Question

A carbon resistor is 5 mm long and has a constant cross section of$\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}{\mathbf{mm}}^{\mathbf{2}}$The conductivity of carbon at room temperature is $\mathit{\sigma }\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{per}\mathbf{}\mathbf{ohm}\mathbf{-}\mathbf{m}$.In a circuit its potential at one end of the resistor is 12 V relative to ground, and at the other end the potential is 15 V. Calculate the resistance Rand the current I (b) A thin copper wire in this circuit is 5 mm long and has a constant cross section of $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}{\mathbf{mm}}^{\mathbf{2}}$.The conductivity of copper at room temperature is $\mathit{\sigma }\mathbf{=}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}{\mathbf{ohm}}^{\mathbf{-}\mathbf{1}}{\mathbf{m}}^{\mathbf{-}\mathbf{1}}$.The copper wire is in series with the carbon resistor, with one end connected to the 15 V end of the carbon resistor, and the current you calculated in part (a) runs through the carbon resistor wire. Calculate the resistance Rof the copper wire and the potential ${\mathit{V}}_{\mathbf{at}\mathbf{}\mathbf{end}}$at the other end of the wire.

You can see that for most purposes a thick copper wire in a circuit would have practically a uniform potential. This is because the small drift speed in a thick, high-conductivity copper wire requires only a very small electric field, and the integral of this very small field creates a very small potential difference along the wire.

Short answer

(i) $R=0.83\mathrm{\Omega }$, (ii)$l=3.61\mathrm{A}$

Step-by-step solution

Given Data

Length of carbon resistor $l=5\mathrm{mm}$

Area $A=0.2{\mathrm{mm}}^2$

${\sigma }_1=3\times {10}^4\mathrm{per}\mathrm{ohm}-\mathrm{m}$

${\sigma }_2=6\times {10}^7\mathrm{per}\mathrm{ohm}-\mathrm{m}$

$V_1=12\mathrm{V},V_2=15\mathrm{V}$$V_1=12\mathrm{V},{\mathrm{V}}_2=15\mathrm{V}$

Concept

When the substance gives opposition to the electric current flow, then it is known as resistance.

Step 3(a) (i): Calculate the resistance R

The resistivity,

$\begin{array}{c}p=\frac{1}{{\sigma }_1}\\ =\frac{1}{3\times {10}^4}\\ =3.33\times {10}^{-5}\mathrm{\Omega m}\end{array}$

Resistance of wire,

$\begin{array}{rcl}R& =& \rho \frac{l}{A}\\ & =& 3.33\times {10}^{-5}\times \frac{5\times {10}^{-3}}{0.2\times {10}^{-6}}\\ & =& 0.83\mathrm{\Omega }\end{array}$

Hence, the resistance is $\begin{array}{rcl}& & 0.83\mathrm{\Omega }\end{array}$

Step 4(a) (ii): Calculate the current 

Potential Difference across the resistor,

$\begin{array}{rcl}∆V& =& V_2-V_1\\ & =& 15-12\\ & =& 3\mathrm{V}\end{array}$

As per Ohms Law,

$\begin{array}{rcl}∆V& =& IR\\ I& =& \frac{∆V}{R}\\ & =& \frac{3}{0.83}\\ & =& 3.61\mathrm{A}\end{array}$

Hence, the current is 3.61 A