# A circuit with a resistance of 6 Omega has a fuse that melts at 8 A. Can a voltage of 42 V be applied to the circuit without blowing the fuse?

Jan 14, 2016

The current flowing in a circuit can be calculated using Ohm's Law: $V = I R$ or $I = \frac{V}{R}$. In this case the current is $7 A$, so the fuse is safe.

#### Explanation:

Ohm's Law relates the current, $I \left(A\right)$, electrical potential difference (voltage), $V \left(V\right)$ and resistance, $R \left(\Omega\right)$ in a circuit.

It is usually stated and remembered as $V = I R$.

Rearranging, $I = \frac{V}{R} = \frac{42}{6} = 7 A$

Since the fuse will blow (melt) if the current in the circuit is $8 A$ and the current is less than that, the fuse will not blow.