Question

A 5 amp fuse wire can withstand a maximum power of 1 W in circuit. The resistance in the fuse wire is :
\[\begin{align}
  & \text{A) 0}\text{.2}\Omega \\
 & \text{B) 5}\Omega \\
 & \text{C) 0}\text{.4}\Omega \\
 & \text{D) 0}\text{.04}\Omega \\
\end{align}\]

Answer 1

Hint: Electric power is given as the product of voltage and the current. From ohm’s law voltage is given as a product of resistance and current and so by using ohm’s law we can get a relation between power and resistance. Once we get the relation between power and resistance we can substitute the given values to find the resistance of the fuse wire.

Formula used:
\[P={{I}^{2}}R\]

Complete answer:
We have given the current and the maximum power for the fuse fire.
We know that power is the work done per unit time and electric power can be given as electric current passing through one point to another or electric charge flowing per unit time. Mathematically it can be given as
 \[P=VI\]
Where V is the voltage and I is the current.
From the Ohm’s Law the voltage is given as product of resistance and current and it can be given as
\[V=IR\]
Now substituting this value of voltage in the equation for power we get
\[\begin{align}
  & P=(IR)I \\
 & P={{I}^{2}}R \\
\end{align}\]
Now according to the question \[I=5A\text{ and }P=1W\], substituting it in the above equation we get,
\[\begin{align}
  & 1W={{(5A)}^{2}}R \\
 & R=\dfrac{1}{25}\Omega \\
 & R=0.04\Omega \\
\end{align}\]

So, the correct answer is “Option D”.

Note:
The fuse should have high resistance and low melting point because if it had low resistance it would break for the high value of current as the role of resistance is to resist the current flowing through the device. And the fuse is used to protect the device from high voltage. So when the voltage is high, the fuse will break and the circuit or device will be protected.