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6 - homework 06 HOFFMA DAVID Due 4:00 am = 23 9 Question 1...

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homework 06 – HOFFMA, DAVID – Due: Feb 25 2008, 4:00 am 1 Question 1, chap 28, sect 4. part 1 of 2 10 points A 23 V battery has an internal resistance r . 1 A 23 V r 10 Ω 90 Ω internal resistance What is the value of r ? Correct answer: 14 Ω (tolerance ± 1 %). Explanation: I 1 I 2 I 3 E r R 2 R 3 internal resistance Let : E = 23 V , R 2 = 10 Ω , R 3 = 90 Ω , and I 1 = 1 A . Since R 2 and R 3 are connected parallel, their equivalent resistance R 23 is 1 R 23 = 1 R 2 + 1 R 3 = R 3 + R 2 R 2 R 3 R 23 = R 2 R 3 R 2 + R 3 = (10 Ω) (90 Ω) 10 Ω + 90 Ω = 9 Ω . Using Ohm’s law, we have E = I 1 r + I 1 R 23 r = E − I 1 R 23 I 1 = 23 V (1 A) (9 Ω) 1 A = 23 Ω 9 Ω = 14 Ω . Question 2, chap 28, sect 4. part 2 of 2 10 points Determine the magnitude of the current through the 90 Ω resistor on the right-hand side of the circuit. Correct answer: 0 . 1 A (tolerance ± 1 %). Explanation: The potential drop across the 90 Ω resistor on the right-hand side of the circuit is E 3 = E − I 1 r = 23 V (1 A) (14 Ω) = 23 V 14 V = 9 V , so the current through the resistor is I 3 = E 3 r 3 = 9 V 90 Ω = 1 10 A = 0 . 1 A , and I 2 = I 1 I 3 = 1 A 1 10 A = 9 10 A = 0 . 9 A . Question 3, chap 28, sect 4. part 1 of 3 10 points 9 . 9 V 1 . 7 V 3 . 1 V I 1 0 . 1 Ω 3 . 1 Ω I 2 6 . 5 Ω I 3 9 . 7 Ω
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homework 06 – HOFFMA, DAVID – Due: Feb 25 2008, 4:00 am 2 Find the current I 1 in the 0 . 1 Ω resistor at the bottom of the circuit between the two power supplies. Correct answer: 1 . 37392 A (tolerance ± 1 %). 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 = 0 . 1 Ω , R B = 3 . 1 Ω , R C = 6 . 5 Ω , R D = 9 . 7 Ω , E 1 = 9 . 9 V , E 2 = 1 . 7 V , and E 3 = 3 . 1 V . Using determinants, I 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 1 1 E 1 + E 2 0 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 0 R D 0 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 0 1 1 E 1 + E 2 0 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 0 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 ) = (9 . 7 Ω) (3 . 1 V 9 . 9 V 1 . 7 V) (6 . 5 Ω) (9 . 9 V + 1 . 7 V) = 157 . 85 V Ω .
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