Question

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.

Answer 1

E = 15 P₀

Step-by-step 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