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?

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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?

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Answer 1 · 2016-01-23T22:25:17

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.

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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?

1 Answer

Explanation:

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A circuit with a resistance of 3 Omega3 Omega has a fuse with a capacity of 4 A4 A. Can a voltage of 2 V2 V be applied to the circuit without blowing the fuse?

1 Answer

Explanation:

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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?