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Electrical Applications 2 by David W. Tyler

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Fuse protection
135
Energy = 1.152 x IO
6
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
2
x 0.02 = 460 800 A
2
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 =