Fa07-MT2-Huang-Soln-correction

Fa07-MT2-Huang-Soln-correction - Solution to #1 of HUANG...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution to #1 of HUANG fall 07 midterm II We can reduce the circuit to a single loop by successively combining parallel and series combinations. We combine R 1 and R 2 , which are in series: R 7 = R 1 + R 2 = 2.8 k Ω + 2.8 k Ω = 5.6 k Ω . We combine R 3 and R 7 , which are in parallel: 1/ R 8 = (1/ R 3 ) + (1/ R 7 ) = [1/(2.8 k Ω )] + [1/(5.6 k Ω )], which gives R 8 = 1.87 k Ω . We combine R 4 and R 8 , which are in series: R 9 = R 4 + R 8 = 2.8 k Ω + 1.87 k Ω = 4.67 k Ω . We combine R 5 and R 9 , which are in parallel: 1/ R 10 = (1/ R 5 ) + (1/ R 9 ) = [1/(2.8 k Ω )] + [1/(4.67 k
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online