Solution to #1 of HUANG fall 07 midterm IIWe can reduce the circuit to a single loop by successively combining parallel and series combinations. We combine R1and R2, which are in series: R7= R1+ R2= 2.8 kΩ+ 2.8 kΩ= 5.6 kΩ. We combine R3and R7, which are in parallel: 1/R8= (1/R3) + (1/R7) = [1/(2.8 kΩ)] + [1/(5.6 kΩ)], which gives R8= 1.87 kΩ. We combine R4and R8, which are in series: R9= R4+ R8= 2.8 kΩ+ 1.87 kΩ= 4.67 kΩ. We combine R5and R9, which are in parallel: 1/R10= (1/R5) + (1/R9) = [1/(2.8 kΩ)] + [1/(4.67 k
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This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at Berkeley.