How to Calculate Conductor Size for Different Types of Loads?

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Calculating conductor size for different load types primarily depends on the current, while voltage and environmental factors also play a role in insulation selection. The minimum conductor cross-section should consider temperature rise and heat dissipation, with voltage drop being a critical factor that may necessitate larger conductors. The voltage drop formula incorporates power factor and reactance, with specific considerations for resistive, inductive, and capacitive loads. In single-phase systems, neutral current matches live current, while in three-phase systems, neutral size is limited to half of the live conductor. Proper calculations are essential for ensuring safety and efficiency in electrical installations.
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How to calculate condoctor size for different types of load like resistive load, inductive load or capacitive load etc and what will be neutral wire size for these load?

Any body please help me.

Thnx
 
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From a conductor standpoint - ONLY current matters, the type of load does not. From an insulation standpoint ( material or type of wire) - in general voltage matters, but temperature and environment help to define the type (THHN for example) .
 
I agree with windadct about importance of total current [absolute value] in order to start with

the minimum conductor cross section from the temperature rise-losses and heat evacuation are

important. However, in my opinion, the voltage drop it is important too and it could require to

increase the cross section.

The voltage drop depends on power factor[cos(fi)] and reactance of the cable also.

DV=R*I*cos(fi)+X*I*sin(fi)

For resistive load cos(fi)=1 [fi=0] and sin(fi)=0 If the circuit is entire inductive or entire capacitive the above formula does not change.

The neutral current is the same as live for single-phase system so you have to double the voltage drop.

In three-phase and neutral system only unbalance current will flow through neutral and the

cross section of the neutral is limited to ½ of live.

For voltage drop you have to multiply the above formula by sqrt(3).

I am positive windadct forgot it simply.:biggrin:
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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