The transformation ratio is defined as the ratio of the secondary voltage to primary voltage. It is denoted by the letter K.
Example 1. A 40 kVA, single phase transformer has 400 turns on the primary and 100 turns on the secondary, The primary is connected to 2000 V, 50 Hz supply. Determine:
i. The secondary voltage on open circuit.
ii. The current flowing through the two windings on full-load.
iii. The maximum value of flux.
Solution. Rating = 40 kVA
Primary turns N1 = 400
Secondary turns, N2 = 100
Primary induced voltage, E1 = V1 = 2000 V
(i) Secondary voltage on open, circuit V2:
Example 2. The no-load ratio required in a single-phase 50 Hz transformer is 6600/600 V. If the maximum value of flux in the core is to be about 0.08 Wb, find the number of turns in each winding.
Solution. Primary E1 = V1 = 6600 V
Secondary, E2 = V2 = 600V
Maximum value of flux ɸmax = 0.08 Wb.
Example 3. A single-phase transformer is connected to a 230 V, 50 Hz supply. The net cross-sectional area of the core is 60 cm2. The number of turns in the primary is 500 and in the secondary 100. Determine:
(i) Transformation ratio.
(ii) E.m.f. induced in secondary winding.
(iii) Maximum value of flux density in the core.
Solution. Primary turns, N1 = 500
Secondary turns, N2 = 100
Primary, E1 = V1 = 230 V
Core area, α = 60 cm2 = 60 × 10-4 m2
(i) Transformation ratio, K:
Hence, K = 0.2. (Ans.)
(ii) Maximum value of flux density, Bmax:
Using the e.m.f. equation, E1 = 4.44fɸmax N1
230 = 4.44 × 50 × ɸmax × 500
Example 4. 3300/300 V single-phase 300 k VA transformer has 1100 primary turns. Find:
i. Transformation ratio.
ii. Secondary turns.
iii. Voltage / turn.
iv. Secondary current when it supplies a load of 200 k W at 0.8 power factor lagging.
Solution, Primary, E1 = 3300 V
N1 = 1100
Secondary, E2 = 300 V
Rating of the transformer = 300 kVA
Output = 200 kW
Example 5. The voltage per turn of a single-phase transformer is 1.1 V. When the primary winding is connected to a 220 V, 50 Hz A.C. supply, the secondary voltage is found to be 550 V. Find:
(i) Primary and secondary turns.
(ii) Core area if the maximum flux density is 1.1 T.
Solution. Voltage per turn = 1.1 V
Primary, E1 = 220 V
Secondary, E2 = 550 V
Max. flux density, Bmax = 1.1 T
Example 6, The core of 1000 kVA, 11000/550 V, 50 Hz, single-phase transformer has a cross-section of 20 cm × 20 cm. If the maximum core density is not to exceed 1.3 tesla, calculate:
(i) The number of h.v. and l.v. turns per phase.
(ii) The e.m.f. per turns.
Assume a stacking factor of 0.9.
A = 20 cm × 20 cm = 400 × 10-4 m2 ;
Bmax = 1.3 Wb/m2,
f = 50 Hz ;
E1 = 11000 V
E2 = 550 V
Stacking factor = 0.9
(i) Number of turns N1 (h.v.) ; N2 (l.v.)
We know that, Flux = flux density × area of cross-section × stacking factor
ɸmax = 1.3 × 400 × 10-4 × 0.9 = 0.0468 Wb
E1 = 4.44fɸmax N1
11000 = 4.44 × 50 × 0.0468 × N1