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
Show Hint
- 4
- 8.4
- 2
- 6.2
The Correct Option is C
Approach Solution - 1
The input power to the motor is given by: \[ P_{input} = V \times I \] where \( V \) is the voltage and \( I \) is the current. Given \( V = 100 \, \text{V} \) and \( I = 1 \, \text{A} \), \[ P_{input} = 100 \, \text{V} \times 1 \, \text{A} = 100 \, \text{W} \] The efficiency \( \eta \) of the motor is given by: \[ \eta = \frac{P_{output}}{P_{input}} \] Given \( \eta = 91.6% = 0.916 \), we can find the output power: \[ P_{output} = \eta \times P_{input} = 0.916 \times 100 \, \text{W} = 91.6 \, \text{W} \] The power loss in the motor is the difference between the input power and the output power: \[ P_{loss} = P_{input} - P_{output} = 100 \, \text{W} - 91.6 \, \text{W} = 8.4 \, \text{W} \] We need to convert the power loss from watts to calories per second (cal/s).
We know that 1 calorie (cal) is equal to 4.184 Joules (J).
Since power is the rate of energy transfer (1 W = 1 J/s), we have: \[ 1 \, \text{W} = 1 \, \text{J/s} = \frac{1}{4.184} \, \text{cal/s} \] So, the power loss in cal/s is: \[ P_{loss} (\text{cal/s}) = 8.4 \, \text{W} \times \frac{1}{4.184} \, \text{cal/s/W} \] \[ P_{loss} (\text{cal/s}) \approx 2.0076 \, \text{cal/s} \] Rounding to the nearest whole number, the loss of power is approximately 2 cal/s.
Approach Solution -2
To find the loss of power in units of cal/s for the motor, we need to understand the relationship between input power, output power, efficiency, and power loss. Here is a step-by-step explanation:
- Firstly, calculate the input power of the motor. The input power \( P_{\text{input}} \) is calculated using the formula: \(P_{\text{input}} = V \times I\). Here, \( V = 100 \text{ V} \) (voltage) and \( I = 1 \text{ A} \) (current).
\(P_{\text{input}} = 100 \times 1 = 100 \text{ W}\). - Next, note the efficiency of the motor, which is given as \( 91.6\% \). Efficiency (\( \eta \)) is calculated as: \(\eta = \frac{P_{\text{output}}}{P_{\text{input}}} \times 100\%\).
- Rearrange the formula to find the output power \( P_{\text{output}} \): \(P_{\text{output}} = \frac{\eta \times P_{\text{input}}}{100}\).
Substitute the given values: \(P_{\text{output}} = \frac{91.6 \times 100}{100} = 91.6 \text{ W}\). - Determine the power loss, which is the difference between the input and output power: \(P_{\text{loss}} = P_{\text{input}} - P_{\text{output}}\).
Calculating this gives: \(P_{\text{loss}} = 100 - 91.6 = 8.4 \text{ W}\). - Convert the power loss in watts to calories per second. We know that \( 1 \text{ W} = 0.239 \text{ cal/s} \).
Thus, \(8.4 \text{ W} = 8.4 \times 0.239 \text{ cal/s}\). - Calculate: \(P_{\text{loss in cal/s}} = 8.4 \times 0.239 = 2.0076 \approx 2 \text{ cal/s}\).
Therefore, the power loss in the motor is approximately 2 cal/s, making option 2 the correct answer.
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