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A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to 500 A/cm2. What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.64 A?

1 Answer

1 vote

Answer:

The diameter of wire should be
4.04 * 10^(-4) m

Step-by-step explanation:

Given:

Current density
J = 500 * 10^(4)
(A)/(m^(2) )

Current
I = 0.64 A

From the formula of current density,


J = (I)/(A)

Where
A = area of cylindrical wire =
\pi r^(2)


\pi r^(2) = (I)/(J)


r^(2) = (I)/(\pi J )


r = \sqrt{(0.64)/(3.14 * 500 * 10^(4) ) }


r = 2.02 * 10^(-4)m

For finding the diameter of wire,


d = 2r


d = 4.04 * 10^(-4)m

Therefore, the diameter of wire should be
4.04 * 10^(-4) m

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