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

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To power a 1.4W, 4.0V bulb with a 9.0V battery, a resistor must be used to drop the excess voltage. The bulb requires 4.0V, meaning the resistor needs to handle a voltage drop of 5.0V. Using the power equation P=IV, the current through the bulb can be calculated, which is approximately 0.35A. Applying Ohm's law (V=IR), the required resistor value can be determined to be about 14.3 ohms. This setup ensures the bulb operates at its rated voltage and power.
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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!
 
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Welcome to PF!

Hi alwaysdazed! Welcome to PF! :smile:
alwaysdazed said:
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. :wink:
 

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