A circuit with a resistance of #6 Omega# has a fuse that melts at #8 A#. Can a voltage of #64 V# be applied to the circuit without blowing the fuse?

1 Answer

Answer:

No,..Check the reason below

Explanation:

The circuit has resistance of #6Omega#. Applying 64 V to the #6Omega# of the circuit would produce a current of

#I = (64 V)/(6Omega) = 10.7 A#

Except that the circuit has a fuse to protect it. And a current of #8A# melts it...

Thus, use the equation #V=I*R#

The formula tells you that the voltage at which the #6Omega# circuit draws #8A# and melts the fuse is therefore: #48V#.

Thus, u get your answer that any voltage greater than #48V# will blow it up.