What Resistor Value Is Needed to Power a 1.4W, 4.0V Bulb with a 9.0V Battery?

[alwaysdazed]

Homework Statement

A flashlight bulb rated at 1.4W and 4.0V is operated by a 9.0V battery. To light the bulb at its rated voltage and power, a resistor is connected in series as shown in the figure. ( i don't know how to paste the picture but its only a resistor and then light bulb in series with a 9v battery attached)

What value should the resistor have?
Express your answer using two significant figures.

Homework Equations

P=IV
V=IR

The Attempt at a Solution

I know that when resistors are in series the current is equal throughout.. so i used P=IV to solve for current, then pluggled that into V=IR using to find R but that is obviously wrong i am going crazy!

 

[tiny-tim]

Welcome to PF!

Hi alwaysdazed! Welcome to PF!

A flashlight bulb rated at 1.4W and 4.0V is operated by a 9.0V battery. To light the bulb at its rated voltage and power, a resistor is connected in series as shown in the figure. ( i don't know how to paste the picture but its only a resistor and then light bulb in series with a 9v battery attached)

What value should the resistor have?

Hint: the flashlight will only take 4.0V …

so you need to use up 5.0V on the reisistor.

use Kirchhoff's rules for potential differences.

 

[nlightner]

As a scientist, it is important to approach problems in a systematic and logical manner. In this scenario, the first step would be to identify the known quantities and the unknown quantity. In this case, the known quantities are the power rating of the bulb (1.4W), the voltage rating of the bulb (4.0V), and the voltage of the battery (9.0V). The unknown quantity is the value of the resistor (R).

Next, we can use the equations P=IV and V=IR to solve for the unknown quantity. Since we are trying to find the value of the resistor, we can rearrange the equations to solve for R. Using the power equation, we can solve for current (I) by dividing both sides by voltage (V), giving us I = P/V. Then, using the voltage equation, we can substitute in the values we know and solve for R. The equation would look like this: R = V/I = (9.0V)/(1.4W/4.0V) = 25.7 Ω.

Therefore, the value of the resistor should be approximately 25.7 Ω in order to light the bulb at its rated voltage and power. It is important to note that this answer is only an approximation, as the actual value of the resistor may vary slightly due to factors such as the internal resistance of the battery and the tolerance of the resistor.