hw06 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 6...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 6. 27.18, 27.68, 27.73; Q29.01.2, Q29.08 27.18. a) From the diagram, it is clear that there are three parallel paths from F to H . The upper and the lower both consist of two 5.0 resistors in series, so the resistance of each is 10 . The middle path has a resistance of 5.0 . Therefore, the equivalent resistance of the three paths is given by: = + + = + + = 10 4 10 2 10 1 10 1 5 1 10 1 10 1 1 FH R , and 4 10 = FH R R FH = 2.5 b) This part is easier if the diagram is re-drawn as shown below. From this sketch, it is apparent that there are two parallel paths from F to G . The lower path has a resistance of 5.0 , but it will take some work to compute the resistance of the upper path. First, calculate the equivalent resistance of the three 5.0 resistors in the upper left of the diagram, and replace them with their equivalent, R’ . To get R’ , note that there are two parallel paths, the upper has two
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/07/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

Page1 / 3

hw06 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 6...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online