Question

A motor operating on 100 V draws a current of 1 A. If the efficiency of the motor is 91.6%, then the loss of power in units of cal/s is

• A. 4
• B. 8.4
• C. 2
• D. 6.2

Hint:

First, calculate the input power to the motor using \( P_{input} = VI \). Then, use the efficiency to find the output power \( P_{output} = \eta P_{input} \). The power loss is \( P_{loss} = P_{input} - P_{output} \). Finally, convert the power loss from watts (J/s) to calories per second (cal/s) using the conversion factor 1 cal = 4.184 J.

Answer 1

The Correct Option is C

The input power to the motor is calculated as \( P_{input} = V \times I \). With \( V = 100 \, \text{V} \) and \( I = 1 \, \text{A} \), the input power is \( P_{input} = 100 \, \text{V} \times 1 \, \text{A} = 100 \, \text{W} \). The motor's efficiency \( \eta \) is given by \( \eta = \frac{P_{output}}{P_{input}} \). Given \( \eta = 91.6\% = 0.916 \), the output power is \( P_{output} = \eta \times P_{input} = 0.916 \times 100 \, \text{W} = 91.6 \, \text{W} \). Power loss in the motor is the difference between input and output power: \( P_{loss} = P_{input} - P_{output} = 100 \, \text{W} - 91.6 \, \text{W} = 8.4 \, \text{W} \). To convert the power loss from watts to calories per second (cal/s), we use the conversion factor 1 calorie (cal) = 4.184 Joules (J). Since 1 W = 1 J/s, then \( 1 \, \text{W} = 1 \, \text{J/s} = \frac{1}{4.184} \, \text{cal/s} \). Therefore, the power loss in cal/s is \( P_{loss} (\text{cal/s}) = 8.4 \, \text{W} \times \frac{1}{4.184} \, \text{cal/s/W} \approx 2.0076 \, \text{cal/s} \). Rounded to the nearest whole number, the power loss is approximately 2 cal/s.