# Regulation of transformer with inductive load

The regulation of transformer and alternators is poor while supplying inductive loads like a motor as compared to resistive loads.

Why is it so?

I do understand the phasor analysis of these loads and desire to know the physical reason behind the poor regulation of lagging loads.

I'm going to use the transformer as an example.

A transformer will have leakage inductance i.e. "spare" inductance in each winding that isn't coupling through the core. This leakage inductance is a series component and can drop voltage across it when load currents are taken. This is a significant reason why regulation is not perfectly 100% in a transformer.

So, if the net leakage inductance is X and the load is (say) 50X and reactive, a simple potential divider is formed and the secondary voltage is: -

$\dfrac{50}{50+1}$ = 98%

But, if the same load current were taken with a resistor equivalent in impedance to 50X then the secondary voltage would be: -

$\dfrac{50}{\sqrt{50^2+1^2}}$ = 99.98%

If the copper losses were as significant as the leakage inductance in their effect, regulation would be the same for inductive and resistive loads.

• I did understand most of what you posted but could not understand this statement. If the copper losses were as significant as the leakage inductance in their effect, regulation would be the same for inductive and resistive loads. Please do explain this statement elaborately. – Adnan Arif Sait Jun 11 '17 at 10:28
• Think about a potential divider formed by an inductor and resistor of the same impedance magnitude. The output isn't fifty percent but seventy one percent. If leakage inductance impedance equalled copper loss resistance then no matter what load you had (resistive or inductive) the dominant component that produces imperfect regulation would be either the leakage reactance or the copper loss resistance. Hence regulation would be the same for both types of load. – Andy aka Jun 11 '17 at 10:40