A circuit with a resistance of #6 Omega# has a fuse that melts at #8 A#. Can a voltage of #12 V# be applied to the circuit without blowing the fuse?

1 Answer
Apr 17, 2016

Answer:

No, you would need a voltage of at least #48V#.

Explanation:

We are interested here in the current, because if the current comes above a certain point the circuit will blow. The equation for current is

#I=V/R#.

We know that #R=6Omega#, and that #12V# are being sent through the circuit, so

#I=(12V)/(6Omega)=2A#

A current of #8A# will blow the circuit, so quite clearly a current of #2A# when the voltage is #12V# won't.

When current is at #8A# or above, then voltage must be

#V>=8A*6Omega#
#V>=48V#

for the circuit to break.