Find the power of the light bulb

[TwinCamGTS]

Homework Statement

In the figure below, you will attach one or more light bulbs of R=30.12 Ω in parallel with the battery (with ε = 20 V, internal resistance r). (i will upload with work and the picture of the wiring diagram)


a.) For an ideal battery (r=0), what would be power dissipated in the first bulb if it was:

connected alone:
P=____ watt

connected as one of four bulbs:
P=____ watt

b.) For a non-ideal battery where r=1.1 Ω,

i) reanswer part a). Also, in each case find the voltage drop that would be measured across the battery terminals:

connected alone:
P=____ watt Vbattery=____ Volt

connected as one of four bulbs:
P=____ watt Vbattery=____ volt


ii) Challenge:
Suppose you wish to dim the first bulb, such that its power output is at or below ___% of its value when connected alone. Find the minimum number of total bulbs you must connect in parallel to make this happen.
Solve for a general expression so you can use it for all the cases below.

50% of its actual value
35% of its actual value
10% of its actual value

Homework Equations

V=I*R

P=(v^2)/r p=I*v p=(i^2)*r

The Attempt at a Solution

i did mostly all of them except the challenge problem. i don't know how to start. I will be grateful if someone can tell me how to do it. Thank you for your time and your help.

 

[NascentOxygen]

Hi TwinCamGTS. Can we get clarification of this statement, and what it is supposed to be telling us:

Voltage=18v current=8amp resistance 3Ω

Additionally, can you verify that the word "parallel" is intended in all its occurrences where you wrote it. (I'm thinking maybe someone actually meant "series".)

Also, is the original question written in English, or have you or your teacher translated it from another language?

 

[CWatters]

when the light bulb is on the circuit by itself, it has the value of all the material here

Voltage=18v current=8amp resistance 3Ω

I agree. There is a problem somewhere.

The 3Ω cannot be the resistance of the bulb because the maximum current it could draw from 18V would be 18/3 = 6A not 8A.

 

[TwinCamGTS]

hi guys, i just re-edit the problem. sorry for the confusion because at first i just need the concept of how to solve the last problem which is to find how many light bulbs need to be added to the circuit on parallel so that the power output will be ____% from its actual value.
thank you for helping me out.

 

[NascentOxygen]

hi guys, i just re-edit the problem. sorry for the confusion because at first i just need the concept of how to solve the last problem which is to find how many light bulbs need to be added to the circuit on parallel so that the power output will be ____% from its actual value.
thank you for helping me out.

I haven't looked closely at this, but it sounds like they expect you to connect something in series with a bulb to lower its brightness. That "something" can be a bunch of spare bulbs connected together to form a small resistance. Apply some maths to get the numbers right.

 

[CWatters]

If the battery is non-ideal (eg has internal resistance) then the voltage will drop each time an additional bulb is added in parallel. This will cause the existing bulbs to dim.

 

[NascentOxygen]

In the part where your answer is 10.1105w, you use R of a single bulb to determine that single bulb's power. You could use the smaller R of all bulbs in parallel, but this will equate to the total power in all bulbs together, and you are not asked to determine total power.

hi guys, i just re-edit the problem. sorry for the confusion because at first i just need the concept of how to solve the last problem which is to find how many light bulbs need to be added to the circuit on parallel so that the power output will be ____% from its actual value.
thank you for helping me out.

Let the voltage across all of the bulbs be E when power is 50% of full brightness. The power in one bulb will be E² /R, so you can solve to find this E for 50% power. Then continue on...

 

[TwinCamGTS]

I use they way you described to solve it. I got 17 lights bulb needed to being the power down to 50% from its actual power. But its wrong. Just wondering, do you try to solve it and do you get the same result? Or maybe i did it wrong. I can't post my work coz I am at work and i can't attach the picture using my phone

 

[NascentOxygen]

I didn't find 17. Perhaps you overlooked the fact that voltage will fall by a different amount as you add each parallel bulb?

 

[TwinCamGTS]

here is how i do it. can you tell me where do i get it wrong?
thanks

 

[NascentOxygen]

Can you work out again the bulb power when a single bulb is connected to the non-ideal battery?

 

[TwinCamGTS]

i could not get it right, its alright. i'll just look at my teacher's answer key.
thanks for helping me