Homework 10 Solutions

# Homework 10 Solutions - lim(kl9356 – Homework 10 –...

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Unformatted text preview: lim (kl9356) – Homework 10 – Weathers – (22202) 1 This print-out should have 11 questions. Multiple-choice 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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Homework 10 Solutions - lim(kl9356 – Homework 10 –...

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