Question

An electrical cable consists of125strands 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?

Short 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-by-step solution

The given data

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

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)

(a) Calculation 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 $6.0\times {10}^{-3}\mathrm{A}$.

(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 $1.59\times {10}^{-8}\mathrm{V}$.

(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 $2.12\times {10}^{-8}\mathrm{\Omega }$.