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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 $\mathit{σ}\mathbf{=}\mathbf{3}\mathbf{×}{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{per}\mathbf{}\mathbf{ohm}\mathbf{-}\mathbf{m}$.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$\mathit{σ}\mathbf{=}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{7}}\mathbf{}{\mathbf{ohm}}^{\mathbf{-}\mathbf{1}}{\mathbf{m}}^{\mathbf{-}\mathbf{1}}$ .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.

Step by step solution

01

Given Data

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

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

${\mathrm{σ}}_1=3×{10}^4\mathrm{per}\mathrm{ohm}-\mathrm{m}$

${\mathrm{σ}}_2=6×{10}^7\mathrm{per}\mathrm{ohm}-\mathrm{m}$

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

02

Concept

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

03

Step 3(b) (i): Calculate the resistance  R  of the copper wire

The resistance of copper wire,

$$\begin{gathered} \begin{array}{l}{R}_2=\frac{1}{{\mathrm{σ}}_2}\end{array}\frac{I_2}{A_2} \\ =\frac{1}{6×{10}^7}\frac{5×{10}^{-3}}{0.2×{10}^{-6}} \\ =4.2×{10}^{-4}\mathrm{Î\copyright } \end{gathered}$$

Hence, the resistance of copper wire is $4.2×{10}^{-4}\mathrm{Î\copyright }$

04

Step 4(b) (ii): Calculate the potential   Vat end at the other end of the wire

Voltage across the copper wire,

$$\begin{gathered} \begin{array}{rcl}{V}_2& =& I_2{R}_2\\ & =& 3.61×4.2×{10}^{-4}\\ & =& 1.5×{10}^{-3}\mathrm{V}\\ & & \left(\text{Referring to subpart}\left(\text{a}\right)\text{of the SID:875865 - 19 - 49P - a} \\ \text{take}{\mathrm{I}}_1\text{Value}\right)\end{array} \end{gathered}$$

Hence the Voltage across the copper wire is .$1.5×{10}^{-3}\mathrm{V}$

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