- A 2 KVA transformer has ironloss of 150 watt & full load copper loss of 250 Watt .The maximum efficiency of the transformer would occur when the total loss is-
When variable loss becomes equal to constant loss,efficiency is maximum.Thus at maximum efficiency of this transformer total loss = 150 X 2 = 300 watt.
- If the rated voltage from the power lines is applied to the primary of a single phase transformer which is operated on no-load, then-
Magnetizing current is usually non-sinusoidal as under no load condition, total current comprises magnetizing current & core loss component. But at the no load condition voltage appeared at the primary is equal to applied or rated voltage. So, input voltage will be sinusoidal but current will be non sinusoidal.
- Generally the output of the transformer is taken out from
The purpose of bushings is to provide proper insulation for the output leads to be taken out from the transformer tank.
- If the output of transformer is up to 66kv then which type of bushing is preferred?
Bushings are generally two types

1. Porcelain type used up to 66 kv

2. Condenser type used higher than 66 kv - Windings in transformer are generally made of
Windings are generally made of high grade copper.For carrying higher currents stranded conductors are used. Bare copper wires are also given enamel coating in addition to the inter turn insulation.
- For large size transformers, which will provide greater heat radiation area?
For small size the transformer, tanks are usually smooth surface but for large sizes to get greater heat radiation area without increasing the cubical capacity in the tank the surface is provided with the radiators or pipes.
- ------------------ will take care of transformer oil which doesn't exposed with atmosphere.
Transformer oil should not be exposed directly to atmosphere because it may absorb moisture and dust from environment in a very short time. To avoid this a breather is provided which contain silica gel or calcium carbide.
- The primary winding of a single phase transformer having 200 turns is excited with the following value

V(t)= 155.5 Sin 120πt+ 15.5 sin 360πt V neglecting the leakage. The rms value of the voltage isrms voltage= √{ (155.5)^{2}/2 + (15.5))^{2}/2 } = 110.5 V - The primary winding of a single phase transformer having 200 turns is excited with the following value

V(t)= 155.5 Sin 120πt+ 15.5 sin 360πt V neglecting the leakage. The frequency value of the core isω = 2 π f

120π = 2 π f ⇒ f = 60 hz - The primary winding of a single phase transformer having 200 turns is excited with the following value

V(t) = 155.5 Sin 120πt + 15.5 sin 360πt volt, neglecting the leakage. The rms value of the core flux isfrom above problems we conclude that V_{rms}= 110.5 V; f = 60 Hz

E = 4.44fΦ_{max}N

110.5 = 4.44X60XΦ_{max}X200

Φ_{max}= 2.07 mwb

Φ_{rms}= Φ_{max}/√2 = 2.07 m ⁄ 1.414 = 1.46 mwb - The no load current of a transformer is 0.282 amp and it legs behind primary voltage by 45°. The working and magnetising components are respectively
Working component = I
_{w}= 0.282 X cos45 = 0.2 A

Magnetising component l_{m}= 0.282 X sin45 = 0.2 A

i.e. option A. - A transformer of rating 2000 kVA, 250 Hz is operated at 50 Hz. Then, kVA rating should be revised to
kVA ∝ f

2000 /new kVA =K X 250/K_{1}X 50

∴ new kVA= 2000 X 50/250 =400 kVA - A 2400V /240 V ,150KVA single phase transformer has a coreloss of 1.8KW at rated voltage .Its equivalent resistance is 1.1 percent .Then the transformer efficiency at 0.8 power factor &fullload is-
here Pculoss/KVA rated=re(p.u) so Pcu loss=0.011*150=1.65 KW .η=V*I*cosφ/(V*I*cosφ+culoss+core loss)*100%=150*0.8/(150*0.8+1.8+1.65)*100%=97.2%/p>
- A single phase transformer when supplied from 220 V ,50 hz eddy current loss of 50 W ,if transformer is connected voltage 430V,50 hz then eddy current loss will be-
Pe(eddy current loss)=Kef
^{2}Bm^{2}(f->frequency,Bm->magnetic flux density) and V=4.44 fBmAn, so Pe1/Pe2=(f1*Bm1/f2Bm2)=(v1/v2)^{2}or,Pe2=Pe1(v2/v1)^{2}=50*(430/220)^{2}=191.01 Watts. - A 2 KVA transformer has iron loss of 150 Watts & full load copper loss of 250 Watt .The maximum efficiency of the transformer would occur when the total loss is-
When variable loss becomes equal to constant loss,efficiency is maximum.Thus at maximum efficiency of this transformer total loss=150*2 = 300 Watt.
- If the rated voltage voltage from the power lines is applied to the primary of a single phase transformer which is operated on noload ,then-
Magnetizing current is usually nonsinusodial at the flux density values normally used & under noload condition ,the total current comprises magnetizing current & coreloss component
- A 200KVA has iron loss of 150 watts and full load Cu loss of 250 watts. The maximum efficiency of the transformer would occur when total loss is
VARIABLE LOSS= CONSTANT LOSS (efficiency maximum).

Total loss=150*2=300W - Recommended preventive maintenance inspection time of bushings of transformer
According to IS-10118 standards maintenance inspection time of bushing is every 12 months.
- In a transformer operation which quantity remains fixed
There is no relative motion between the coils of the turns of the transformer, the frequency of the induced voltage in secondary coil is same as that in primary coil.
- EMF equation a transformer is equal to
EMF EQUATION of a transformer is E=4.44 &phi f T WHERE f= frequency in hz, &phi = flux in weber, T= no. of turns.

Design with by SARU TECH