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In my circuit,I have a resistive load(Lamp).(Rlamp=1 ohm).At the ends of the lamp is voltage V(t) in the picture.In the picture, there is the first 12 seconds of the signal.But you may think that periodically the signal goes on for hours.The signal is already periodic.

My V(t) signal progresses for hours as 0.75 seconds on,5 seconds off.(Let assume V(t)=I_lamp(t),R_lamp=1 ohm)

The general formula for fuse selection at room temperature is: enter image description here

In summary,what should I write instead of "Normal Operating Current" in formula for calculation?(rms value or avarage value or another think)What is the my "Normal Operating Current"? enter image description here

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    \$\begingroup\$ Why are you fitting a fuse; what does it protect? \$\endgroup\$
    – Andy aka
    Jan 7 at 12:54
  • \$\begingroup\$ I'd say \$\frac{\text{Normal (ON) Operating Current}}{0.75}\$ since the fuse must support the typical operating current while on, with a margin. But this can be complicated if the load is reactive (not just a purely-resistive lamp.) \$\endgroup\$
    – rdtsc
    Jan 7 at 13:08
  • \$\begingroup\$ The fuse protects the wire, not the load. \$\endgroup\$
    – Kartman
    Jan 7 at 13:35
  • \$\begingroup\$ The main problem here is not the function of the fuse. I'm looking for how to find the operational operating current. \$\endgroup\$ Jan 8 at 9:38
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I think choosing the fuse, you should account current which flows during your 0.75 seconds ON period. Effectively assuming your sine wave is there all the time. So, I think Normal Operating Current = Vrms/R. Again during this 0.75 seconds.

RMS or Peak to Peak voltage was discussed thoroughly here: Does the rating of an AC fuse point to the amplitude or rms value of the current?

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