Q51P - Fundamentals Of Physics

Question

Wire C and wire D are made from different materials and have length ${\mathit{L}}_{\mathbf{c}}\mathbf{=}{\mathit{L}}_{\mathbf{D}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}$.  The resistivity and diameter of wire C are $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\Omega }\mathbf{.}\mathbf{m}$ and 1.00 mm, and those of wire D are . 0.25 mm and 0.50 mm. The wires are joined as shown in Figure, and a current of 2.0 A is set up in them. What is the electric potential difference between (a) points 1 and 2 and (b) points 2 and 3? What is the rate at which energy is dissipated between (c) points 1 and 2 and (d) points 2 and 3?

                  

Step-by-Step Solution

Answer

1. The electric potential difference between points 1 and 2 is 5.1 V.
2. The electric potential difference between points 2 and 3 is 10 V.
3. The rate of energy dissipation between points 1 and 2 is 10 W.
4. The rate of energy dissipation between points 2 and 3 is 20 W.

Step 1: Identification of the given data

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

Step 2: Understanding the concept of the potential difference and energy dissipation rate

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,

$A={\mathrm{\pi r}}^2$                                                                                                                       … (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)

 

Step 3: (a) Calculation of the potential difference between points 1 and 2

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

$R_c=p_c\frac{L_c}{{\mathrm{\pi r}}_{\mathrm{c}}^2}$

 

Substitute the values in the above equation.

$$\begin{gathered} R_c=2.0\times {10}^{-6}\Omega .m\times \frac{1.0}{\mathrm{\pi }\times {\left(5.0\times {10}^{-4}\mathrm{m}\right)}^2} \\ =2.55\Omega \end{gathered}$$

 

Now, according to Ohm’s law, the electric 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.

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

 

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

Step 4: (b) Calculation of the potential difference between points 2 and 3

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

$R_D=p_D\frac{L_D}{{\mathrm{\pi r}}_{\mathrm{D}}^2}$

 

Substitute the values in the above equation.

$$\begin{gathered} R_D=1.0\times {10}^{-6}\mathrm{\Omega }.\mathrm{m}\times \frac{1.0 \mathrm{m}}{\mathrm{\pi }\times {\left(2.5\times {10}^{-4}\mathrm{m}\right)}^2} \\ =5.09 \mathrm{\Omega } \end{gathered}$$

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.

$$\begin{gathered} \left|{\mathrm{V}}_2-{\mathrm{V}}_3\right|=2.0\mathrm{A}\times 5.09 \mathrm{\Omega } \\ =10.2 \mathrm{V} \\ \approx 10\hspace{0.17em} \mathrm{V} \end{gathered}$$

 

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

Step 5: (c) Calculation of the rate of dissipated energy between points 1 and 2

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

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

 

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

Step 6: (d) Calculation of the rate of dissipated energy between points 2 and 3

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

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

 

Hence, the value of the power is 20 W.