6 - homework 06 – HOFFMA, DAVID – Due: Feb 25 2008,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 = . 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 . 1 Ω 3 . 1 Ω I 2 6 . 5 Ω I 3 9 . 7 Ω 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 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...
View Full Document

This note was uploaded on 03/19/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 11

6 - homework 06 – HOFFMA, DAVID – Due: Feb 25 2008,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online