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Taking the simplest circuit: battery and resistors.

If I connect lots of resistors in parallel, wouldn't that increase the current to an extent that it would be technically be very similar to shorting the circuit?

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Yes. The equivalent resistance for $n$ equal resistors of value $R$ connected in parallel is $R(n)=\frac{R}{n}$. As $n \to \infty$ then $R(n) \to 0$, provided that $R$ is finite.

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If you have $N$ resistors in parallel, all of which have a resistance of $R$, the total equivalent resistance will be $$ \left( \frac 1 R + \cdots + \frac 1 R \right)^{-1} = \left( \frac N R \right)^{-1} = \frac R N \;. $$ So yes, if you take a sufficiently large amount of (identical) resistors in parallel, it's the same as not having any resistors at all.

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No, a short circuit needn't be an overload. There are circumstances (like in current transformers) where no load, however small in resistance, is an overload. There are ideal signal sources that are voltage sources (i.e. low impedance), and sources that are current sources (i.e. high output impedance), and sources that are of known impedance (50 ohm RF wiring, and 110 ohm digital differential wiring, depend on that).

When something is an overload, it means that it is outside the specified intended load limits. Sometimes, that means a HIGH resistance is an overload (and a current source will overvoltage and damage the insulation). Low resistance can be an overload if the source is such low impedance that destructive currents flow. Even that, though, isn't an overload if the intention is an explosive squib.

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Yes it is. Voltage taking another path is essentially the same as not enough voltage.

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protected by Qmechanic Jan 29 '17 at 6:21

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