the sign of the battery and the direction of the current in the figure. Us- ing the lower circuit in the figure, we get E 2 + I 2 R 2 = 0 (2) so I 2 =-E R 2 =- 16 . 2 V 23 . 7 Ω =- . 683544 A . Then, for the upper circuit E 1- I 2 R 2- I 1 R 1 = 0 . (3) E 1 + E 2- I 1 R 1 = 0 . I 1 = E 1 + E 2 R 1 = 6 . 4 V + 16 . 2 V 7 . 9 Ω = 2 . 86076 A . Alternate Solution: Using the outside loop-E 1-E 2 + I 1 R 1 = 0 (4) I 1 = E 1 + E 2 R 1 . 002 (part 2 of 2) 10 points Find the current through the 16 . 2 V battery at the bottom of the circuit. Correct answer: 3 . 5443 A. Explanation: From Eg.(1), and using the loop equation E 2- I 2 R 2 = 0 to solve for I 2 , I 3 = I 1- I 2 = 2 . 86076 A- (- . 683544 A) = 3 . 5443 A . Alternate Solution: We have three equa- tions and three unknowns, the three currents. Rewriting the node equation and the two loop equations, we have I 1- I 2- I 3 = 0 (1) R 2 I 2 =-E 2 (2) R 1 I 1 + R 2 I 2 = E 1 (3) Husain, Zeena – Exam 2 – Due: Mar 9 2004, 11:00 pm – Inst: Sonia Paban 2 Using determinants, I 1 = fl fl fl fl fl fl- 1- 1-E 2 R 2 E 1 R 2 fl fl fl fl fl fl fl fl fl fl fl fl 1- 1- 1 R 2 R 1 R 2 fl fl fl fl fl fl Using the first row, the numerator is fl fl fl fl fl fl- 1- 1-E 2 R 2 E 1 R 2 fl fl fl fl fl fl = 0- (- 1) fl fl fl fl-E 2 E 1 fl fl fl fl + (- 1) fl fl fl fl-E 2 R 2 E 1 R 2 fl fl fl fl = 0 + 1(0- 0) + (- 1)(-E 2 R 2-E 1 R 2 ) = E 1 R 2 + E 2 R 2 = R 2 ( E 1 + E 2 ) . Using the first column, the denominator is fl fl fl fl fl fl 1- 1- 1 R 2 R 1 R 2 fl fl fl fl fl fl = (1) fl fl fl fl R 2 R 2 fl fl fl fl- 0 + R 1 fl fl fl fl- 1- 1 R 2 fl fl fl fl = (1)(0)- 0 + R 1 (0 + R 2 ) = R 1 R 2 . Thus I 1 = R 2 ( E 1 + E 2 ) R 1 R 2 = E 1 + E 2 R 1 = 6 . 4 V + 16 . 2 V 7 . 9 Ω = 2 . 86076 A . From Eq. (2) I 2 =- E 2 R 2 =- 16 . 2 V 23 . 7 Ω =- . 683544 A . From Eq.(1) I 3 = I 1- I 2 = 2 . 86076 A- (- . 683544 A) = 3 . 5443 A ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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09exam2 - Husain, Zeena – Exam 2 – Due: Mar 9 2004,...

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