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 at a reduced voltage and the power drawn by the bulb is 36 W, what is the operating voltage of the bulb?

Answer 1

The voltage operating is the 85,20563362

Step-by-step explanation:

Power is the relation between Voltage and Current so knowing the resistance is going to be constant the equation of power can be just replacing in voltage terms:

1.
`P=I*V`

using also Law OHM

2.
`V= I*R`

`I = (V)/(R)`

Replacing 2 in the equation 1 so all the data are going to be in Voltage terms:

1.
`P= (V)/(R) *V`

`P= (V^(2) )/(R)` ⇒
`R= (V^(2) )/(P)`

`R= (110^(2) v )/(60 w)`

`R= 201,6666667` Ω

So the resistance is constant so the current is going to be the same at the other Power 36 W:

`P= (V^(2) )/(R)`

`V^(2) = R*P`

`V=√(R*P)`

`V=√(201,6666667*36 )`

`V=√(7260)` ⇒
`V= 85,20563362`