{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

10 - roofner(bar784 Homework 10 Weathers(22202 This...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
roofner (bar784) – 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. 1 . 52 Ω R R E S Determine the value of R . Correct answer: 2 . 1496 Ω. Explanation: r R R E S Let : r = 1 . 52 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 (1 . 52 Ω) = 2 . 1496 Ω 002 10.0 points Consider the combination of resistors shown in the figure. 3 . 3 Ω 1 Ω 4 Ω 9 . 8 Ω 7 Ω 2 . 4 Ω 3 . 2 Ω a b What is the resistance between point a and point b ? Correct answer: 6 . 63187 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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
roofner (bar784) – Homework 10 – Weathers – (22202) 2 Let : R 1 = 3 . 3 Ω , R 2 = 1 Ω , R 3 = 4 Ω , R 4 = 9 . 8 Ω , R 5 = 7 Ω , R 6 = 2 . 4 Ω , and R 7 = 3 . 2 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 Step1: The three resistors on the right are all in series, so R 367 = R 3 + R 6 + R 7 = (4 Ω) + (2 . 4 Ω) + (3 . 2 Ω) = 9 . 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 Ω) (9 . 6 Ω) 16 . 6 Ω = 4 . 04819 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 Ω) + (4 . 04819 Ω) = 5 . 04819 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 . 8 Ω) (5 . 04819 Ω) 14 . 8482 Ω = 3 . 33187 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 = 3 . 3 Ω + 3 . 33187 Ω = 6 . 63187 Ω .
Image of page 2
roofner (bar784) – Homework 10 – Weathers – (22202) 3 003 10.0 points 3 Ω 4 Ω 6 Ω 9 Ω 10 Ω 2 Ω 1 Ω 2 Ω 17 V 21 V 33 V Find the magnitude of the current in the 17 V cell. Correct answer: 0 . 523282 A. Explanation: i 1 R 1 i 3 R 2 i 1 R 3 i 3 R 4 i 2 R 5 r 1 r 2 r 3 E 1 E 2 E 3 Let : E 1 = 17 V , E 2 = 21 V , E 3 = 33 V , R 1 = 3 Ω , R 2 = 4 Ω , R 3 = 6 Ω , R 4 = 9 Ω , R 5 = 10 Ω , r 1 = 2 Ω , r 2 = 1 Ω , and r 3 = 2 Ω .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern