Question:
An engine delivers \(1000\) watt of power with \(80%\) efficiency. The input power is
An engine delivers \(1000\) watt of power with \(80%\) efficiency. The input power is
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Efficiency \(=\frac{\text{output}}{\text{input}}\). So input is always greater than output when efficiency is less than \(100%\).
Updated On: May 7, 2026
- \(800\,W\)
- \(1000\,W\)
- \(1250\,W\)
- \(1500\,W\)
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The Correct Option is C
Solution and Explanation
Concept:
Efficiency is given by: \[ \eta=\frac{\text{Output power}}{\text{Input power}}\times 100 \]
Step 1: Given output power: \[ P_{out}=1000\,W \]
Step 2: Efficiency: \[ \eta=80%=0.8 \]
Step 3: Use: \[ \eta=\frac{P_{out}}{P_{in}} \]
Step 4: Therefore: \[ P_{in}=\frac{P_{out}}{\eta} \] \[ P_{in}=\frac{1000}{0.8} \] \[ P_{in}=1250\,W \] \[ \boxed{1250\,W} \]
Efficiency is given by: \[ \eta=\frac{\text{Output power}}{\text{Input power}}\times 100 \]
Step 1: Given output power: \[ P_{out}=1000\,W \]
Step 2: Efficiency: \[ \eta=80%=0.8 \]
Step 3: Use: \[ \eta=\frac{P_{out}}{P_{in}} \]
Step 4: Therefore: \[ P_{in}=\frac{P_{out}}{\eta} \] \[ P_{in}=\frac{1000}{0.8} \] \[ P_{in}=1250\,W \] \[ \boxed{1250\,W} \]
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