Question

An electric motor operates on a 50V supply and draws a current of 12 A. If the motor yields a mechanical power of 300 W, what is the percentage efficiency of the motor?

Answer 1

The percentage efficiency of the motor is 50%.

Explanation:

The power of the motor, operates on a voltage supply and draws

a current is given as

$P_{in}  = V\times I$

Here, V is the voltage and I is the current.

Given V =50 V and I =12 A.

Substitute the given values, we get

$P_{in}  = 50V\times 12 A$

$P_{in}  = 600 W.$

The percentage efficiency of the motor is given as

$efficiency=\frac{P_{out} }{P_{in} } \times100\%$

As $P_{out} =300W$.

So,

$efficiency=\frac{300W }{600W} } \times100=50\%$

Thus, the percentage efficiency of the motor is 50%.

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Answer 2

Motor is 50% efficient.

Explanation:

Input voltage of the electric motor = 50 V

Input current = 12 A.

As per power law equation, we have P = I × V where P is the power, I is the current and V is the voltage.

Therefore, input power = input current * input voltage = 12 * 50 = 600 W

Given that Output power = 300 W

Percentage of efficiency of the motor = Output power / input power = 300 / 600 * 100 = 0.5 * 100 = 50%