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Q.

How is the efficiency of a transformer calculated .. ???

Tags: electronics, science, technology
Asked by Manoj Joshi, 10 Apr '13 12:18 pm
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Answers (5)

1.

For an transformer of 500W
Copper Loss = Ip2Rp + Is2Rs
P = VI I = P/V
For the primary:
500/230 = 2.27 Amps
For the secondary:
500/24 = 20.83 Amps
Copper loss for primary:
2.272 x 0.05 = 0.258 W
Copper loss for secondary: ...more
Source: Physics forum
Answered by Josna, 10 Apr '13 01:15 pm

 
  
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2.

Assuming 500 Watts is the transformer`s rated input power.
Set the sum of the powers in the primary equal to 500 Watts and solve for the primary current.
(primary wire resistance loss) + (Vp x Ip) + (core loss) = 500W
.05 (Ip squared) + (230V x Ip) + .07 Watts = 500 Watts
.05 (Ip squared) + (230 Ip) - 499.93 = 0
Ip = 2.173 Amps

Is = (2.173A) X (230V / 24V) = 20.821 Amps

Output power of secondary = (secondary current) X [(secondary Voltage) - (Voltage drop across the resistance of the wire in the secondary)] ...more
Answered by Piaa, 10 Apr '13 12:20 pm

 
  
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3.

There is a formula for that
Answered by chikku, 10 Apr '13 12:54 pm

 
  
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4.

Efficiency = (Output/Input)*100
output=(x)(KVA * Power Factor)
input=(x)(KVA * Power Factor)+(X)^2 copperlosses+core losses

the volue of x depend on load
x=1 for full load
x=0.75 for (3/4)th load
x=0.5 for half load
x=0.25 for quarter load
Answered by rajan, 10 Apr '13 12:22 pm

 
  
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5.

Assuming 500 Watts is the transformer`s rated input power.
Set the sum of the powers in the primary equal to 500 Watts and solve for the primary current.
(primary wire resistance loss) + (Vp x Ip) + (core loss) = 500W
.05 (Ip squared) + (230V x Ip) + .07 Watts = 500 Watts
.05 (Ip squared) + (230 Ip) - 499.93 = 0
Ip = 2.173 Amps

Is = (2.173A) X (230V / 24V) = 20.821 Amps

Output power of secondary = (secondary current) X [(secondary Voltage) - (Voltage drop across the resistance of the wire in the secondary)] ...more
Answered by Quest, 10 Apr '13 09:38 pm

 
  
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