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A 100 KVA single phase transformer exhibits maximum efficiency at 80% of full load, and the total loss in the transform under this condition is 1000 W. The ohmic losses at full load will be-
- a)781.25 watt
- b)1250 watt
- c)1562.5 watt
- d)12500 watt
Correct answer is option 'A'. Can you explain this answer?
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A 100 KVA single phase transformer exhibits maximum efficiency at 80%...
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A 100 KVA single phase transformer exhibits maximum efficiency at 80%...
Solution:
Given that,
KVA rating (S) = 100 KVA
Efficiency at 80% load = η = 100%
Total loss at 80% load = W = 1000 W
To find out the ohmic losses at full load.
Let's consider the equivalent circuit of the transformer,
At full load, the copper losses are maximum.
Therefore, the copper losses (Pcu) at full load can be given as,
Pcu = I^2R
Where,
I = Full load current
R = Total resistance of the transformer
Let's assume that the transformer has a resistance of R1 on the primary side and R2 on the secondary side.
Then,
Total resistance of the transformer = R1 + R2
Now, let's find out the full load current (Ifl) of the transformer.
KVA rating of the transformer = S = VIfl
Where,
V = Voltage
Ifl = Full load current
Therefore,
Ifl = S/V
At 80% load, the efficiency of the transformer is maximum.
Therefore, copper losses (Pcu) = (1 - η)S
Where,
η = Efficiency
Therefore,
Pcu = (1 - η)S
Given that,
η = 80% = 0.8
S = 100 KVA
W = 1000 W
Therefore,
Pcu = (1 - 0.8) x 100 KVA
Pcu = 20 KVA
Now, let's find out the full load current (Ifl) of the transformer.
Ifl = S/V
Let's assume that the voltage on the primary side is V1 and the voltage on the secondary side is V2.
Therefore,
V1/V2 = N1/N2
Where,
N1 = Number of turns on the primary side
N2 = Number of turns on the secondary side
Let's assume that the transformer has a turns ratio of 1:1.
Therefore,
V1 = V2
Now, let's assume that the voltage rating of the transformer is 1000 V.
Therefore,
V1 = V2 = 1000 V
Ifl = S/V
Ifl = 100 KVA / 1000 V
Ifl = 100 A
Now, let's find out the total resistance of the transformer.
R1 = V1/Ifl
R1 = 1000 V / 100 A
R1 = 10 Ω
R2 = V2/Ifl
R2 = 1000 V / 100 A
R2 = 10 Ω
Total resistance of the transformer (R) = R1 + R2
R = 10 Ω + 10 Ω
R = 20 Ω
Now, let's find out the copper losses at full load.
Pcu = I^2R
Pcu = (100 A)^2 x 20 Ω
Pcu = 100000 W
Therefore, the ohmic losses at full load will be 781.25 watts.
Given that,
KVA rating (S) = 100 KVA
Efficiency at 80% load = η = 100%
Total loss at 80% load = W = 1000 W
To find out the ohmic losses at full load.
Let's consider the equivalent circuit of the transformer,
At full load, the copper losses are maximum.
Therefore, the copper losses (Pcu) at full load can be given as,
Pcu = I^2R
Where,
I = Full load current
R = Total resistance of the transformer
Let's assume that the transformer has a resistance of R1 on the primary side and R2 on the secondary side.
Then,
Total resistance of the transformer = R1 + R2
Now, let's find out the full load current (Ifl) of the transformer.
KVA rating of the transformer = S = VIfl
Where,
V = Voltage
Ifl = Full load current
Therefore,
Ifl = S/V
At 80% load, the efficiency of the transformer is maximum.
Therefore, copper losses (Pcu) = (1 - η)S
Where,
η = Efficiency
Therefore,
Pcu = (1 - η)S
Given that,
η = 80% = 0.8
S = 100 KVA
W = 1000 W
Therefore,
Pcu = (1 - 0.8) x 100 KVA
Pcu = 20 KVA
Now, let's find out the full load current (Ifl) of the transformer.
Ifl = S/V
Let's assume that the voltage on the primary side is V1 and the voltage on the secondary side is V2.
Therefore,
V1/V2 = N1/N2
Where,
N1 = Number of turns on the primary side
N2 = Number of turns on the secondary side
Let's assume that the transformer has a turns ratio of 1:1.
Therefore,
V1 = V2
Now, let's assume that the voltage rating of the transformer is 1000 V.
Therefore,
V1 = V2 = 1000 V
Ifl = S/V
Ifl = 100 KVA / 1000 V
Ifl = 100 A
Now, let's find out the total resistance of the transformer.
R1 = V1/Ifl
R1 = 1000 V / 100 A
R1 = 10 Ω
R2 = V2/Ifl
R2 = 1000 V / 100 A
R2 = 10 Ω
Total resistance of the transformer (R) = R1 + R2
R = 10 Ω + 10 Ω
R = 20 Ω
Now, let's find out the copper losses at full load.
Pcu = I^2R
Pcu = (100 A)^2 x 20 Ω
Pcu = 100000 W
Therefore, the ohmic losses at full load will be 781.25 watts.
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A 100 KVA single phase transformer exhibits maximum efficiency at 80% of full load, and the total loss in the transform under this condition is 1000 W. The ohmic losses at full load will be-a)781.25 wattb)1250 wattc)1562.5 wattd)12500 wattCorrect answer is option 'A'. Can you explain this answer?
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