# HW6 - oliver (jmo673) – HW06 – Tsoi – (59090) 1 This...

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Unformatted text preview: oliver (jmo673) – HW06 – Tsoi – (59090) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points E S 1 c d a b 8Ω 2 Ω 6 Ω 9 Ω 4 Ω Find the equivalent resistance between a and b , ( i.e. , R eq = R ab ). Correct answer: 4 . 86765 Ω. Explanation: Let : R 1 = 2 Ω , R 2 = 6 Ω , R 3 = 8 Ω , R 4 = 9 Ω , and R 5 = 4 Ω . Let the current subscripts be identical to the resistance subscripts, i.e. , current I 1 runs through resistor R 1 , etc . E S 1 c d a b I 3 R 3 I 1 R 1 I 2 R 2 I 4 R 4 I 5 R 5 By Kirchhoff’s Laws, summationdisplay V = 0 around a closed loop and summationdisplay I = 0 at a circuit junction. At nodes c and d , we have I 4 = I 1 − I 3 and I 5 = I 2 + I 3 . For loop acda , we have − R 1 I 1 − R 3 I 3 + R 2 I 2 = 0 For loop cbdc , we have − R 4 I 4 + R 5 I 5 + R 3 I 3 = 0 − R 4 ( I 1 − I 3 ) + R 5 ( I 2 + I 3 ) + R 3 I 3 = 0 − R 4 I 1 + R 5 I 2 + ( R 3 + R 4 + R 5 ) I 3 = 0 . For loop acba , we have R 1 I 1 + R 4 I 4 = E R 1 I 1 + R 4 ( I 1 − I 3 ) = E R 1 I 1 + R 4 I 1 − R 4 I 3 = E . The loop equations yield three equations and three unknowns (the three currents I 1 , I 2 , and I 3 ): (2 Ω) I 1 + ( − 6 Ω) I 2 + (8 Ω) I 3 = 0 ( − 9 Ω) I 1 + (4 Ω) I 2 + (21 Ω) I 3 = 0 (11 Ω) I 1 + (0 Ω) I 2 + ( − 9 Ω) I 3 = E . with determinants solutions I 1 = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle − 6 Ω 8 Ω 4 Ω 21 Ω E 0 Ω − 9 Ω vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 2 Ω − 6 Ω 8 Ω − 9 Ω 4 Ω 21 Ω 11 Ω 0 Ω − 9 Ω vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Expanding the numerator on the first col- umn, D 1 = 0 − 0 + E vextendsingle vextendsingle vextendsingle vextendsingle − 6 Ω 8 Ω 4 Ω 21 Ω vextendsingle vextendsingle vextendsingle vextendsingle = E [( − 6 Ω) (21 Ω) − (4 Ω) (8 Ω)] = − 158 E · Ω 2 . Expanding the denominator along the bot- tom row, D = (11 Ω) vextendsingle vextendsingle vextendsingle vextendsingle − 6 Ω 8 Ω − 6 Ω 21 Ω vextendsingle vextendsingle vextendsingle vextendsingle − 0 + ( − 9 Ω) vextendsingle vextendsingle vextendsingle vextendsingle 2 Ω − 6 Ω − 9 Ω 4 Ω vextendsingle vextendsingle vextendsingle vextendsingle = (11 Ω) [( − 6 Ω) (21 Ω) − (4 Ω) (8 Ω)] + ( − 9 Ω) [(2 Ω) (4 Ω) − ( − 9 Ω) ( − 6 Ω)] = − 1324 Ω 3 . oliver (jmo673) – HW06 – Tsoi – (59090) 2 Thus I 1 = − 158 E · Ω 2 − 1324 Ω 3 = 0 . 119335 E / Ω ....
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## This note was uploaded on 10/21/2009 for the course PHY 59090 taught by Professor Tsoi during the Fall '09 term at University of Texas at Austin.

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HW6 - oliver (jmo673) – HW06 – Tsoi – (59090) 1 This...

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