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How much energy does the light bulb dissipate in 1 minmin when half that voltage is supplied? Express your answer in kilojoules to two significant figures.

Respuesta :

Answer:

E = 15 P₀

Explanation:

The power dissipated in a light bulb is

       P = V I

       V = I R

       P = V² / R

the power is defined by

       P = W / t

work equals energy

       P = E / t

we substitute

       V² / R = E / t

        E = V² t / R

let's reduce the time to SI units

         t = 1 min = 60 s

let's calculate the dissipated energy

     

In the exercise it does not indicate the nominal voltage of the bulb, but in general this voltage is V₀= 120 V

         

The applied voltage is half the nominal voltage

      V = V₀ / 2

      V = 120/2 = 60 V

       

       E = (V₀ / 2)² t / R

       E = ¼ t   V₀² / R

       E = ¼ 60 P₀

       E = 15 P₀

Many times the nominal power (P₀) is written on the box of the bulb

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