They are looking for the d.c resistance as well as the a.c resistance

Question

A single phase 8kv, 50hz, 50cm transmission line consists of three aluminum conductors with 3cm diameter. If the 50hz of the line is 5% greater than the D.C resistance, calculate the (i) D.C resistance, (ii) A.C resistance

Answer 1

Hi Mersaydess:

 

   The resistance of a wire under direct current can be obtained from the expression:

 

R_DC = (ρL)/A

 

   where

 

A = cross-section area of the wire

L = length of the wire

ρ = resistivity of wire material (based on type of material)

 

These are the three physical factors that affect resistance in a wire.

 

Resisitvity is what is called an empirical quantity, meaning it has to be measured experimentally for various materials, and cannot be derived from more fundamental principles.  We usually look values like that up in a table or online.  For aluminum, I find a value of 2.65x10^-8 Ω•m (ohm-meter is the unit of resistivity, so that ρL/A has units of resistance, as it should from the formula above).

 

Note that aluminum has a very small resistivity, which corresponds to it be a very good electrical conductor.

 

Since the wire has a diameter, its cross-section is a circle, with diameter 0.03 m (in all cases, I have to be sure my lengths are in m, to match the units of the resistivity value I looked up).  The cross-sectional area of the wire, A, is thus A = π(d²)/4 or πr².

 

Now we have ρ, A, and L for each wire.  We can use the formula above to determine R, just by putting in values.

 

I obtained a value for a 50-cm wire of aluminum with 3-cm diameter:

 

R_DC = 1.87x10^-5 Ω  (this is reasonable, as I would want to design the resistance of transmission wires to be pretty small)

 

Of course, you would have to triple this to get the total R_DC for three aluminum wires of this kind.

 

If the AC "resistance" is 5% greater, then that means it is 105% of the DC resistance just calculated, so that 

 

R_AC = 1.05(R_DC), and you can then solve for R_AC

 

Hope this helps!  I am available for online tutoring if you have any more detailed questions about this or other physics material.