O'Reilly logo

Electrical Applications 2 by David W. Tyler

Stay ahead with the world's most comprehensive technology and business learning platform.

With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.

Start Free Trial

No credit card required

Fuse protection
Energy = 1.152 x IO
x 0.02 = 23 040 joules.
Now suppose that the resistance of the fuse itself is 0.01 Ω and that
of the circuit is 0.04 Ω. Consider the quantity Pt.
Pt = 4800
x 0.02 = 460 800 A
s. Since this quantity does not con-
tain the resistances of the two parts of the circuit, it applies both to
the fuse and to the cable.
Energy used in clearing the fuse = PRt = (Pt) x R = 460 800 x 0.01
= 4608 joules.
Notice that this is the actual energy used in melting the fuse element.
Energy supplied to the cable = PRt =