Question

A 3HP motor requires 2.4 kW to drive it. Its efficiency is about:
A. 90%
B. 75%
C. 60%
D. 50%

Answer 1

Hint: The efficiency of the electrical instruments is the ratio of output electric power to the input electric power. Use \[1\,hp = 746\,W\] to convert the output power into watts. To determine the percentage efficiency, multiply the final answer by 100.

Formula used:
\[\eta = \dfrac{{{P_o}}}{{{P_i}}}\]
Here, \[{P_o}\] is the output power and \[{P_i}\] is the input power.

Complete step by step answer:
We know that the efficiency of any electrical instrument is the ratio of output power to input power.
\[\eta = \dfrac{{{P_o}}}{{{P_i}}}\]

Here, \[{P_o}\] is the output power and \[{P_i}\] is the input power.

We have given that the output power of the motor is 3hp. We also know that \[1\,hp = 746\,W\].
Therefore, the output of the electrical motor is,
\[3 \times 746\,W = 2238\,W\]

We have given that the input power to drive the 3hp motor is 2.4 kW. Therefore, we can calculate the efficiency of the motor by substituting \[2238\,W\] for \[{P_o}\] and 2.4 kW for \[{P_i}\] in the above equation for efficiency.
\[\eta = \dfrac{{2238\,W}}{{2.4\,kW}} = \dfrac{{2238\,W}}{{2.4 \times {{10}^3}\,W}}\]
\[ \Rightarrow \eta = 0.93\]

We have to calculate the percentage efficiency of the motor. Therefore, multiply by 100 to the above quantity.
\[\eta = 0.93 \times 100\]
\[ \Rightarrow \eta = 93\% \approx 90\% \]

So, the correct answer is “Option A”.

Additional Information:
You may wonder if the input of the motor is 2.4 kW and output is 2.238 kW, then where 0.16 kW of input power has gone missing. This is the loss in the input power due to heating of the parts of the electrical motor. For the motors with higher efficiency, this loss is negligible.

Note:
Note that to determine the efficiency of the electrical instruments, both input power and output power must have the same unit. The efficiency is the dimensionless quantity. Hp is a mechanical unit of power used in large industries and it is an abbreviation of horse power given by James Watt. Watt is the unit of power for small electrical components like electric bulbs.