Question:medium

An engine delivers \(1000\) watt of power with \(80%\) efficiency. The input power is

Show Hint

Efficiency \(=\frac{\text{output}}{\text{input}}\). So input is always greater than output when efficiency is less than \(100%\).
  • \(800\,W\)
  • \(1000\,W\)
  • \(1250\,W\)
  • \(1500\,W\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
Efficiency of a machine (like an engine) is the ratio of the useful output power to the total input power. It is usually expressed as a percentage.
Step 2: Key Formula or Approach:
The formula for efficiency ($\eta$) is: \[ \eta = \frac{\text{Output Power}}{\text{Input Power}} \] We are given the output power and the efficiency, and we need to find the input power. We can rearrange the formula: \[ \text{Input Power} = \frac{\text{Output Power}}{\eta} \] Step 3: Detailed Explanation:
Given: - Output Power = 1000 W. - Efficiency, $\eta = 80% = \frac{80}{100} = 0.8$. Using the rearranged formula: \[ \text{Input Power} = \frac{1000}{0.8} \] \[ \text{Input Power} = \frac{1000}{8/10} = \frac{1000 \times 10}{8} = \frac{10000}{8} \] \[ \text{Input Power} = 1250 \text{ W} \] Step 4: Final Answer:
The input power required by the engine is 1250 W. Therefore, option (C) is correct.
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