# A 20.0-cm-long hollow nichrome tube of inner diameter 1.70 mm, outer diameter 4.40 mm is...

## Question:

A 20.0-cm-long hollow nichrome tube of inner diameter 1.70 mm, outer diameter 4.40 mm is connected to a 2.00 V battery. What is the current in the tube?

## Current:

In a circuit, the current flow varies with the electric potential directly, and with the resistance, it varies indirectly. Ampere meter is a device in the electric world which is used to measure current in ampere.

• The tube length is: {eq}L = 20\,{\rm{cm}} = 0.20\;{\rm{m}} {/eq}
• The inner diameter of the tube is: {eq}{d_i} = 1.70\;{\rm{mm}} = 0.0017\;{\rm{m}} {/eq}
• The outer diameter of the tube is: {eq}{d_o} = 4.40\;{\rm{mm}} = 0.0044\;{\rm{m}} {/eq}
• The electric potential of the battery is: {eq}V = 2\;{\rm{V}} {/eq}

The area of the tube is calculated as,

{eq}A = \dfrac{\pi }{4}\left( {{d_o}^2 - {d_i}^2} \right) {/eq}

Substitute the given values in above equation.

{eq}\begin{align*} A &= \dfrac{\pi }{4}\left( {{{\left( {0.0044} \right)}^2} - {{\left( {0.0017} \right)}^2}} \right)\\ &= 1.5 \times {10^{ - 5}} - 0.23 \times {10^{ - 5}}\\ &= 1.27 \times {10^{ - 5}}\;{{\rm{m}}^2} \end{align*} {/eq}

The expression for the resistance is,

{eq}R = \dfrac{{\rho L}}{A} {/eq}

Here, {eq}\rho = 1.5 \times {10^{ - 6}}\;\Omega \cdot {\rm{m}}. {/eq}

Substitute the given values in above equation.

{eq}\begin{align*} R &= \dfrac{{1.5 \times {{10}^{ - 6}}\;\Omega \cdot {\rm{m}} \times 0.20\;{\rm{m}}}}{{1.27 \times {{10}^{ - 5}}\;{{\rm{m}}^2}}}\\ &= 0.024\Omega \end{align*} {/eq}

The expression used to calculate the current is,

{eq}i = \dfrac{V}{R} {/eq}

Substitute the given values in above equation.

{eq}\begin{align*} i &= \dfrac{{2\;{\rm{V}}}}{{0.024\Omega }}\\ &= 83.33\;{\rm{A}} \end{align*} {/eq}

Thus, the current in the tube is {eq}83.33\;{\rm{A}}. {/eq}