Question

Q4. A copper transmission cable 100 km long and 10.0 cm in diameter carries a current of 125 A. The copper resistivity is 1.72 × 10-8 Ω.m.

(a) What is the potential drop across the cable?

(b) How much electrical energy is dissipated as thermal energy per hour?

(c) What temperature increase is required to increase the cable resistance by 5.00% ? [Thermal resistivity coefficient of copper at 25 deg C is 0.00393 (°C)-1]

Answer 1

the voltage drop occurs when resistance is experienced by the current flowing through the copper wire. Hence Ohms law can be used to find the voltage drop across the entire length of the transmission line. According to Ohm's Law

  

where

V is the voltage (V)

I is the current (A)

R is the resistance () . But resistance depends on length (l) and area cross-section (A) of the conductor, given by

Where,

is the resistivity of the material (copper)

Therefore voltage drop can be modified as

  

a)

Given

I = 125 A

= 1.72x10-8

l = 100km = 100x103 m

Diameter of the cable, d = 10cm

Therefore radius of the cable, r = d/2 = 5cm = 0.05m

Therefore area cross section of the cable, A = = (3.14)*(0.05)2 = 7.853x10-3 m2

Now voltage drop is given by,

V = 27.38 V

b)

When current is passes through a conductor, the electrical energy is lost in the form of heat. Joule's law stats that the Quantity of heat (H) or energy (E) developed in a current-carrying current is equal to the product of the square of electric current flowing through the conduct, resistance of the conductor and the time of flow of electric current through the conductor.

  

From ohm's law substitute for R,

  

  

where,

V is the voltage (V)

I is the current (A)

t is the time (s)

Giiven,

I =125 A

t = 1 hour = 60x60 seconds = 3600s

We found out that V = 27.38V

Therefore the amount of thermal energy lost per hour is given by

  

E = 12321x103 J = 12.321x106 J = 12.321 MJ (Ans.)

c)

if R1 and R2 are the resistance of a conductor at t10C and t20C respectively, then the temperature coefficient of resistance is given by,

Given,

Let R1 = R be the resistance at t1 = 270C

(cable resistance increased by 5%)

  

On rearranging,

t = 41.070C

Hence to increase the resistance by 5% temperature has to be increased from 270C to 41.070C.

ie, an increase of (41.070C - 270C) = 14.070C is required