Q31P - Fundamentals Of Physics

Question

An electrical cable consists of 125 strands of fine wire, each having $\mathbf{2}\mathbf{.}\mathbf{65}\mathbf{ }\mathit{\mu }\mathit{\Omega }$ resistance. The same potential difference is applied between the ends of all the strands and results in a total current of 0.750 A .  (a) What is the current in each strand?  (b) What is the applied potential difference?  (c) What is the resistance of the cable?

Step-by-Step Solution

Answer

1. The current in each strand is $6.0\times {10}^{-3} \mathrm{A}$ .
2. The applied potential difference is $1.59\times {10}^{-8} \mathrm{V}$ .
3. The resistance of the cable is $2.12\times {10}^{-8} \mathrm{\Omega }$ .

Step 1: The given data

1. The cable consists of 125 strands
2. Total current, $l=0.750 \mathrm{A}$ 
3. Resistance of each strand, $R=2.65 \mathrm{\mu \Omega } \mathrm{or} 2.65\times {10}^{-6} \mathrm{\Omega }$

Step 2: Understanding the concept of Ohm’s law

Under the assumption that all physical parameters and temperatures stay constant, Ohm's law asserts that the voltage across a conductor is precisely proportional to the current flowing through it. We can find the current, potential difference and resistance by using the given values and Ohm’s law.

 

Formulae:

The equation of current in each strand, $i=\frac{\mathrm{Total} \mathrm{current}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{strands}}$                         …(i)

The Voltage equation using the Ohm’s law, $V=iR$                                                         …(ii)

The resistance of the cable, $R_{Total}=\frac{\mathrm{Resistance} \mathrm{of} \mathrm{each} \mathrm{strand}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{strands}}$                              …(iii)

Step 3: (a) C alculation of the current in each strand

Using the given data in equation (i), we can get the current value in each strand as follows:

$$\begin{gathered} i=0.750\mathrm{A}/125 \\ =6.0\times {10}^{-3} \mathrm{A} \end{gathered}$$ 

Hence, the value of the current is $6.0\times {10}^{-3} \mathrm{A}$ .

Step 4: (b) Calculation of the applied potential difference

Using the given values, the applied potential difference in this case can be calculated using equation (ii) as follows:

$$\begin{gathered} V=\left(6.0\times {10}^{-3} \mathrm{A}\right)\left(2.65\times {10}^{-6} \mathrm{\Omega }\right) \\ =1.59\times {10}^{-8} \mathrm{V} \end{gathered}$$ 

Hence, the value of the potential difference is $1.59\times {10}^{-8} \mathrm{V}$ .

Step 5: (c) Calculation of the resistance of the cable

Resistance of the cable is calculated using the given data in equation (iii) as follows:

$$\begin{gathered} R_{Total}=\frac{2.65\times {10}^{-6} \mathrm{\Omega }}{125} \\ =2.12\times {10}^{-8} \mathrm{\Omega } \end{gathered}$$

Hence, the value of the resistance is $2.12\times {10}^{-8} \mathrm{\Omega }$ .