Q20P - Fundamentals Of Physics

Question

A certain wire has a resistance R. What is the resistance of a second wire, made of the same material, that is half as long and has half the diameter?

Step-by-Step Solution

Answer

The resistance of the second wire, made of the same material is 2R.

Step 1: The given data

a)    Resistance of the first wire is R 

b)    Length of the second wire, $L_2=\frac{L_1}{2}$ 

c)    Diameter is, $D_2=\frac{D_1}{2}$

Step 2: Understanding the concept of resistance

The voltage is directly proportional to the current flowing through the circuit. The constant of proportionality is called resistance. The resistance is opposition to the flow of current. 

 

We have to use the concept of resistance. Using the equation of resistance in terms of resistivity, length, and cross-sectional area, we can find the resistance.

 

Formulae:

The resistance of the wire, $R=\frac{\mathrm{\rho L}}{A}$                                                                                …(i)

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

The cross-sectional area of the wire in terms of diameter, $A=\pi {\left(\frac{D}{2}\right)}^2$                         …(ii)

Here,D is the diameter of the wire of the conductor.

Step 3: Calculation of the resistance of the second wire

Using the given values in equation (i), we can get the resistances of the two wires as follows:

$$\begin{gathered} R_1=\frac{{\mathrm{\rho L}}_1}{A_1} \\ {R}_2=\frac{{\mathrm{\rho L}}_2}{A_2} \end{gathered}$$                                                                                                                        …(iii)

                                                                                                                      …(iv)

Now, dividing equations (iv) by (iii) and using the given data, we can get the resistance of the second wire as follows:

$$\begin{gathered} \frac{R_2}{R_1}=\frac{\left(\frac{{\mathrm{\rho L}}_2}{A_2}\right)}{\left(\frac{{\mathrm{\rho L}}_1}{A_1}\right)} \\ =\frac{\left({\displaystyle \frac{\left({\displaystyle \frac{L_1}{2}}\right)}{A_2}}\right)}{\left(\frac{L_1}{A_1}\right)}=\frac{\left({\displaystyle \frac{1}{2A_2}}\right)}{\left(\frac{1}{A_2}\right)} \\ =\frac{A_1}{2A_2} \\ =\frac{\pi {\left({\displaystyle \frac{D1}{2}}\right)}^2}{2\pi {\left(\frac{D_2}{2}\right)}^2} \\ =\frac{D_1^2}{2D_2^2} \\ =\frac{D_1^2}{2\left({\displaystyle \frac{D_1}{2}}\right)} \\ {R}_2=2R_1 \\ =2R \left(\because R_1=R\right) \end{gathered}$$ 

Hence, the value of the resistance of the wire is 2R .