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?

Question

1391 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-17T11:46:45

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

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.

A circuit with a resistance of 6 Omega6 Omega has a fuse that melts at 8 A8 A. Can a voltage of 12 V12 V be applied to the circuit without blowing the fuse?