Question

A$20$-cm-long hollow nichrome tube of inner diameter $2.8\mathrm{mm}$, outer diameter$3.0\mathrm{mm}$is connected, at its ends, to a $3.0\mathrm{V}$battery. What is the current in the tube?

Short answer

The current in the tube is$I=36A.$

Step-by-step solution

Given information

Nichrome tube length $=20cm$

Nichrome tube inner diameter$=2.8mm$

Nichrome tube outer diameter$=3.0mm$

Voltage$=3.0V$

Explanation

Use the $I=\frac{V}{R}\to \left(1\right)$formula,

The area of the tube is provided by the area of the outer circle $D=3.0\mathrm{mm}$minus the area of the inner circle $d=2.8\mathrm{mm}$:

$A=\left(0.{003}^2\times \pi -0.{0028}^2\times \pi \right)$

role="math" localid="1649165023302" $A=3.64\times {10}^{-6}{m}^2$

Find the resistance of this tube,

$R=\frac{(\rho \left.\times L\right)}{A}$

Where $L$is the length of the wire and $A$is the outer area minus the inner area and rho is the resistivity of nichrome.

role="math" localid="1649165124828" $R=\left[\frac{\left(1.5\Omega m\times {10}^{-6}\times 0.02m\right)}{3.64\times {10}^{-6}{m}^2}\right]$

$R=0.0823\Omega$

Substitute the value in equation (1)

$I=\frac{3V}{0.0823\Omega }$

$I=36.44A$

Final Answer

Hence, the current in the tube is$36A.$