HW13-solutions

# HW13-solutions - karani(syk437 HW13 keto(58945 This...

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karani (syk437) – HW13 – keto – (58945) 1 This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points 9 . 2 V 1 . 8 V 3 . 7 V I 1 1 . 9 Ω 2 . 7 Ω I 2 7 . 3 Ω I 3 8 . 6 Ω ±ind the current I 1 in the 1 . 9 Ω resistor at the bottom oF the circuit between the two power supplies. Correct answer: 1 . 05268 A. Explanation: E 1 E 2 E 3 I 1 R A R B I 2 R C I 3 R D At a junction (Conservation oF Charge) I 1 + I 2 - I 3 = 0 . (1) Kirchho²’s law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchho²’s law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 1 . 9 Ω , R B = 2 . 7 Ω , R C = 7 . 3 Ω , R D = 8 . 6 Ω , E 1 = 9 . 2 V , E 2 = 1 . 8 V , and E 3 = 3 . 7 V . Using determinants, I 1 = v v v v v v 0 1 - 1 E 1 + E 2 0 R D E 3 R C R D v v v v v v v v v v v v 1 1 - 1 R A + R B 0 R D 0 R C R D v v v v v v Expanding along the frst row, the numera- tor is D 1 = v v v v v v 0 1 - 1 E 1 + E 2 0 R D E 3 R C R D v v v v v v = 0 - 1 v v v v E 1 + E 2 R D E 3 R D v v v v + ( - 1) v v v v E 1 + E 2 0 E 3 R C v v v v = - [( E 1 + E 2 ) R D - E 3 R D ] - [ R C ( E 1 + E 2 ) - 0] = R D ( E 3 - E 1 - E 2 ) - R C ( E 1 + E 2 ) = (8 . 6 Ω) (3 . 7 V - 9 . 2 V - 1 . 8 V) - (7 . 3 Ω) (9 . 2 V + 1 . 8 V) = - 143 . 08 V Ω .

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