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I am trying to do a simple project in which I am trying to show that when a high voltage battery is connected in a circuit then the fuse will blow and protect other appliances.

In detail, I have four 3 V bulbs which I want to connect parallel to each other, a 0.6 ampere fuse, a fuse holder, and three batteries which are of voltage 1.5 V, 6 V, and 9 V, respectively. How can I blow the fuse?

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  • \$\begingroup\$ You can add the voltage required and use a short to simulate the fuze blowing as it will cause more current to flow than the .6 amperes if the short + the currents of the bulbs is more than .6 amperes. \$\endgroup\$
    – deathismyfriend
    Commented Jul 13, 2014 at 21:46

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To get useful data points it's helpful to look at what the fuse will likely do from a datasheet:

enter image description here

At the standard temperature (usually 25°C) and 1A, a 600mA-rated fuse will typically blow in about 15 seconds. The time will be less if the ambient temperature is higher, and greater if the temperature is cooler than 25°C, but even a +/-10°C ambient change will only change the effective current rating by about -/+5% (see below).

enter image description here

You can measure the current through the light bulbs without the fuse and see if you can get current values that make sense from the perspective of the fuse characteristics. Putting an overvoltage on an incandescent lamp reduces its life by something like the 13th power of the voltage.. so you may burn out the lamps (keep in mind that the fuse element is thicker and will respond more quickly than the fuse) but you may be able to get useful results. Even if the life of a 1000 hour bulb is reduced 10,000:1 it will still operate for a number of minutes.

Complicating factors:

  • Currents in the 1A range are more than what a PP3 9V alkaline battery may be able to supply, and it may significantly reduce the output voltage of even a fresh high-quality AA alkaline cell.

  • The lamp filaments will increase in resistance when they get hotter, so with over-voltage they won't give you as much current as you might expect.

If you want to do a useful experiment, you should monitor the current (for sure) and also the voltage.

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  • \$\begingroup\$ That was something I was wondering about myself: the steady state resistance of the bulb compared to the nominal value. I was expecting it to be higher but thought that since the nature of a bulb is to get hot then perhaps the change would be small. I suspect the design of a bulb, and the resistance characterizing it, is likely given at steady-state since that is the only useful value, right? \$\endgroup\$
    – sherrellbc
    Commented Jul 13, 2014 at 14:17
  • \$\begingroup\$ @sherrellbc Yes, they'll not likely give the data for any other situation. There's some old GE data that can be found on the net that has some typical characteristics for various types of bulbs. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 13, 2014 at 14:19
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You can blow the fuse by allowing more than 0.6A to flow through the circuit.

The voltage doesn't blow the fuse, but the current.

To be able to predict the current for a specific voltage you need to know the resistance of your bulbs, then calculate the total resistance of the bulbs in parallel, then apply Ohm's Law to calculate the current.

You're far more likely to blow the bulbs by applying > 3V to them than to blow the fuse.

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  • \$\begingroup\$ That's interesting. Is there some sort of an E-field limit to the builds? What happens in that case? \$\endgroup\$
    – sherrellbc
    Commented Jul 13, 2014 at 14:09
  • \$\begingroup\$ ...? I understood the words, but not how they were joined together. What are you trying to ask? \$\endgroup\$
    – Majenko
    Commented Jul 13, 2014 at 14:10
  • \$\begingroup\$ I meant "bulbs" rather than "builds". For example a 3V bulb and placing 9V across the filament. You said that a higher voltage was likely to destroy the bulbs before any over-current would. \$\endgroup\$
    – sherrellbc
    Commented Jul 13, 2014 at 14:19
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    \$\begingroup\$ No, the higher voltage would be more likely to blow the bulb before the resulting increase in current would blow the fuse. A bulb is basically a fuse held on the verge of blowing by a vacuum. You are more likely to blow the fuse by adding more bulbs in parallel. \$\endgroup\$
    – Majenko
    Commented Jul 13, 2014 at 14:21
  • \$\begingroup\$ That's an interesting way to describe a bulb. I've never much thought about it but that makes good sense. How exactly does the over-voltage destroy the bulb? My comment was with regards to E = -potentialDifference/distance being too large. \$\endgroup\$
    – sherrellbc
    Commented Jul 13, 2014 at 14:33
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As mentioned, you are more than likely going to damage the bulb before enough current will flow to blow the fuse.

You might want to consider a crowbar circuit.

enter image description here

http://en.wikipedia.org/wiki/Crowbar_(circuit)

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  • \$\begingroup\$ Wonderful circuit and a perfect use of the TRIAC. \$\endgroup\$
    – sherrellbc
    Commented Jul 13, 2014 at 14:29
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You did not state the current of the bulbs, so we can't predict whether they (alone or collectively) will blow the fuse.

Hence the only sure way left to blow the fuse is to connect it directly (no lamps) to one of the batteries (assuming they can deliver enough current, which might no be the case, especially for the 9V battery).

Note that a 0.6A fuse means that 0.6A will not blow the fuse, you will need more current. How much more depends ons on tempertaure, duration, the nature of the fuse, etc.

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What about a high-wattage Zener across the bulb? Simpler and does the same thing (in this case) as the crowbar.

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