Question

The parameters measured for a 220V/110V, 50 Hz, single-phase transformer are:

Self inductance of primary winding = 45 mH

Self inductance of secondary winding = 30 mH

Mutual inductance between primary and secondary windings = 20 mH

Using the above parameters, the leakage (L_l1, L_l2) and magnetizing (L_m) inductances at referred to primary side in the equivalent circuit respectively, are
1. 5 mH, 20 mH and 40 mH
2. 5 mH, 80 mH and 40 mH
3. 25 mH, 10 mH and 20 mH
4. 45 mH, 30 mH and 20 mH

Answer 1

Correct Answer - Option 2 : 5 mH, 80 mH and 40 mH

Concept:

The leakage inductance is defined as the ratio of no of turns multiplied by leakage flux divided by current through it.

The magnetizing inductance is the difference of self-inductance of the primary winding and primary leakage inductance

Let N₁ = primary turns,

N₂ = secondary turns

Let a current I₁, flowing through the primary produce a flux ϕ₁, of which part ϕ links with the secondary also while remaining (ϕ₁ – ϕ) is the leakage flux.

Self-inductance of primary \({L_1} = \frac{{{N_1}{\phi _1}}}{{{I_1}}}\)

And mutual inductance between primary and secondary, \(M = \frac{{{N_2}\phi }}{{{I_1}}}\)

The leakage inductance of the primary

\({L_{l1}} = \frac{{{N_1}\left( {{\phi _1} - \phi } \right)}}{{{I_1}}} = \frac{{{N_1}{\phi _1}}}{{{I_1}}} - \frac{{{N_1}\phi }}{{{I_1}}}\) 

\( = {L_1} - \left( {\frac{{\left( {{N_2}\phi } \right)}}{2}} \right)\left( {\frac{{{N_1}}}{{{N_2}}}} \right)\) 

\({L_{l1}} = {L_1} - \left( {\frac{{{N_1}}}{{{N_2}}}} \right)M\)

Similarly,

\({L_{l2}} = {L_2} - \left( {\frac{{{N_2}}}{{{N_1}}}} \right)M\)

Calculation:

L₁ = 45 mH, L₂ = 30 mH, M = 20 mH

L_l1 = 40 – 2(20) = 5 mH

L_l2 = 30 – 0.5 (20) = 20mH

Leakage inductance of secondary referred to primary = 4L_l2 = 80 mH

And the magnetizing inductance L_m (referred to primary) = L₁ – L_l1 = 45 mH – 5 mH = 40 mH