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A fuse wire repeatedly gets burnt when used with a good heater. It is advised to use a fuse wire of:

  • A) More length
  • B) Less radius
  • C) Less length
  • D) More radius

I think the correct answer is option(B) as Heat ∝ 1/R when the voltage is constant(here it is being used on 220V), as the fuse wire is being burnt we should increase it's resistance(by decreasing the area[or the radius]) as to reduce the amount of heat and prevent it from being burnt.

The textbook from where I got this question from shows that option(D) is the correct answer and I am not able to understand why because increasing the radius will decrease the resistance and increase the heat and just burn the fuse wire even more??

Please tell whether my thinking is correct or incorrect. Also I am just a beginner so it would be helpful if you were to answer in simple terms. Thanks for reading my question!

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2 Answers 2

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Oh no! This is a terrible question. You should always use a fuse of the designed amperage, never lower and especially never higher. A lower amperage fuse will break the circuit during safe operation and a higher amperage fuse will not break the circuit during unsafe operation!

So the only correct answer is “E) use only the designed amperage fuses”

Frankly it is a bit unconscionable to teach students about electricity using examples that implicitly encourage unsafe practices

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  • $\begingroup$ But if the fuse is blowing out then doesn't that mean that a fuse of higher rating should be used, but my doubt with this thing is that if we use a fuse of higher rating for eg- 10A instead of a 5A one then the fuse wire of the 10A Fuse will be thicker than the one of 5A, but here's the confusing part that the thicker wire will have less resistance allowing more current to pass through and hence with the formula H = I^2 * R * t what can we say about heat because on one hand the resistance is less than case of 5A rated wire but the current is now more as compared to the same wire. $\endgroup$
    – Bhavya Jain
    Commented Feb 11, 2023 at 16:40
  • $\begingroup$ So which of the two fuse wires will produce more heat because in one case the resistance is more but the current is less and in the other case the resistance is less but the current is more?? $\endgroup$
    – Bhavya Jain
    Commented Feb 11, 2023 at 16:42
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    $\begingroup$ @BhavyaJain said “But if the fuse is blowing out then doesn't that mean that a fuse of higher rating should be used?” No, that is unsafe. If the designed amperage fuse is blowing out then it means you are trying to operate it in an unsafe manner. E.g. perhaps you are connecting it to a 220 V supply instead of the 110 V it was designed for. Using a bigger fuse may cause the device to catch on fire. $\endgroup$
    – Dale
    Commented Feb 11, 2023 at 17:37
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    $\begingroup$ Oh okay I understand the concept now, I thank you a lot for spending your time to answer out my doubt while informing me about the correct thing as well. Thanks very much! $\endgroup$
    – Bhavya Jain
    Commented Feb 12, 2023 at 10:27
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(a) The resistance of a short piece of wire (the fuse) will be much less than that of the heater. Therefore the current through the fuse will be determined by the heater's resistance; the fuse's resistance (as long as it is much less than the heater's) will have almost no effect on the current, $I$. So the power dissipated by the fuse will be given by $$P=I^2R_{\text{fuse}}$$ in which $I$ can be considered constant. Therefore the lower the resistance of the fuse the less the power dissipated.

(b) A wire of greater radius, $r$, will have a lower resistance. As well as the power dissipated in it (that is internal thermal energy released in it per second) being less, its greater surface area will mean that at a given temperature, more heat per second will escape to the surroundings. Both these factors will lower the equilibrium temperature of the wire, and make it less likely to melt.

At equilibrium, Electrical power dissipated = Heat given out per second, so, assuming that the heat given out per second is proportional to the wire's surface area, $2\pi r l$, and to its excess temperature, $\Delta T$, above its surroundings, $$I^2 \frac{\rho l}{\pi r^2} =k\ 2\pi r l \Delta T$$ in which $\rho$ is the resistivity of the metal of the wire and $k$ is a constant.

(c) Fuses prevent excessive currents, not by burning, but by melting (to fuse means to melt).

(d) If a fuse is chosen just because it doesn't melt when used with a non-faulty heater, it may cause a serious hazard. This is because it may not 'blow' (melt) if the heater does develop a fault. The fuse needs to have the correct rating for the appliance. For example, a heater of power 1.0 kW for use in the UK where the mains voltage is approximately 230 V should take a current of 4.3 A so should be fitted with a 5 A fuse (one that will melt when 5 A or more passes through it).

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  • $\begingroup$ Thanks to you a lot for clarifying my doubt so clearly and taking your time out to answer :) $\endgroup$
    – Bhavya Jain
    Commented Feb 11, 2023 at 19:02
  • $\begingroup$ Thank you for your kind comment. I agree, incidentally, with Dale's remark: the question you were set is indeed a terrible question! $\endgroup$
    – Philip Wood
    Commented Feb 11, 2023 at 19:09

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