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Unformatted text preview: lim (kl9356) – Homework 10 – Weathers – (22202) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The power dissipated in the combination of the resistors labeled R in the circuit does not depend on whether the switch is opened or closed. Neglect the internal resistance of the volt age source. . 99 Ω R R E S Determine the value of R . Correct answer: 1 . 40007 Ω. Explanation: r R R E S Let : r = 0 . 99 Ω . If the switch is open, I = V r + R and the power for the resistor R is P = V 2 R ( r + R ) 2 . If the switch is closed, I = V r + R 2 and P = V 2 parenleftbigg R 2 parenrightbigg parenleftbigg r + R 2 parenrightbigg 2 Then V 2 R ( r + R ) 2 = V 2 R 2 parenleftbigg r + R 2 parenrightbigg 2 2 r 2 + 2 r R + R 2 2 = R 2 + 2 r R + r 2 r 2 = 1 2 R 2 R = √ 2 r = √ 2 (0 . 99 Ω) = 1 . 40007 Ω 002 10.0 points Consider the combination of resistors shown in the figure. 4 . 3 Ω 1 . 5 Ω 3 . 7 Ω 9Ω 7 . 9Ω 2 . 3Ω 4 . 6 Ω a b What is the resistance between point a and point b ? Correct answer: 7 . 90952 Ω. Explanation: Let’s redraw the figure i R 1 i R 2 i R 5 i R 4 i R 3 i R 6 i R 7 a b lim (kl9356) – Homework 10 – Weathers – (22202) 2 Let : R 1 = 4 . 3 Ω , R 2 = 1 . 5 Ω , R 3 = 3 . 7 Ω , R 4 = 9 Ω , R 5 = 7 . 9 Ω , R 6 = 2 . 3 Ω , and R 7 = 4 . 6 Ω . Basic Concepts: • Equivalent resistance. • Ohm’s Law. There are two rules for adding up resis tances. If the resistances are in series, then R series = R 1 + R 2 + R 3 + ··· + R n . If the resistances are parallel, then 1 R parallel = 1 R 1 + 1 R 2 + 1 R 3 + ··· + 1 R n . Solution: The key to a complex arrange ments of resistors like this is to split the prob lem up into smaller parts where either all the resistors are in series, or all of them are in parallel. It is easier to visualize the problem if you redraw the circuit each time you add them. i R 1 i R 2 i R 5 i R 4 i R 367 a b Step 1: The three resistors on the right are all in series, so R 367 = R 3 + R 6 + R 7 = (3 . 7 Ω) + (2 . 3 Ω) + (4 . 6 Ω) = 10 . 6 Ω . i R 1 i R 2 i R 3675 i R 4 a b Step 2: R 5 and R 367 are connected paral lel, so R 3675 = parenleftbigg 1 R 5 + 1 R 367 parenrightbigg 1 = R 5 R 367 R 5 + R 367 = (7 . 9 Ω) (10 . 6 Ω) 18 . 5 Ω = 4 . 52649 Ω . i R 1 i R 36752 i R 4 a b Step 3: R 2 and R 3675 are in series, so R 23675 = R 2 + R 3675 = (1 . 5 Ω) + (4 . 52649 Ω) = 6 . 02649 Ω . Step 4: R 23675 and R 4 are parallel, so R 236754 = parenleftbigg 1 R 4 + 1 R 23675 parenrightbigg 1 = R 4 R 23675 R 4 + R 23675 = (9 Ω) (6 . 02649 Ω) 15 . 0265 Ω = 3 . 60952 Ω . i R 1 i R 367524 a b Step 5: Finally, R 1 and R 236754 are in se ries, so the equivalent resistance of the circuit is R eq = R 1 + R 236754 = 4 . 3 Ω + 3 . 60952 Ω = 7 . 90952 Ω . lim (kl9356) – Homework 10 – Weathers – (22202) 3 003 10.0 points 2 Ω 5 Ω 6 Ω 9 Ω 11 Ω 2 Ω 1 Ω 2 Ω 16 V 27 V 31 V...
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This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Magnetism, Power, Work

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