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240J of electric energy flows into a certain light bulb every second, and some energy is transformed into light energy while some ‘waste’ thermal energy is also generated. If 200J of thermal energy is generated each second, what is the efficiency of this bulb?

Respuesta :

asuwafohjames

The efficiency of the bulb is 16.67%.

What is efficiency?

Efficiency is the ratio of the useful power output to the total power input expressed in percentage.

To calculate the efficiency of the bulb, we use the formula below.

⇒ Formula:

  • E = (Pi/Po)×100........... Equation 1

Where:

  • Pi = Power input into the bulb
  • Po = Power output of the bulb

From the question,

Given:

  • Pi = 240 J/s
  • Po = Power input-power loss = 240-200 = 40 J/s

Substitute these values into equation 1

  • E = {(240-200)/240}100
  • E = (40/240)100
  • E = 16.67%

Hence, the efficiency of the bulb is 16.67%.

Learn more about efficiency here: https://brainly.com/question/15418098

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