F2A(237) * -F2A(37) * 0F2A(6) * 1F2A(7) * 2F2A(5) * 3F2A(2) * 4F2A(4) * 5F2A(1) * 6F2A(5) * 9F2A(1) * AF2A(12) * BF2A(17) * CF2A(13) * DF2A(37) * EF2A(21) * FF2A(4) * GF2A(5) * HF2A(12) * IF2A(2) * JF2A(5) * MF2A(1) * RF2A(5) * SF2A(2) * UF2A(1) * WF2A(1) So if a fuse is rated for 12V DC and 20 A, this would be equal to 240 watts. If a different voltage is supplied, will this change the current at which the fuse will break? Does the fuse technically 'blow' at 240 watts? It's the watts dissipated in the fuse itself not the watts in the system. Therefore since the fuse has resistance (R) it's the current, which provides that power I F2A*(2112) * F2A-*(46) * F2A1*(82) * F2A2*(19) * F2A3*(5) * F2A4*(1) * F2A5*(24) * F2A6*(1) * F2A8*(2) * F2AA*(6) * F2AF*(2) * F2AP*(17) * F2AS*(21) * F2A_*(4)

F2AL250V Datasheet, PDF - Alldatasheet F2AL250V Datasheet, PDF - Alldatasheet

It is the current that a fuse is able to interrupt without being destroyed or causing an electric arc with unacceptable duration. The capacity of a fuse to operate between the lowest and the Rated Breaking Current code could be: