# How do I calculate the power output of the lightbulb?

## Homework Statement

How do I find the answer for part C?

3) A battery with an Emf of 12 volts is hooked up to a 125 ohm resistor and an ammeter. The ammeter measures a current of 75 mA.
a) Calculate the internal resistance of the battery and the actual voltage across the battery.
b) Calculate the power dissipated in the resistor and the power generated by the battery.
c) The battery is now disconnected from the 125 ohm resistor and connected to a 60 watt light bulb.
What will be the actual power output of the light bulb?

## Homework Equations

Vbattery = Emf - IRinternal
V = IR = I2R
P = iV

## The Attempt at a Solution

a) Vbattery = .075(125) = 9.375
Rinternal = (9.375 - 12) / -0.075 = 35ohms

b) Presistor = (0.075)2125 = 0.703 watts
Pbattery = 0.075(9.375) = 0.703 watts

c)I thought the power supplied to the lightbulb would be 0.703 watts, but it isn't... can anyone explain why?

SteamKing
Staff Emeritus
Homework Helper
I think you are assuming that the internal resistance of the light bulb is also 125 ohms.

The voltage of the battery and the internal resistance of it doesn't change right? So wouldn't the current be the same regardless if the 125 ohm resistor is gone? I tried to use P = IV to calculate the power output for the lightbulb

SteamKing
Staff Emeritus
Homework Helper
What if the 125 ohm resistor were replaced with a 50 ohm resistor? How would that change the current?

The current would increase to .188 amps to make up for the decrease in resistance if the voltage of the battery is the same.

CWatters
Homework Helper
Gold Member
If the 125 was replaced by a 50 Ohm the current becomes... 12/(35+50)=0.14A

For c) If the nominal specification for the bulb is 12V 60W then what will it's resistance be?

the resistance would be R = (12)^2/60 = 2.4 ohms?

CWatters
Homework Helper
Gold Member
So replace the 125R with 2.4R. Calculate the current and then the power in the 2.4 ohm resistor/bulb.

So, I get I = 12/(2.4+35) = 0.321 amps. P = I^2R = (.321)^2(2.4) = 0.247 watts. I got the wrong answer. The answer is actually 0.46 watts.

CWatters
Homework Helper
Gold Member
I agree with your figure for the power (eg 0.247W). I can't see how they get 0.46W.

Anyone else want to comment?

Edit: Does the question mention the nominal voltage rating of the bulb? I can't see how to solve it unless you assume it's a 12V bulb.

Last edited:
I'd take a guess at 120 V rating for the bulb after U.S. standard - gives a nice round number for its resistance and my napkin math shows an output of around 0.46 W.

CWatters
Homework Helper
Gold Member
Have just run the numbers and I agree...

Bulb resistance = 1202/60 = 240 Ohms

Current would be..

12/(35+240) = 43.6mA

Power would be

(43.6 * 10-3)2 * 240 = 0.457W

It's a badly worded question when the answer depends on which country you live in :-)

how would I have known that the voltage rating of the lightbulb is 120 V if it isn't stated in the question? Is there no other way to solve part c given what is in the question?

cepheid
Staff Emeritus
Gold Member
how would I have known that the voltage rating of the lightbulb is 120 V if it isn't stated in the question? Is there no other way to solve part c given what is in the question?
Since 120 V is what's used for AC mains electricity (i.e. wall power) in North America, usually when a bulb has a "60 W rating", it's meant that if you plug it into wall power, it will consume 60 W of power. From this you can infer its resistance. That's why 120 V was meant to be assumed. Obviously if you plugged it into a supply at some other voltage, a different amount of power would be drawn. That having been said, I totally agree that problems in which you are meant to assume information that isn't given are BAD.

CWatters