# 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?

Jun 23, 2018

No,..Check the reason below

#### Explanation:

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

$I = \frac{64 V}{6 \Omega} = 10.7 A$

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

Thus, use the equation $V = I \cdot R$

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

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