Hello esteemed members. I appreciate that my question will strike some as being somewhat childishly simple, but it genuine perplexes me! I have repaired several laptops before by removing a faulty capacitor that was short-circuiting the motherboard, whereupon the laptop would then work. For me, it begs the question that, if the current can bypass where the capcitor was when it has been removed, why didn't it just bypass it when it was present and faulty,? I can't get my head round the premise that if the other side of the capacitor is connected to ground, then why doesn't the current simply bypass it, in the way that it obviously bypasses the space where it used to be?
Why doesn't current just bypass faulty capacitors.
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Answer selected by Spudgun at 04-01-2025, 05:30 PM.
When there are multiple paths of a circuit, the amount of current that flows is inversely proportional to the amount of resistance on a given path. This is sometimes said as "current takes the path of least resistance". Not strictly true, it'd be more accurate to say "the path with the most current has the least resistance". When one of the paths has a "short circuit" it means the resistance on that path is nearly zero, so nearly all the current will go through that path. By removing the shorted capacitor, you open that leg (and make the resistance essentially "infinite", so no current can flow on that path). Think of the shorted capacitor as a closed switch. Removing it opens that switch.
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When there are multiple paths of a circuit, the amount of current that flows is inversely proportional to the amount of resistance on a given path. This is sometimes said as "current takes the path of least resistance". Not strictly true, it'd be more accurate to say "the path with the most current has the least resistance". When one of the paths has a "short circuit" it means the resistance on that path is nearly zero, so nearly all the current will go through that path. By removing the shorted capacitor, you open that leg (and make the resistance essentially "infinite", so no current can flow on that path). Think of the shorted capacitor as a closed switch. Removing it opens that switch.
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When there are multiple paths of a circuit, the amount of current that flows is inversely proportional to the amount of resistance on a given path. This is sometimes said as "current takes the path of least resistance". Not strictly true, it'd be more accurate to say "the path with the most current has the least resistance". When one of the paths has a "short circuit" it means the resistance on that path is nearly zero, so nearly all the current will go through that path. By removing the shorted capacitor, you open that leg (and make the resistance essentially "infinite", so no current can flow on that path). Think of the shorted capacitor as a closed switch. Removing it opens that switch.Comment
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