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

1. Apr 21, 2013

### jawhnay

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
Vbattery = Emf - IRinternal
V = IR = I2R
P = iV

3. 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?

2. Apr 21, 2013

### SteamKing

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

3. Apr 21, 2013

### jawhnay

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

4. Apr 22, 2013

### SteamKing

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

5. Apr 22, 2013

### jawhnay

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

6. Apr 22, 2013

### CWatters

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?

7. Apr 22, 2013

### jawhnay

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

8. Apr 22, 2013

### CWatters

So replace the 125R with 2.4R. Calculate the current and then the power in the 2.4 ohm resistor/bulb.

9. Apr 22, 2013

### jawhnay

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.

10. Apr 22, 2013

### CWatters

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: Apr 22, 2013
11. Apr 22, 2013

### milesyoung

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.

12. Apr 22, 2013

### CWatters

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 :-)

13. Apr 22, 2013

### jawhnay

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?

14. Apr 22, 2013

### cepheid

Staff Emeritus
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.

15. Apr 23, 2013

### CWatters

jawhnay - out of interest what country are you in?

16. Apr 25, 2013

### jawhnay

@cwatters I'm from the united states.