Consider a transformer shown in Fig.1 indicating various voltages and currents.
 |
| Fig. 1 Ratios of transformer |
1. Voltage Ratio
We known from the e.m.f. equations of a transformer that
E1 = 4.44 f Φm N1 and E2 = 4.44 f Φm N2
Taking ratio of the two equations we get,
This ratio of secondary induced e.m.f. to primary induced e.m.f. is known as voltage transformation ratio denoted as K,
Thus,
- If N2 > N1 i.e. K > 1, E2 > E1 we get then the transformer is called step-up transformer.
- If N2 < N1 i.e. K < 1, we get E2 < E1 then the transformer is called step-down transformer.
- If = i.e. K= 1, we get E2 = E1 then the transformer is called isolation transformer or 1:1 transformer.
2. Concept of Ideal Transformer
A transformer is said to be ideal if it satisfies following properties :i) It has no losses.
ii) Its windings have zero resistance.
iii) Leakage flux is zero i.e. 100% flux produced by primary links with the secondary.
iv) Permeability of core is so high that negligible current is required to establish the flux in it.
Key point : For an ideal transformer, the primary applied voltage V1 is same as the primary induced e.m.f. V2 as there are no voltage drops.
Similarly the secondary induced e.m.f. E2 is also same as the terminal voltage V2 across the load. Hence for an ideal transformer we can write,
No transformer is ideal in practice but the value of E1 is almost equal to V1 for properly designed transformer.
3. Current ratio
For an ideal transformer there are no losses. Hence the product of primary voltage V1 and primary current I1, is same as the product of secondary voltage V2 and the secondary current I2.
So V1 I1 = input VA and V2 I2 = output VAFor an ideal transformer,
V1 I1 = V2 I2
Key point : Hence the currents are in the inverse ratio of the voltage transformation ratio.
4. Voltage ampere rating
When electrical power is transferred from primary winding to secondary there are few power losses in between. These power losses appear in the form of heat which increase the temperature of the device.Now this temperature must be maintained below certain limiting values as it is always harmful from insulation point of view. As current is the main cause in producing heat, the output maximum rating is generally specified as the product of output voltage and output current i.e.V2 I2. This always indicates that when transformer is operated under this specified rating, its temperature rise will not be excessive. The copper loss (I2R) in the transformer depends on the current 'I' through the winding while the iron or core loss depends on the voltage 'V' as frequency of operation is constant. None of these losses depend on the power factor (cos Φ) of the load. Hence losses decide the temperature and hence the rating of the transformer. As losses depend on V and I only, the rating of the transformer is specified as a product of these two parameters VxI.
On both sides, primary and secondary VA rating remains same. This rating is generally expresses in KVA (kilo volt amperes rating).
Now V1 /V2 = I2 /I1 = K... V1 I1 = V2 I2
If V1 and V2 are the terminal voltages of primary and secondary then from specified KVA rating we can decide full load currents of primary and secondary, I1 and I2. This is the safe maximum current limit which may carry, keeping temperature rise below its limiting value.
Key point : The full load primary and secondary currents indicate the safe maximum values of currents which transformer windings can carry.
Example 1 : A single phase, 50 Hz transformer has 80 turns on the primary winding and 400 turns on the secondary winding. The net cross-sectional area of the core is 200 cm2. If the primary winding is connected at a 240 V , 50 Hz supply, determine :
i) The e.m.f. induced in the secondary winding.ii) The maximum value of the flux density in the core.
Solution
N1 = 80 , f = 50 Hz , N2 = 400 , a = 200 cm2 = 200 x 10-4 cm2
E1 = 240
K = N2 /N1 = 400/80 = 5/1
... K =E2 /E1 = E2 /240= 5/1
E2 = 5 x 240 = 1200 V
Now E1 = 4.44 f Φm N1
240 = 4.44 x 50 x Φm x 80
... Φm = 240/(4.44 x 50 x 80) = 0.01351 Wb
... Bm = Φm /a = 0.01351/(200 x 10-4) = 0.6756 Wb/m2
Example 2 : For a single phase transformer having primary and secondary turns of 440 and 880 respectively, determine the transformer KVA rating if half load secondary current is 7.5 A and maximum value of core flux is 2.25 Wb.
SolutionN1 = 440 , N2 = 880 , (I2)H.L. = 7.5 A,
fm = 2.25 mWb , E2 = 4.44 Φm f N2
Assuming f = 50 Hz,
... E2 = 4.44 x 2.25 x 10-3x 50x880 = 439.56 V
(I2)F.L. = KVA rating / E2
And (I2)H.L. = 0.5 (I2)F.L.
... (I2)H.L. = 0.5 x (KVA rating /E2 )
... 7.5 = 0.5 x (KVA rating / 439.56)
... KVA rating = 2 x 7.5 x 439.56 x 10-3
= 6.5934 KVA .....(10-3 for KVA)
Example 3 : A single phase transformer has 350 primary and 1050 secondary turns. The primary is connected to 400 V, 50 Hz a.c. supply. If the net cross-sectional area of the core is 50 cm2, calculate i) The maximum value of the flux density in the core ii) The induced e.m.f. in the secondary winding.
Solution The given value are,
N1 = 350 turns, N2 = 1050 turns
V1 = 400 V , A = 50 cm2= 50 x 10-4 m2
The e.m.f. of the transformer is,
E1 = 4.44 f Φm N1
E1 = 4.44 Bm A f N1 as Φm = Bm A
Flux density Bm = E1 / (4.44 A f N1)
= 400 / (4.44 x 50 x 10-4 x50 x 350) assume E1 = V1
= 1.0296 Wb/m2
K = N2 /N1 = 1050/350 = 3
And K = E2 /E1 = 3
... E2 = 3 x E1 = 3 x 400 = 1200 V
Sponsored links :
A transformer is said to be ideal if it satisfies following properties :
ReplyDeletei) It has no losses.
ii) Its windings have zero resistance.
iii) Leakage flux is zero i.e. 100% flux produced by primary links with the secondary.
iv) Permeability of core is so high that negligible current is required to establish the flux in it.
Power Transformers in India | Transformer Manufacturer in India