91Ó°ÊÓ

The resistivity of a metal increases slightly with increased temperature. This can be expressed as $\mathrm{ÒÏ}={\mathrm{ÒÏ}}_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]$, where $T_0$ is a reference temperature, usually $20°C$, and a is the temperature coefficient of resistivity

We know that the resistance is given by

$R=\mathrm{ÒÏ}\frac{L}{A}$

$R=$resistance, $L=$Length, $A=$Cross-section area , $T=$Temperature.

Substitute the given expression,

$\mathrm{ÒÏ}={\mathrm{ÒÏ}}_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]$

Gives us,

$R={\mathrm{ÒÏ}}_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]\frac{L}{A}$

Appling Ohm's Law we have find intensity,

role="math" localid="1648705844053" $I=\frac{U}{R}=\frac{UA}{\mathrm{ÒÏ}L}=\frac{UA}{{\mathrm{ÒÏ}}_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]L}$.

Let us recall what the binomial approximation is:

$(1+x{)}^n≈1+nx$

Consider everything in the square brackets apart from the one as the $x$in the above approximation

We can consider our expression in brackets,

${\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]}^{-1}$

As the $-1$is what puts it into the denominator. So, let us write the intensity again as

$I=\frac{UA{\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]}^{-1}}{{\mathrm{ÒÏ}}_{0}L}$

We can apply the binomial approximation,

We get,

$I≈\frac{UA\left[1-\mathrm{Î\pm }\left(T-T_0\right)\right]}{{\mathrm{ÒÏ}}_{0}L}$.

For copper, $\mathrm{Î\pm }=3.9×{10}^{-3}{{}^{∘}\mathrm{C}}^{-1}$. Suppose a $2.5-m$-long, $0.40-mm$-diameter copper wire is connected across the terminals of a $1.5V$ ideal battery.

We need to do before applying numerical values is substitute the expression for the area when the latter is a circle with the diameter known:

$A=\frac{\mathrm{Ï€}{D}^2}{4}$

We will get that the intensity is

$I≈\frac{\mathrm{Ï€}U{D}^2\left[1-\mathrm{Î\pm }\left(T-T_0\right)\right]}{4{\mathrm{ÒÏ}}_{0}L}$

Substituting the numerical values,

We get,

$I≈\frac{\mathrm{π}×1.5×0.{0004}^2×[1-(20-20\left)\right]}{4×1.7×{10}^{-8}×2.5}=4.4\mathrm{A}$

The current in the wire at${20}^{o}C$found to be$4.4A$

The resistivity of a metal increases slightly with increased temperature. This can be expressed as $\mathrm{ÒÏ}={\mathrm{ÒÏ}}_0\left[1+\mathrm{Î\pm }\left(T-T_0\right)\right]$, where$T_0$ is a reference temperature, usually $20°C$, and a is the temperature coefficient of resistivity.

What we are examining is the rate of change of the intensity with changing temperature; that is, the derivative with respect to the temperature of our expression for the intensity. This will be

$\frac{dI}{dT}=\frac{d}{dT}\frac{\mathrm{Ï€}U{D}^2\left[1-\mathrm{Î\pm }\left(T-T_0\right)\right]}{4{\mathrm{ÒÏ}}_{0}L}=-\frac{\mathrm{Ï€}\mathrm{Î\pm }U{D}^2}{4{\mathrm{ÒÏ}}_{0}L}$

Given numerical case, the answer will be,

$-\frac{\mathrm{π}×3.9×{10}^{-9}×1.5×0.{0004}^2}{4×1.7×{10}^{-8}×2.5}=15.5\mathrm{mA}/{}^{∘}\mathrm{C}$

Therefore, The rate, in A/°C, at which the current changes with temperature as the wire heats up found to be$15.5\mathrm{mA}{/}^{∘}\mathrm{C}$