Question

A cylindrical copper cable carries a current of 1200 A. There is a potential difference of 1.6×10−2 V between two points on the cable that are 0.24 m apart. What is the radius of the cable?

Answer with explaination......​

Answer 1

Answer:

The radius of the cable is = 0.99cm.

Explanation:

By Ohm's Law,

U=RI

I(Current) =1200A

U(Potential Difference) = 1.6⋅10−2V

Resistance:

R=UI=1.6⋅10−21200

=1.61.2⋅10−2−3

=1.33⋅10−5Ω

By applying,

R=ρla

Length of the cable is l=0.24m

Resistivity of copper is ρ=1.7⋅10−8Ωm

Cross section is a=ρ⋅lR

a=1.7⋅10−8⋅0.241.3310−5

=(1.7⋅0.241.33)⋅10−810−5

=0.31⋅10−3

Radius of the cable:

a=πr2

r=√aπ=√0.31⋅10−3π

=0.99⋅10−2m

=0.99cm.

Answer 2

Answer:

The radius of the cable is 0.99 cm.

SOLUTION :

Apply ohm's law,

Here, U = RI.

Current, I = 1200 A.

Potential difference, V = $1.6 \times 10^{-2} v$

Resistance, R is

$\bf R = \dfrac {U}{I}$

→ $\dfrac {1.6 \times 10^{-2}{1200}}}$

→ $\bf \dfrac {1.6}{1.2} \times 10^{-2-3}$

= $\bf 1.33 \times 10^{-5} \ohm.$

Now,

$R \: = \: \rho \dfrac {l}{a}$

The length of the cable, l = 0.24 m.

Resistivity of the copper is $1.7 \times 10^{-8} \ohm \: m$.

The cross section is $\bf a \: = \: \dfrac {\rho \times l}{R}$

→ $a \: = \: \dfrac {1.7 \times 10^{-8} \times 0.24}{1.3310^{-5}$

→ $\bf \bigg( 1.77 \times \dfrac {0.24}{1.33} )\bigg) \times \dfrac {10^{-8}}{10^{-5}}$

= $0.31 \times 10^{-3}$

The radius of the cable is calculated with,

★ $a \: = \: \pi r^{2}$

$r \: = \: \displaystyle \sqrt {0.31 \times 10^{-3}{\pi}}$

$\bf 0.99 \times 10^{-2} m$

= $0.99 \: cm$

$\therefore$ The radius of the cable is 0.99 cm.