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hw6 - homework 06 LIN KEVIN Due 4:00 am c Question 1 chap...

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homework 06 – LIN, KEVIN – Due: Feb 24 2008, 4:00 am 1 Question 1, chap 28, sect 2. part 1 of 2 10 points A battery with an emf of 5 . 3 V and internal resistance of 0 . 36 Ω is connected across a load resistor R . If the current in the circuit is 1 . 99 A, what is the value of R ? Correct answer: 2 . 30332 Ω (tolerance ± 1 %). Explanation: Let : E = 5 . 3 V , I = 1 . 99 A , and R i = 0 . 36 Ω . The electromotive force E is given by E = I ( R + R i ) R = E I R i = 5 . 3 V 1 . 99 A 0 . 36 Ω = 2 . 30332 Ω . Question 2, chap 28, sect 2. part 2 of 2 10 points What power is dissipated in the internal resistance of the battery? Correct answer: 1 . 42564 W (tolerance ± 1 %). Explanation: The power dissipation due to the internal resistance is P = I 2 R i = (1 . 99 A) 2 (0 . 36 Ω) = 1 . 42564 W . Question 3, chap 28, sect 4. part 1 of 2 10 points Four resistors are connected as shown in the figure. 92 V S 1 c d a b 22 Ω 32 Ω 57 Ω 84 Ω Find the resistance between points a and b . Correct answer: 38 . 1928 Ω (tolerance ± 1 %). Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 22 Ω , R 2 = 32 Ω , R 3 = 57 Ω , R 4 = 84 Ω , and E B = 92 V . Ohm’s law is V = I R . A good rule of thumb is to eliminate junc- tions connected by zero resistance. E B c a b R 1 R 2 R 3 R 4 The parallel connection of R 1 and R 2 gives the equivalent resistance 1 R 12 = 1 R 1 + 1 R 2 = R 2 + R 1 R 1 R 2 R 12 = R 1 R 2 R 1 + R 2
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homework 06 – LIN, KEVIN – Due: Feb 24 2008, 4:00 am 2 = (22 Ω) (32 Ω) 22 Ω + 32 Ω = 13 . 037 Ω . E c a b R 12 R 3 R 4 The series connection of R 12 and R 3 gives the equivalent resistance R 123 = R 12 + R 3 = 13 . 037 Ω + 57 Ω = 70 . 037 Ω . E B a b R 123 R 4 The parallel connection of R 123 and R 4 gives the equivalent resistance 1 R ab = 1 R 123 + 1 R 4 = R 4 + R 123 R 123 R 4 R ab = R 123 R 4 R 123 + R 4 = (70 . 037 Ω) (84 Ω) 70 . 037 Ω + 84 Ω = 38 . 1928 Ω . or combining the above steps, the equivalent resistance is R ab = parenleftbigg R 1 R 2 R 1 + R 2 + R 3 parenrightbigg R 4 R 1 R 2 R 1 + R 2 + R 3 + R 4 = bracketleftbigg (22 Ω) (32 Ω) 22 Ω + 32 Ω + 57 Ω bracketrightbigg (84 Ω) (22 Ω) (32 Ω) 22 Ω + 32 Ω + 57 Ω + 84 Ω = 38 . 1928 Ω . Question 4, chap 28, sect 4. part 2 of 2 10 points What is the current through the 57 Ω resis- tor? Correct answer: 1 . 31359 A (tolerance ± 1 %). Explanation: The voltages across R 12 and R 3 , respec- tively, (the voltage between a and b ) is V ab = V 12 + V 3 = I ( R 12 + R 3 ) = I R 123 . where I = I 12 = I 3 is the current through either resistor R 12 or R 3 . Hence, the current through R 3 is I 3 = V ab R 12 + R 3 = V ab R 123 = 92 V 70 . 037 Ω = 1 . 31359 A . Question 5, chap 28, sect 4. part 1 of 1 10 points The following diagram shows a closed elec- trical circuit. The ammeter in the center of the resistive network reads zero amperes. E S 1 A 4 Ω 6 Ω 2 Ω R x Find the electric resistance R X . 1. R X = 5 Ω . 2. R X = 20 Ω .
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homework 06 – LIN, KEVIN – Due: Feb 24 2008, 4:00 am 3 3. R X = 4 Ω . 4. R X = 8 Ω . 5. R X = 6 Ω . 6. R X = 17 Ω . 7. R X = 19 Ω . 8. R X = 10 Ω . 9. R X = 3 Ω . correct 10. R X = 11 Ω . Explanation: E S 1 u x y I u R 1 I u R 2 I R 3 I R x I A 0 A Let : R 1 = 4 Ω , R 2 = 6 Ω , and R 3 = 2 Ω .
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