Physics - Resistance AS question(s)

Okay, so here is the question:
" The filament bulb of a torch is rated as 720 mW. Another torch with similar inut voltage and power contains LEDs instead of a filament bulb. The power input to the LED is 0.090W.
a) Does the filament bulb or the LED have higher resistance
b) Both torches contain similar batteries capable of 13000J. For how long after the filament bulb has run out does the LED torch continue to emit light?
c) How more efficient are the LEDs than the filament bulb?
d) suggest why
"
Okay, so for a) I thought of using the P=VI formula
so our 720mW = 0.72 w
our LED = 0.092 w
so for our Filament bulb, its power is higher than our LEDs power, but seeing as they have the same voltage (similar) the I on the filament must be higher.
therefore R = V/I
& seeing that the I on the filament is larger, the LED must have higher R, correct?

Now for b)
I presume this is a Q=I*t question from first glance but can't see how that would work.
J related to work,right?
so we could do
P=w/t
given that we have been supplied with the power of the filament & LED

Filament P= 0.72W
so
0.72=13000/t
(re-arrange)
t= 13000/0.72
t= 18055.5(rec)
t=18056 seconds
t= 1.8*10^4

LED P = 0.09W
so
0.09 = 13000/t
(re-arrange)
13000/0.09=t
t=144444.4(rec)
t=144444 seconds
t= 1.44 * 10^5

so Delta time = (1.44*10^5) - (1.8*10^4) = 126000
so 126000/60 =2100 minutes
2100 minutes / 60 = 35 hours
35/24 = 1.45 days longer
Does that look right to you guys?

c)
"How much more efficient is the LED than filament bulb?"
Brain immediately jumps to:
Efficiency = (output/input)*100
but I'm not sure how to do that. I guess I would use the
emf = energy converted/charge
but I dont have that info.
I dont really have a clue for this one, so I'll take a stab in the dark and do:
(1.44*10^5) / (1.8 * 10^4)
=8
Seems like a nice number, hesitant to use this as answer as it wants "how much more efficient" is it opposed to "how many times is it more efficient".
Maybe I'm overthinking it.

d)
I presume this is because its Power is lower?
I cant figure out how to word it.
The current on the Filament is higher and as a result has a higher power consumption. This is due to both the LED and bulb producing the same Pd and as a result the higher I in the filament bulb results in a larger power due to P = v*I
I understand this in terms of the formula however not logically, hopefully some of you will understand what I am about to say:
Your V = Pd which is how much energy is "used"/converted between two points, in this case the LED or bulb, however I have a difficult time logically proceeding from here as surely if both the LED and bulb have the same pd, the same energy is being used they surely require the same amount of energy? & infact neither has a larger power?
Or is your pd the amount of energy needed to "push" 1 electron through, and due to the filament bulb having a larger current, essentially more electrons are being pushed, each requiring the pd which results in an overall larger energy consumption.
For example, the current on the led means only 1 electron needs to pass through per second, whereas the bulb needs 10 electrons per second. All the a-for mentioned electrons require the same energy to be pushed through the bulb/led, but because there is more of them in the filament bulb, the power is 10x higher and will expend the battery 10 x quicker as there are 10x more electrons in the filament bulb that need to be "pushed" through it.
To summarise, I am asking is the pd the energy PER electron or overall for the entire voltage passing through.
e.g.
if your v (pd) = 1
is it 1 volt per electron/charge carrier
or 1 volt for your whole current.

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get more responses. :redface:

Hopefully someone will be able to get back to you :h:

Reply 2

Joinedup

for D I think they're expecting you to suggest that there's less heat produced when running an LED than there is with a filament bulb of the same visible light output - that's about it really.