Equation is - E = 4.44ƒNΦm Volts.
When,
E = coils generated voltage.
ƒ = Supply Frequency
N = number of screw coil.
Weber Flux of mutual Φm = coil units.
However, the AC voltage applied transformer primary Coil sinusoidal. The results mutual Flux, Ep secondary voltage generated primary coil. coils generates AC voltage Es sinusoidal.
Generates an average voltage
Eav = N × Φm ÷ t Volts.
When,
N = number of screw coils.
Weber flux of mutual Φm = coil units.
t = time in seconds flux change
There sinusoidal Φm AC,
If, Φm gero points be up to the maximum for the second time when 1/4ƒ
ƒ = Supply Frequency.
Therefore, transformer coil
Eav = N × Φm ÷ t Volts
Again, Eeƒƒ / Eave = 1.11
This equation is a simple equation, and both primary and secondary coil is applied.
The
Ep = 4.44ƒNp Φm Volts ..............
Es = 4.44ƒNs Φm Volts ...............
When,
Np = number of primary coil screw.
Ns = secondary coil of the screw.
Transformers coil voltage value of the screw, and mutual flux Frequency depends on the supply.
For example of mathematics : -
1. A 50Hz transformer primary voltage of the screw 300 and 5000, determining the tax –
b. If the 230 V secondary voltage of the secondary screw.
Are,
ƒ = 50Hz has been invoked
Np = 5000 Qm =?
Ep = 2300 Volt Ns =?
Es = 230 Volt
A. Ep = 4.44 ƒNpΦ
Φ = Ep / 4.44ƒNp
= 2300/4.44 × 50 × 5000
= 2.09 mweb. Ans.
B. Es = 4.44 ƒNsΦ
Ns = Es / 4.44ƒΦ
= 230 / 4.44 × 50 × 2.09mweb
= 500 Ans.
2. 50KVA, 11KV/230V, 50c / s Transformer 5000 primary turn there. Cores transverse area of 85cm ² What is the maximum flux density?
Are,Rating = 50KVA
Ep = 11KV = 11000 V
Es = 230 V
Acore = 85 Cm ²
Ep = 4.44 ƒNpΦm
Φm = Ep / 4.44ƒNp
βm = Φm / Acore
= Ep / 4.44 ƒNpAcore
= 11000 / 4.44 × 50 × 5000 × 85 web / cm ²
= 0.000117 web / cm ²
= 0.11 M web / cm ² Ans.
read it emf equation of transformer
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