A circuit with a resistance of 3 Omega has a fuse with a capacity of 4 A. Can a voltage of 2 V be applied to the circuit without blowing the fuse?

Jan 23, 2016

The current in a circuit where a voltage of $2$ $V$ passes through a resistance of $3$ $\Omega$ is given by: $I = \frac{V}{R} = \frac{2}{3}$ $A$. This is well short of the capacity of the fuse, so the fuse will not blow.

Explanation:

Ohm's Law relates voltage $V$ $\left(V\right)$, current $I$ $\left(A\right)$ and resistance $R$ $\left(\Omega\right)$:

$V = I R$

In this case we want to know the current, so we rearrange to make $I$ the subject:

$I = \frac{V}{R} = \frac{2}{3}$ $A$

The current flowing in the circuit is $\frac{2}{3}$ $A$. A fuse is designed to 'blow' (burn out) if the current in the circuit is more than its rated value, in this case $4$ $A$. This is to protect the rest of the circuit from excessive current which generates heat.

The current in this circuit, $\frac{2}{3}$ $A$, is considerably less than $4$ $A$, so the fuse will not blow.