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8. Ella wants to replace old bulbs of her home with new LED bulbs. She buys LED bulb that has inbuilt AC to DC converter of output 5V and 2.5A. a. If the LED is 75% efficient, what would be the power dissipated? b. Find the work done by the LED when it is connected to the power supply and left for 1hour 15 minutes.​

Sagot :

The power dissipated by the LED is 20 Watts and the work done for 1 hour 15 minutes is 56.25 kJ.

What is electrical power?

Electrical power is the rate at which electrical work is done.

  • Electrical power = voltage × current

The LED is 75% efficient means that 75% of power dissipated by the LED is converted to light.

Total power dissipated = 5 × 2.5 = 12.5 Watts

  • Work done = power × time (in seconds)

Work done = 12.5 × (1 × 3600 + 15 × 60)

Work done = 56250 J = 56.25 kJ

Therefore, the power dissipated by the LED is 20 Watts and the work done for 1 hour 15 minutes is 56.25 kJ.

Learn more about electrical power and work done at: https://brainly.com/question/23901751

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