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Okay from an engineering perspective it is obvious that short circuits happen because of electrons taking the path of least resistance, thus bypassing the resistor. Just like how a short circuit in a simple lightbulb current (by this I mean a current with a single lightbulb and a single battery) causes the lightbulb to go out. However, from a mathematical point of view, this concept confuses me. Since there is technically an electric current going through both the lightbulb and "short" wire the resistance of the resistor-wire system should be an incredibly small number. (given that the "short" wire has a tiny amount of resistance) Given this logic and V=IR, we can conclude that the electric current is magnitudes stronger than before. So based on this, the lightbulb should stay on, but it shouldn't. Am I missing something?

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  • $\begingroup$ Please clarify your question by adding a diagram of the circuit before and after the short circuit is implemented. It would help if you labeled any loops and specified which current you're referencing, as different loops will generally carry different currents. $\endgroup$ – Chemomechanics Feb 26 '18 at 22:48
  • $\begingroup$ @Chemomechanics I would expect I don’t really know how. I mean the circuit I talking about only involves a lightbulb and a battery with wires connecting them. The short circuit is just a wire connected to the circuit in a way so the electrons can bypass the lightbulb $\endgroup$ – Dannnnnnn Feb 26 '18 at 22:51
  • $\begingroup$ Well, from the "mathematical point of view", you're right. In the real world, though, that small resistance short circuit will cause the voltage across the bulb to drop and/or burst or melt the short circuit resistance because so much power is being dissipated by it, etc.. $\endgroup$ – Samuel Weir Feb 26 '18 at 22:55
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You ignore the fact that the lightbulb's resistance is in parallel with that of the short circuit. And both of these are in series with the battery's internal resistance.

Let's make the assumption that the battery can supply the large current through the short without dropping its voltage. Then, the total current supplied by the battery is $$I=\frac{V}{(R_i+\frac{R_sR_l}{R_l +R_s})}$$

In this, $R_i$ is the battery's internal resistance, $R_l$ that of the lightbulb and $R_s$ that of the short. If $R_s$ is very small, then effectively $I=V/R_i$.

As $R_s$ is much smaller than $R_l$, just about all of $I$ will flow through the short, and virtually no current flows through the bulb, because they each get a current that is inversely proportional to their resistance.

Hence, your bulb will be off. Of course, it all depends on the relative values of the various resistors.

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