Practice Midterm2

Practice Midterm2 - thwing(jrt2329 – Practice Exam 02 –...

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Unformatted text preview: thwing (jrt2329) – Practice Exam 02 – Yao – (59110) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 9 . 5 V 2 V 3 . 1 V I 1 1 . 2 Ω 3 Ω I 2 5 . 4 Ω I 3 8 . 4 Ω Find the current I 1 in the 1 . 2 Ω resistor at the bottom of the circuit between the two power supplies. Correct answer: 1 . 28397 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) Kirchhoff’s law on the large outside loop gives ( R A + R B ) I 1 + R D I 3 = E 1 + E 2 . (2) Kirchhoff’s law on the right-hand small loop gives R C I 2 + R D I 3 = E 3 . (3) Let : R A = 1 . 2 Ω , R B = 3 Ω , R C = 5 . 4 Ω , R D = 8 . 4 Ω , E 1 = 9 . 5 V , E 2 = 2 V , and E 3 = 3 . 1 V . Using determinants, I 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1- 1 R A + R B R D R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Expanding along the first row, the numera- tor is D 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 E 1 + E 2 R D E 3 R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 0- 1 vextendsingle vextendsingle vextendsingle vextendsingle E 1 + E 2 R D E 3 R D vextendsingle vextendsingle vextendsingle vextendsingle + (- 1) vextendsingle vextendsingle vextendsingle vextendsingle E 1 + E 2 E 3 R C vextendsingle vextendsingle vextendsingle vextendsingle =- [( 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 . 4 Ω) (3 . 1 V- 9 . 5 V- 2 V)- (5 . 4 Ω) (9 . 5 V + 2 V) =- 132 . 66 V Ω . Expanding along the first column, the de- nominator is D = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1- 1 R A + R B R D R C R D vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 1 vextendsingle vextendsingle vextendsingle vextendsingle R D R C R D vextendsingle vextendsingle vextendsingle vextendsingle- ( R A + R B ) vextendsingle vextendsingle vextendsingle vextendsingle 1- 1 R C R D vextendsingle vextendsingle vextendsingle vextendsingle + 0 = 0- R C R D- ( R A + R B ) ( R D + R C ) = (5 . 4 Ω) (8 . 4 Ω)- (1 . 2 Ω + 3 Ω) (8 . 4 Ω + 5 . 4 Ω) =- 103 . 32 Ω 2 , and I 1 = D 1 D =- 132 . 66 V Ω- 103 . 32 Ω 2 = 1 . 28397 A . thwing (jrt2329) – Practice Exam 02 – Yao – (59110) 2 002 (part 1 of 3) 10.0 points A battery with an internal resistance is con- nected to two resistors in series....
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Practice Midterm2 - thwing(jrt2329 – Practice Exam 02 –...

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