Question

The voltage and power ratings of a particular light bulb, which are its normal operating values, are 110 V and 60 W. Assume the resistance of the filament of the bulb is constant and is independent of operating conditions. If the light bulb is operated with a current that is 50% of the current rating of the bulb, what is the actual power drawn by the bulb

Answer 1

P = 14.85 W

Step-by-step explanation:

First, we find the initial current value of the bulb by using the formula of power:

`P = VI`

where,

P = Power = 60 W

V = Voltage = 110 V

I = current = ?

Therefore,

`60\ W = (110\ V)(I)\\\\I = (60\ W)/(110\ V)\\\\I = 0.54\ A`

Now we use Ohm's Law to find the resistance of bulb:

V = IR

110 V = (0.54 A)R

R = 201.67 Ω

Now the current becomes half in actual:

I = 0.27 A

but the resistance remains same:

R = 201.67 Ω

hence, from Ohm's Law:

V = (0.27 A)(201.67 Ω)

V = 55 V

Therefore, the actual power drawn by the bulb will be:

P = VI = (55 V)(0.27 A)

P = 14.85 W