hw6 - homework 06 LIN, KEVIN Due: Feb 24 2008, 4:00 am 1...

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Unformatted text preview: 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 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 (tolerance 1 %). Explanation: Let : E = 5 . 3 V , I = 1 . 99 A , and R i = 0 . 36 . The electromotive force E is given by E = I ( R + R i ) R = E I R i = 5 . 3 V 1 . 99 A . 36 = 2 . 30332 . 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 ) = 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 2 2 32 5 7 8 4 Find the resistance between points a and b . Correct answer: 38 . 1928 (tolerance 1 %). Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 22 , R 2 = 32 , R 3 = 57 , R 4 = 84 , and E B = 92 V . Ohms 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 homework 06 LIN, KEVIN Due: Feb 24 2008, 4:00 am 2 = (22 ) (32 ) 22 + 32 = 13 . 037 . 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 + 57 = 70 . 037 . 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 ) (84 ) 70 . 037 + 84 = 38 . 1928 . 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 ) (32 ) 22 + 32 + 57 bracketrightbigg (84 ) (22 ) (32 ) 22 + 32 + 57 + 84 = 38 . 1928 . Question 4, chap 28, sect 4. part 2 of 2 10 points What is the current through the 57 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 = 1 . 31359 A ....
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This note was uploaded on 03/03/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw6 - homework 06 LIN, KEVIN Due: Feb 24 2008, 4:00 am 1...

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