Will a capacitor save a battery from a short circuit?

Question

Do it yourself

I want to make a device that will notify me if a fuse blows with a sound and light signal. The dimensions of the device are quite minimal. There is a fuse, the manufacturer claims that it works from 0.5A and voltage from 2V - and as I understand it, the upper ceiling of current consumption is not limited, well, that is. at the moment of short circuit it can use more than 0.5A. The device, in theory, should be powered by 3V tablets and shine with diodes and squeak if the fuse burns out, but the capacity of the tablets is very small and I'm afraid that at the moment of Kz the fuse will simply suck out the entire battery and there will be nothing to shine and squeak, even if you take, suppose the tablets with a capacity of 0.7A (there are none, but suppose there are).
Question:
1) can capacitors be used to trip the fuse? I understand that he gives the entire charge instantly?
2) is it possible to somehow do it in such a way as to power the capacitor from a finger battery, but the battery has a voltage of 1.5V, but it is necessary for the conder to give, let's say, 0.7A 3V
3) what gives the current faster - a battery or a capacitor?
(in radio electronics I am at the level of the most basic concepts)

Answer 1

read about the principle of operation of the analog comparator.
2 inputs per device, before and after the fuse. when the fuse blows at the input "after", the voltage drops, is detected, at the output - to taste. the same can be done in logic.
if we take modern cmos chips - consumption of microamps, in standby mode, one tablet can be expected to work for years.
similarly, you can do it on any controller if you have a familiar programmer. a six-legged 4*4mm pic10f/pic12f will perform the same functions with the same (or less) consumption but with more flexibility

Answer 2

1. In electronics, you don’t have any concepts at all.
2. The fuse is triggered when a current passes through it above a certain threshold value (in your case it is 0.5A). In this case, the voltage does not matter, because. fuse resistance should tend to zero.
3. To install some kind of "alert device" on a penny flashlight, which, in the most favorable scenario, will be disproportionately more expensive, to put it mildly, strange. About the fact that this device will never work, if only because even 3 "tablets" connected in parallel are not able to deliver a current of 0.5 A, which means that the fuse will not work in any case