Solved Problem on Electric Current

A wire made of nickel has 50 m long and 0.5 mm in diameter. between the ends of this wire is applied a potential difference of 110 volts. The electrical resistivity of nickel is 0.342 Ω.mm².m⁻¹. Determine:
a) The electrical conductivity of nickel;
b) the resistance of wire;
c) the conductance of it;
d) the intensity of the current;
e) The absorbed power.
f) the energy absorbed in 1 hour;

Problem data:

Wire length: L = 50 m;
Wire diameter: d = 0.5 mm;
Potential difference between the ends of the wire: U = 110 V;
Electrical resistivity of nickel: ρ = 0.342 Ω.mm².m⁻¹.

Problem diagram:

Figure 1

Solution

First, we will convert the diameter of the wire given in millimeters (mm) to meters (m), the nickel resistivity given in ohm-square millimeter per meter (Ω.mm².m⁻¹) to ohm-meter (Ω.m) and, in the item (f), the interval of time in hours (h) to seconds (s), used in the International System of Units (SI).

\[ \begin{gather} d=0.5\;\cancel{\text{mm}}\times \frac{10^{-3}\;\text{m}}{1\;\cancel{\text{mm}}}=5\times 10^{-1}\times 10^{-3}\;\text{m}=5\times 10^{-4}\;\text{m}\\[10pt] \rho=0.342\;\frac{\Omega.\cancel{\text{mm}^{2}}}{\text{m}}\times \frac{\left(10^{-3}\;\text{m}\right)^{2}}{1\;\cancel{\text{mm}^{2}}}=3.42\times 10^{-1}\;\frac{\Omega}{\cancel{\text{m}}}\times 10^{-6}\;\text{m}^{\cancel{2}}=3.42\times 10^{-7}\;\Omega\text{m}\\[10pt] 1\;\text{hora}=3600\;\text{segundos} \end{gather} \]

a) Conductivity σ is given by

\[ \bbox[#99CCFF,10px] {\sigma =\frac{1}{\rho }} \]

\[ \begin{gather} \sigma =\frac{1}{3.42\times 10^{-7}}\\ \sigma =0.292\times 10^{7}\\ \sigma =2.92\times 10^{-1}\times 10^{7} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\sigma =2.92\times 10^{6}\;\text{S/m}} \]

b) The resistance R of the wire is

\[ \begin{gather} \bbox[#99CCFF,10px] {R=\rho \frac{L}{A}} \tag{I} \end{gather} \]

where A is the cross-section area of the wire, the area a circle is given by

\[ \bbox[#99CCFF,10px] {A=\pi r^{2}} \]

the radius of the wire will be half of the diameter given in the problem \( r=\dfrac{d}{2} \)

\[ \begin{gather} A=\pi \left(\frac{d}{2}\right)^{2} \tag{II} \end{gather} \]

substituting the expression (II) into (I), the resistance will be

\[ R=\rho \frac{L}{\pi \left(\dfrac{d}{2}\right)^{2}} \]

substituting the data of the problem and assuming π = 3.14, we have

\[ \begin{gather} R=3.42\times 10^{-7}\times \frac{50}{3.14\times \left(\dfrac{5\times 10^{-4}}{2}\right)^{2}}\\ R=3.42\times 10^{-7}\times \frac{50}{3.14\times \dfrac{25\times 10^{-8}}{4}}\\ R=\frac{4\times 3.42\times 10^{-7}\times 50}{3.14\times 25\times 10^{-8}}\\ R=\frac{4\times 3.42\times 2}{3.14}\times 10^{-7}\times 10^{8}\\ R=\frac{13.68}{3.14}\times 10 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {R=87.1\;\Omega} \]

c) The conductance G is given by

\[ \bbox[#99CCFF,10px] {G=\frac{1}{R}} \]

\[ G=\frac{1}{87.1} \]

\[ \bbox[#FFCCCC,10px] {G=0.011\;\text{S}} \]

d) Using Ohm's Law, we have

\[ \bbox[#99CCFF,10px] {U=Ri} \]

\[ \begin{gather} i=\frac{U}{R}\\ i=\frac{110}{67.1} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {i=1.26\;\text{A}} \]

e) The power absorbed by the wire is given by

\[ \bbox[#99CCFF,10px] {P=Ri^{2}} \]

\[ \begin{gather} P=87.1\times (1.26)^{2}\\ P=87.1\times 1.59 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {P=138.5\;\text{W}} \]

f) The energy is given by

\[ \bbox[#99CCFF,10px] {\Delta E=P\Delta t} \]

\[ \Delta E=138.5\times 3600 \]

\[ \bbox[#FFCCCC,10px] {\Delta E=498600\;\text{J}} \]

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