Question

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?

Answer 1

The current in a circuit where a voltage of `2` `V` passes through a resistance of `3` `Omega` is given by: `I=V/R = 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` `(V)`, current `I` `(A)` and resistance `R` `(Omega)`:

`V=IR`

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

`I=V/R = 2/3` `A`

The current flowing in the circuit is `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, `2/3` `A`, is considerably less than `4` `A`, so the fuse will not blow.