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If you short circuit a wire, it will heat up and possibly melt or catch fire. If you add a resistor neither the resistor nor the wire will light on fire (if the resistor is good enough). I know that there would be less current with higher resistance but since there is more resistance the lack of current would be made up for with more heat generated, so why would there not be more heat generated that would light the resistor on fire or melt it when you add the resistor?

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  • $\begingroup$ Remember that power is $I^2 R$ so it is linear in the resistance but quadratic in the current, plus a resistor is made of different material than the wires and has a different heat capacity $\endgroup$ – Triatticus Apr 4 '18 at 22:16
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Let's start with the relation $P=IV$ which says that power is proportional to the current times the voltage. Let's take as an example 2 resistors, one of resistance $R_1=100\Omega$ and one of resistance $R_2=0.001\Omega$. If we keep the same voltage (say $V=10V$) across both resistors, using the relation $I=V/R$, we can clearly see that the power generated over $R_1$ is $P_1=I_1V_1=(10V)^2/(100\Omega)=1W$, which is a factor of $10000$ less than the power generated over $R_2$ which is $P_2=I_2V_2=(10V)^2/(0.01\Omega)=10000W$. The "short circuit" with tiny resistance generates massively more power than the regular resistor. If our voltage source is strong enough to to keep up with the "short circuit" we will get massive (possibly explosive) power generated.

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  • $\begingroup$ So basically reducing resistance increases current and therefore increases heat generation by resistance, more than it reduces heat generation because of the lack of resistance? $\endgroup$ – user180969 Apr 9 '18 at 1:51
  • $\begingroup$ Indeed, that is usually the case. You can see that also in the expression $P=I^2R$ which shows explicitly that if you reduce resistance by the same factor as you increase current, the power actually increases. $\endgroup$ – enumaris Apr 9 '18 at 16:00
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I know that there would be less current with higher resistance but since there is more resistance the lack of current would be made up for with more heat generated

I don't actually follow your reasoning here but stipulate, for simplicity, the voltage source, across which this wire and then wire plus resistor is connected, delivers a constant voltage $V$. Further, let the (small) resistance of the wire be $r$ and the resistance of the resistor be $R$.

The power delivered by the voltage source is just the product of the voltage $V$ and the current $I$ where, by Ohm's law,

$$I = \frac{V}{r}$$

for just the wire and

$$I = \frac{V}{r + R}$$

for the wire plus resistor.

Thus, the power dissipated by just the wire is greater than the power dissipated by the wire plus resistor

$$\frac{V^2}{r} \gt \frac{V^2}{r + R}$$

i.e., the wire alone will get hotter than the wire plus resistor.

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