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What is the formula of the kVA input if the rating of the machine is given in horsepower?
kVA input = horse power / (0.746 * efficiency * power factor)
kVA input = horse power * 0.746 * efficiency * power factor
kVA input = (horse power * 0.746) / (efficiency * power factor)
kVA input = (horse power * 0.746 * efficiency) / power factor
kVA input = (horse power * 0.746) / (efficiency * power factor)
The kVA input of a machine is the apparent power drawn from the source, calculated by dividing the real power output (converted to Watts) by the product of efficiency and power factor ┬╖ Since 1 HP is equivalent to 746 Watts, the electrical input power (in VA) is obtained by dividing the mechanical output in Watts by (\eta \times$$\cos$$\phi).
The kVA input of a machine is the apparent power drawn from the source, calculated by dividing the real power output (converted to Watts) by the product of efficiency and power factor ┬╖ Since 1 HP is equivalent to 746 Watts, the electrical input power (in VA) is obtained by dividing the mechanical output in Watts by (\eta \times$$\cos$$\phi).
Pin(kW)тАЛ=╬╖HP├Ч0.746тАЛ тАФ Conversion from HP to kW input
kVAinputтАЛ=cos╧ХPin(kW)тАЛтАЛ тАФ Relationship between kW and kVA
In an induction motor, the mechanical power output is PoutтАЛ=HP├Ч746. To account for losses, the input power in Watts is ╬╖PoutтАЛтАЛ. Since Pinput(Watts)тАЛ=V├ЧI├Чcos╧Х, the apparent power S (in kVA) is defined as cos╧Х├Ч1000Pinput(Watts)тАЛтАЛ. Combining these gives the final formula.
1 HP is standardized as 746 Watts in the British system.
The efficiency (╬╖) represents the ratio of mechanical output to electrical input.
Power factor (cos╧Х) accounts for the phase difference between voltage and current in AC machines.
kVA is the unit of apparent power, representing the total capacity required by the machine.
Provides a direct method to estimate current requirements for sizing circuit breakers.
Useful for calculating the total load on a distribution transformer.
Does not account for non-sinusoidal supply conditions or harmonics.
Efficiency and power factor vary with machine loading, making the value dynamic.
Sizing electrical feeders for induction motors.
Calculating demand loads in industrial power system design.
In competitive exams, always ensure the conversion factor (0.746) is applied to the horsepower first.
Option A is incorrect because the power factor and efficiency are in the denominator multiplied, but the structure is mathematically inconsistent with the derivation.
Option B represents real power multiplication rather than division, which would drastically underestimate the kVA.
C is correct тАФ The formula correctly represents the conversion of mechanical output (HP) to electrical apparent power (kVA) by accounting for both efficiency losses and the phase displacement (power factor).
Always verify if the question asks for kVA or kW; kW = kVA ├Ч power factor ┬╖ In induction motors, the input is always higher than the output due to copper and iron losses.