A circuit with a resistance of $6 Omega$ has a fuse that melts at $8 A$ . Can a voltage of $64 V$ be applied to the circuit without blowing the fuse?

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Answer 1 · 2018-06-23T16:46:36

No,..Check the reason below

The circuit has resistance of $6Omega$ . Applying 64 V to the $6Omega$ of the circuit would produce a current of

$I = (64 V)/(6Omega) = 10.7 A$

Except that the circuit has a fuse to protect it. And a current of $8A$ melts it...

Thus, use the equation $V=I*R$

The formula tells you that the voltage at which the $6Omega$ circuit draws $8A$ and melts the fuse is therefore: $48V$ .

Thus, u get your answer that any voltage greater than $48V$ will blow it up.

A circuit with a resistance of 6 Omega6 Omega has a fuse that melts at 8 A8 A. Can a voltage of 64 V64 V be applied to the circuit without blowing the fuse?