Question

The efficiency of an electric motor is 80%. If it supplies 6 kW of rotational output power, calculate the amount of electrical input power consumed. (A)7.0kW (B)7.5 kW (C)8.0 kW (D)8.5 kW

Answer 1

(B)

`\eta=7.5kW`

Step-by-step explanation:

It is actually an easy problem, the energy efficiency is a dimensionless number, which is the report that indicates what can be recovered profitably from the machine from what has been spent to make it work. It is defined as:

`\eta=(P_o_u_t)/(P_i_n)`

Where:

`P_o_u_t=` Output power.

`P_i_n=` Input power.

Replacing the data given by the problem:

`0.8=(6000)/(P_i_n)`

Isolating
`P_i_n`

`P_i_n=(6000)/(0.8) =7500W=7.5kW`