Question 503: AC fundamentals

Question

The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at 1000 V is 12 kW . The resistance of the primary is $0.2 \Omega$ and that of the secondary is $2 \Omega$. If the efficiency of the transformer is $80 \%$, then the power loss in the primary is

Accepted Answer

The given problem involves a transformer with a primary coil of 400 turns and a secondary coil of 2000 turns. The secondary side has a power output of 12 kW at 1000 V. The resistances are specified as 0.2 ÃÂ© for the primary and 2 ÃÂ© for the secondary. Given that the efficiency of the transformer is 80%, we will find the power loss in the primary winding.
First, identify the relationship between the number of turns and the voltages in the primary and secondary:
$ \frac{E_p}{E_s} = \frac{N_p}{N_s} $
Substitute the known values to find the primary voltage $ E_p $:
$ E_p = \frac{E_s 	imes N_p}{N_s} = \frac{1000 	imes 400}{2000} = 200 \, 	ext{V} $
With 80% efficiency, the input power can be calculated as follows:
$ 	ext{Input power} = 12 \, 	ext{kW} 	imes \frac{100}{80} = 15 \, 	ext{kW} $
Now, calculate the current in the primary $ I_p $:
$ I_p = \frac{	ext{Input power}}{	ext{Input voltage}} = \frac{15 \, 	ext{kW}}{200 \, 	ext{V}} = \frac{15 	imes 1000}{200} = 75 \, 	ext{A} $
Finally, determine the power loss in the primary using the formula for power loss due to resistance:
$ 	ext{Power loss in primary} = I_p^2 	imes R_p = (75)^2 	imes 0.2 = 1125 \, 	ext{W} $
Thus, the power loss in the primary is 1125 W.

Answer

288 W

Answer

1125 W

Answer

2000 W

Answer

725 W

AC fundamentals

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