# Component failure analysis

Discussion in 'General Electronics Chat' started by p75213, May 24, 2011.

1. ### p75213 Thread Starter Active Member

May 24, 2011
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0
I have been going through the "Series Parallel Combination Circuits" - "Component Failure Analysis" chapter.
http://www.allaboutcircuits.com/vol_1/chpt_7/4.html
Can somebody explain how R1//R2 resistance can be less when R2 is shorted. As this is a parallel circuit and the total // resistance diminishes with each resistor. I would have thought the total // resistance would increase if R2 is shorted. ie. R1R2=(1/R1+1/R2)^-1 would become R1R2=(1/R1)^-1

2. ### KJ6EAD Senior Member

Apr 30, 2011
1,426
364
When R2 is shorted, it's resistance approaches 0. When you invert this value, you get a divide by 0 error, but of course an infinitesimally small value divided into any larger value approaches ∞. When you add ∞ to the other inverted resistor value you still have ∞ as the sum. Now when you invert this you get an infinitesimally small value that for practical purposes we can call 0.

The value of R1//R2 when R2 is 0 is 0. 0 < any other resistance value.

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