91Ó°ÊÓ

The given data can be listed below as:

• The lengths are$L_c=L_D=1.0\mathrm{m}$.
• The resistivity of wire C is$p_w=2.0×{10}^{-6}\mathrm{Î\copyright }.\mathrm{m}$.
• The resistivity of wire D is$p_D=1.0×{10}^{-6}\mathrm{Î\copyright }.\mathrm{m}$.
• The radius of wire C is$r_c=0.50\mathrm{mm}=5.0×{10}^{-4}\mathrm{m}$.
• The radius of wire D is$r_D=0.25\mathrm{mm}=2.5×{10}^{-4}\mathrm{m}$.
• The current flowing is i = 2.0 A.

In this problem, we need to use the equation of Ohm’s law and the below equation (i) and (iii) to find the resistance and the potential difference. To calculate the power, we can use the equation (iv).

Formulae:

The resistance of the material is given as,

$R=p\frac{L}{A}$ … (i)

The cross-sectional area of the wire is given as,

$A={\mathrm{Ï€°ù}}^2$ … (ii)

The voltage equation according to Ohm’s law is given as,

$V=iR$ … (iii)

The power transferred during the dissipation is given as,

$P=i^{2}R$ … (iv)

Substituting the given values and equation (ii) in equation (i), we can get the potential difference between the points 1 and 2 as follows:

$R_c=p_c\frac{L_c}{{\mathrm{Ï€°ù}}_{\mathrm{c}}^2}$

Substitute the values in the above equation.

$$\begin{gathered} R_c=2.0×{10}^{-6}\mathrm{Î\copyright }.m×\frac{1.0}{\mathrm{Ï€}×{\left(5.0×{10}^{-4}\mathrm{m}\right)}^2} \\ =2.55\mathrm{Î\copyright } \end{gathered}$$

Now, according to Ohm’s law, theelectric potential difference between points 1 and 2 can be calculated using the given data in equation (iii) as follows:

$\left|V_1-V_2\right|=V_c$

Substitute the values in the above equation.

$$\begin{gathered} \left|V_1-V_2\right|=2.0\mathrm{A}×2.55\mathrm{Î\copyright } \\ =5.1\mathrm{V} \end{gathered}$$

Hence, the value of the potential difference is 5.1 V.

Substituting the given values and equation (ii) in equation (i), we can get the potential difference between the points 2 and 3 as follows:

$R_D=p_D\frac{L_D}{{\mathrm{Ï€°ù}}_{\mathrm{D}}^2}$

Substitute the values in the above equation.

$$\begin{gathered} R_D=1.0×{10}^{-6}\mathrm{Î\copyright }.\mathrm{m}×\frac{1.0\mathrm{m}}{\mathrm{Ï€}×{\left(2.5×{10}^{-4}\mathrm{m}\right)}^2} \\ =5.09\mathrm{Î\copyright } \end{gathered}$$

Thus, according to Ohm’s law,electric potential difference between points 2 and 3 can be calculated using the given data in equation (iii) as follows:

$\left|V_2-V_3\right|=V_D$

Substitute the values in the above equation.

$$\begin{gathered} \left|{\mathrm{V}}_2-{\mathrm{V}}_3\right|=2.0\mathrm{A}×5.09\mathrm{Î\copyright } \\ =10.2\mathrm{V} \\ ≈10 \mathrm{V} \end{gathered}$$

Hence, the value of the potential difference is 10 V.

Using the above values in equation (iv), we can get the rate of dissipated energy between the points 1 and 2 as follows:

$$\begin{gathered} P_c={\left(2.0\mathrm{A}\right)}^2×2.55\mathrm{Î\copyright } \\ =10.2\mathrm{W} \\ ~10\mathrm{W} \end{gathered}$$

Hence, the value of the dissipated power is 10 W.

Using the above values in equation (iv), we can get the rate of dissipated energy between the points 2 and 3 as follows:

$$\begin{gathered} P_D={\left(2.0\mathrm{A}\right)}^2×5.09\mathrm{Î\copyright } \\ =20.4\mathrm{W} \\ ~20\mathrm{W} \end{gathered}$$

Hence, the value of the power is 20 W.