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PHY303L Solutions%206

# PHY303L Solutions%206 - homework 06 FAUSAK TAYLOR Due 4:00...

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homework 06 – FAUSAK, TAYLOR – Due: Feb 25 2008, 4:00 am 1 Question 1, chap 28, sect 4. part 1 of 2 10 points A 28 V battery has an internal resistance r . 1 A 28 V r 42 Ω 56 Ω internal resistance What is the value of r ? Correct answer: 4 Ω (tolerance ± 1 %). Explanation: I 1 I 2 I 3 E r R 2 R 3 internal resistance Let : E = 28 V , R 2 = 42 Ω , R 3 = 56 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 = (42 Ω) (56 Ω) 42 Ω + 56 Ω = 24 Ω . Using Ohm’s law, we have E = I 1 r + I 1 R 23 r = E − I 1 R 23 I 1 = 28 V (1 A) (24 Ω) 1 A = 28 Ω 24 Ω = 4 Ω . Question 2, chap 28, sect 4. part 2 of 2 10 points Determine the magnitude of the current through the 56 Ω resistor on the right-hand side of the circuit. Correct answer: 0 . 428571 A (tolerance ± 1 %). Explanation: The potential drop across the 56 Ω resistor on the right-hand side of the circuit is E 3 = E − I 1 r = 28 V (1 A) (4 Ω) = 28 V 4 V = 24 V , so the current through the resistor is I 3 = E 3 r 3 = 24 V 56 Ω = 3 7 A = 0 . 428571 A , and I 2 = I 1 I 3 = 1 A 3 7 A = 4 7 A = 0 . 571429 A . Question 3, chap 28, sect 4. part 1 of 3 10 points 7 . 2 V 1 V 4 . 7 V I 1 1 . 3 Ω 2 Ω I 2 7 . 7 Ω I 3 9 . 7 Ω

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homework 06 – FAUSAK, TAYLOR – Due: Feb 25 2008, 4:00 am 2 Find the current I 1 in the 1 . 3 Ω resistor at the bottom of the circuit between the two power supplies. Correct answer: 0 . 734918 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 = 1 . 3 Ω , R B = 2 Ω , R C = 7 . 7 Ω , R D = 9 . 7 Ω , E 1 = 7 . 2 V , E 2 = 1 V , and E 3 = 4 . 7 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) (4 . 7 V 7 . 2 V 1 V) (7 . 7 Ω) (7 . 2 V + 1 V) = 97 . 09 V Ω .
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