# 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?

Mar 11, 2018

#### Answer:

Yes, it can be applied.

#### Explanation:

First, we need to find the voltage at which the fuse melts.

Thus, we use Ohm's law here, which states that

$V = I R$

The fuse melts at $8 \setminus \text{A}$ with a resistance of $6 \setminus \Omega$, so we must plug in those values into the equation.

We get:

$V = 8 \setminus \text{A} \cdot 6 \setminus \Omega$

$= 48 \setminus \text{V}$

So, the fuse will melt at $48 \setminus \text{V}$ or more.

Since $42 \setminus \text{V"<48 \ "V}$, then the fuse will not melt when a voltage of $42$ volts is passed through it.

So, a voltage of $42 \setminus \text{V}$ can be applied on the circuit without blowing the fuse.