A heating element is made by maintaining a potential difference of 75.0Vacross the length of a Nichrome wire that has a$\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}{\mathbf{m}}^{\mathbf{2}}$ cross section. Nichrome has a resistivity of$\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{\Omega m}$.(a) If the element dissipates, what is its length?(b) If 100 Vis used to obtain the same dissipation rate, what should the length be?

1. The potential difference, V = 75.0V
2. The cross-sectional area ofthelength of the Nichrome wire is $\mathrm{A}=2.60\times {10}^{-6}{\mathrm{m}}^2$.
3. The resistivity of the Nichrome wire is $p=5.00\times {10}^{-7}\mathrm{\Omega m}$.
4. The power dissipation is P = 5000W.
5. The new potential difference, V' = 100 V

The power or rate of energy dissipation, in an electrical device across which a potential difference is equal to the product of current and potential difference. It can also be defined as energy transferred per unit time.

By using the equations 26-28 and 26-16, we can find the length of the Nichrome wire, when the element dissipates . Since, L is directly proportional to the square of the potential difference, using the formula for new length; we can find the new length.

Formulae:

The value of the dissipated power, $P=\frac{V^2}{R}$ …(i)

Here, P is the power, R is the resistance, V is the potential difference.

The resistance of the material, $R=p\frac{L}{A}$ …(ii)

Here, R is resistance, L is the length of the wire, p is the resistivity, A is area of cross-section.

The length relation to the potential difference,$L\text{'}=L{\left(\frac{V\text{'}}{V}\right)}^2$ …(iii)

Since,$L\propto V^2$

We can get the length of the Nichrome wire by rearranging the equation after substituting equation (ii) in equation (i) as follows:

$$\begin{gathered} L=\frac{A{V}^2}{pP} \\ =\frac{2.60\times {10}^{-6}{\mathrm{m}}^2\times {(75.0\mathrm{V})}^2}{5.00\times {10}^{-7}\mathrm{\Omega m}\times 5000\mathrm{W}} \\ =5.85\mathrm{m} \end{gathered}$$

Hence, the length of the wire is 5.85m.

Using the given data and above length in equation (iii), we can get the value of the new length of the Nichrome wire as follows:

$$\begin{gathered} L\text{'}=5.85\mathrm{m}\times {\left(\frac{100\mathrm{V}}{75.0\mathrm{V}}\right)}^2 \\ =10.4\mathrm{m} \end{gathered}$$

Hence, the value of the length of the wire is 10.4m.