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) 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: 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) 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?