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Unformatted text preview: homework 06 – SMITH, TAYLOR – 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 12 . 9 V and inter nal resistance of 1 . 02 Ω is connected across a load resistor R . If the current in the circuit is 1 . 98 A, what is the value of R ? Correct answer: 5 . 49515 Ω (tolerance ± 1 %). Explanation: Let : E = 12 . 9 V , I = 1 . 98 A , and R i = 1 . 02 Ω . The electromotive force E is given by E = I ( R + R i ) R = E I − R i = 12 . 9 V 1 . 98 A − 1 . 02 Ω = 5 . 49515 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: 3 . 99881 W (tolerance ± 1 %). Explanation: The power dissipation due to the internal resistance is P = I 2 R i = (1 . 98 A) 2 (1 . 02 Ω) = 3 . 99881 W . Question 3, chap 28, sect 4. part 1 of 2 10 points Four resistors are connected as shown in the figure. 91 V S 1 c d a b 2 1 Ω 40Ω 5 1 Ω 8 3 Ω Find the resistance between points a and b . Correct answer: 36 . 3804 Ω (tolerance ± 1 %). Explanation: E B S 1 c d a b R 1 R 2 R 3 R 4 Let : R 1 = 21 Ω , R 2 = 40 Ω , R 3 = 51 Ω , R 4 = 83 Ω , and E B = 91 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 homework 06 – SMITH, TAYLOR – Due: Feb 24 2008, 4:00 am 2 = (21 Ω) (40 Ω) 21 Ω + 40 Ω = 13 . 7705 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 . 7705 Ω + 51 Ω = 64 . 7705 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 = (64 . 7705 Ω) (83 Ω) 64 . 7705 Ω + 83 Ω = 36 . 3804 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 (21 Ω) (40 Ω) 21 Ω + 40 Ω + 51 Ω bracketrightbigg (83 Ω) (21 Ω) (40 Ω) 21 Ω + 40 Ω + 51 Ω + 83 Ω = 36 . 3804 Ω . Question 4, chap 28, sect 4. part 2 of 2 10 points What is the current through the 51 Ω resis tor? Correct answer: 1 . 40496 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 = 91 V 64 . 7705 Ω = 1 . 40496 A ....
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This homework help was uploaded on 04/10/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Current, Resistance, Work

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