A circuit with a resistance of 6 Omega has a fuse that melts at 8 A. Can a voltage of 49 V be applied to the circuit without blowing the fuse?

$\text{Current"= "voltage"/"Resistance}$
i.e. $I = \frac{V}{R}$
so current is $\frac{49}{6}$= $8.16$ $A$
As current in circuit is $8.16$ $A$ fuse will blow out.