Energy dissipation over RC type setup

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Noodles141

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Ok i have this problem pic below
View attachment 68291
i can get everything except question (d) i've tried some integration of some of the know formula, well the ones given to us and combining some but nothing gives me the right answer of 23mJ. Would appreciate some help very much.
 
Hi,

The energy is 22.948mJ to be more exact, so that's the right result.

You know the energy in the cap before the start due to the voltage, you know the energy in the cap after time t=1.25ms which is due to the voltage decrease over that time period (calculate the voltage across the cap after the time period), so you know the energy that was lost from the cap during that time, and that's the energy dissipated by the resistor.
 
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Ok thank you very much was getting the value of the difference between the energy in the cap before and after switch overs. should've actually seen this since i was integrating but wasn't doing it for the interval for some stupid reason. These questions have been troubling me for some time thank you very much.
 
Hi again,


The radius R of a particle in a magnetic field is:
R=m*v/(q*B)

but it might be nice if you gave the answers they give you and also how you first approached these problems.
 
sorry forgot to post answer with it. its 5 micro m, only want the answer to b if i can get help and understand b then i think i will be able to do c. I tried using that equation R=m*v/q*B but dont you need to velocity of the electron. I may be missing something since i've never seen meV before. I also thought of using the formula F=q*v*B then once i have that i can find the acceleration towards the centre and use one of the equations for circular motion and find the radius but then again I need the velocity.
 
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Hi again,

Yes, and the v here would be the 137MeV divided by the speed of light which is around 300e6.
And the radius would tell you the separation distance, and i'm getting 5e-12 meters not 5e-6.
Note that small m must really be a capital M which stands for "Mega" or 137e6 and not 137e-3.
Do you have a physics book somewhere you are using?
Yes you can use the force too and so you can double check your result that way equating mv^2/R=F.

[LATER]
I just checked and using the second method i still get 5e-12 meters not 5e-6 meters.
You should verify these two results.

[NOTE]
See following posts for corrections.
 
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ok i got it finaly. I didnt follow you on the whole dividing by the speed of light thing but i checked out what MeV actually is and went s abit of a different route. I used the formula of kinetic energy K=1/2*m*v^2 wher K = the 137 meV and it must be meV not MeV since it doesnt work out if it isnt. to convert meV i took (137*10^-3)*(1.6*10^-19) then put that into r=m*v/q*B and i got 2.5μm then x2 and bam 5μm.
 
Hello again,


So what did you use numerically for the mass m and the velocity v?
 
Hi,


Yes, that must be correct as we get 4.99e-6 which is very close to the required 5e-6. So they must have meant 0.137 not 137e6 as i had thought.

I am happy that you figured it out
 
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