A transformer of efficiency 90% draws an input power of 4 kW. An elect...
Efficiency, η = Po/Pi with usual notations. Therefore, the output power is given by
Po = η Pi = 0.9 ×4000 W = 3600 W.
In the problem nothing is mentioned about the power factor cos φ. At your level the load is usually assumed to be effectively resistive so that the power factor may be assumed to be unity. Therefore we have
Po = Io2R where R is the load resistance (or, the load impedence in this case).
This gives R = Po /Io2 = 3600 /62 = 100 Ω.
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A transformer of efficiency 90% draws an input power of 4 kW. An elect...
A transformer of efficiency 90% draws an input power of 4 kW. An elect...
Solution:
Given data:
Efficiency of transformer, η = 90% = 0.9
Input power, P1 = 4 kW
Current drawn by the electrical appliance, I2 = 6 A
To find: Impedance of the device, Z2
We know that,
Efficiency of transformer, η = Output power / Input power
Or, Output power = η x Input power
Or, P2 = 0.9 x 4 kW
Or, P2 = 3.6 kW
We also know that,
Output power, P2 = V2 x I2
Therefore,
V2 = P2 / I2
Or, V2 = 3.6 kW / 6 A
Or, V2 = 600 V
Now,
Impedance of the device, Z2 = V2 / I2
Or, Z2 = 600 V / 6 A
Or, Z2 = 100 Ω
Therefore, the impedance of the device is 100 Ω.
Answer: (d) 100 Ω
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