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05:28
@JohnRennie Hi
I was studying about Wheatstone bridge and have few doubts
Hi :-)
What's the question?
First is that how current flows in ideal wire since its resistance is 0
so there is no potential difference
Electrons are particles and have a mass, and Newton's first law applies to them. If you accelerate an electron up to some velocity by applying an electric field to it then if there is no resistance that electron will just carry on moving at a constant speed. Yes?
And current is just moving electrons.
So in an ideal wire where there is no resistance the electrons move at a constant speed i.e. there is a constant current.
05:35
it is analogous to how we accelerate massless pulley in multiple pulley problems?
You mean the torque must always be zero with a massless pulley?
It's not really the same as electrons do have a mass, though it's a very small mass.
Like since its mass=0 net force on it must be zero but using F=ma a can be any number even without force
It's simply that electrons have momentum so if you start them flowing and if there is no resistance to that flow their momentum keeps them flowing.
I need to drop out for a couple of minutes to make my first coffee of the day. Back soon.
05:39
oh ok
I'm back! :-)
Ok
So about Wheatstone bridge or any circuit whenever there is 0V across why do we remove capacitor or resistor? It makes sense if resistance is high to stop momentum of electron but what if it small or ideal capacitor with 0 resistance
I mean current can still flow through that segment but it doesn't
Electrons do have a momentum, but it's a very, very small momentum. The drift velocity of electrons is typically less than 1 mm/s and their mass is about 10⁻³¹ kg.
So it's very, very easy to stop an electron!
Even a tiny resistance will do it.
I suppose you could argue about what happens with a truly ideal wire, but in real life there is no such thing.
Oo right so like in ideal capacitance we still take a pinch of practicality?
Yes :-)
05:51
Also 1 more thing wait
OK ... ?
When we say wheatstone bridge is balanced it means there is no potential difference across the central capacitor
and remove it
So is it like when ΔV=0 then c1/C2=C3/C4 or when C1/C2=C3/C4 then ΔV=0 which is correct?
Well both are true. I guess you're wondering about which is cause and which is effect i.e. does C₁/C₂ = C₃/C₄ cause ΔV = 0 or is it the other way round?
Yes?
05:58
yes
like in this image even if all capacitors have same capacitance C we can't remove the central one or can we?
I would say C₁/C₂ = C₃/C₄ causes ΔV = 0
@NOTEBook If all the capacitors are the same the voltage V' = 0
So you can remove the central capacitor without changing anything.
but if apply a battery of V' volt there so how V'=0
Oh, you mean you have a battery connected where you've written V'?
I thought it meant V' was the measured voltage...
@JohnRennie yes sorry
Obviously then the voltage across the central capacitor is V' not zero.
Adding the extra battery will change all the voltages. I'd have to sit down and do a Kirchoff calculation to find out how all the voltages changed.
06:07
Oh so derivation will change this time and ΔV≠0 even if C1/C2=C3/C4?
ok ok i should have tried that i asked without trying to derive like that my bad
Thank you!
You're welcome :-)
06:39
@JohnRennie Urgent help needed T_T
Hi :-)
What's up?
I was just using vs code with vite and react
Please say it can be fixed
If you press F2 it will go into the bios and that will show what disks the laptop thinks are attached.
So I should press f2?
Or should I force restart?
T_T
Yes, go into the bios.
F2
There's probably a diagnostic test built into the bios that you can run.
Either the hard disk has failed (worst case!), or something has damaged the boot sector
So go into the bios and see if it shows the disk present.
06:43
Run the "Storage tests"
Hard Drive/SSS Test
Storage tests (emmc, flash)
Back
which one to choose?
Hard drive test
06:45
quick chevk or extensive check
Quick check
Well that's good news, the disk is present and working.
@JohnRennie Passed
Back to main menu?
Yes. Exit and reboot and see what happens.
06:49
@Wolgwang Good! The disk is working.
Exit now?
@JohnRennie
Yes. When you exit the laptop will restart.
See if it restarts or if you get the same error.
It didn't
I am back here
Reboot the laptop (press ctrl-alt-delete or just turn it off and on again) then when it is restarting press F9 and it should show you the boot menu.
06:54
It is charging
Should i unplug it or let it be?
You can leave it plugged in
It won't make any difference
ctrl alt delete didn't work
Turn off then on I guess
it did
@JohnRennie using the power button?
06:58
hp logo
and it is loading
Hurrayyy
It is working
OK :-)
Thank you sooo muchhhh
You're welcome :-)
How old is the laptop?
That was a rollercoaster ride T_T. My heart was pounding
@JohnRennie 4 years
Not very old then ...
Wait and see if the problem recurs.
07:02
Hmm
@JohnRennie How are you though? 😅
I'm good :-)
How are you? What are you doing these days? Working as a programmer?
Take care!
@JohnRennie I am well.
End sem just ended.
Trying to learn React and backend
Ah, I must admit I've never used any of the frameworks.
:)
Do you still books or play games?
I still read lots of science fiction books :-)
07:09
Ah! You still have that Arm chair?
You used to spend the evening on it.
Yes :-)
It's my favourite way to pass the time. Sat in my chair with a good book :-)
And what about bicycling?
Yes, I still go cycling every day, though it's been raining a lot here in the UK recently and I don't like cycling in the rain.
07:12
Do you still wake up around 4am ? O_O
I guess you're at home chilling now. How much holiday do you have before the next semester starts?
@JohnRennie Nah. I am still at Dorm. I am working on some project under a professor and he is unpredictable. Call me at random times to visit him T_T. The next sem starts from 1Jan
Wow, you don't get much time off!
We normally get 2 months break after even semester.
How is it going? Are you enjoying college?
07:16
@JohnRennie Not really. Everyone is just interested in getting placed after graduation and just give exams for the sake of it. They are not really interested in tech or something. But such is life!
Ah, OK :-(
@mo-_- Hi :-)
@JohnRennie Hi :-)
To find the power dissipated by the individual resistors, I had to do $P = I_t \cdot R_i$ ?
Where $I_t$ is the total current and $R_i$ are $R_1, R_2,...$
Power dissipated in a resistor is P = I²R
Right, I wrote it wrong
where I is the current though the resistor and R is the resistance.
07:22
Ah so I don't have to use the Total current
The total power dissipated in all the resistors is just the sum of the powers dissipated in each resistor.
Oh okay
@mo-_- Correct. The I in I²R is the current in that resistor.
Ok okay, instead I didn't understand what was meant by the difference in potential "at the ends" of the resistors $R_2$ and $R_3$
I mean, I have to use this def ∆V = V(R2) - V(R3) ?
or the opposite
I'm not sure what you are asking ...
Is this the question we looked at yesterday?
07:27
21 hours ago, by mo-_-
Given the circuit in the figure, determine the current intensity through each resistor, the potential difference across resistors $R_2$ and $R_3$, the power dissipated by each resistor, and the total power supplied by the generator.
@JohnRennie yes
21 hours ago, by mo-_-
Given the circuit in the figure, determine the current intensity through each resistor, the potential difference across resistors $R_2$ and $R_3$, the power dissipated by each resistor, and the total power supplied by the generator.
Yes :)
When it asks for the potential difference across a resistor it is asking for:
ΔV = IR
where I is the current through the resistor and R is the resistance.
i.e. the difference in the potential between the two ends of the resistor.
Oh ok so we use Ohm's law
20 hours ago, by John Rennie
user image
@mo-_- Yes, so the potential difference across R₂ is given by ΔV₂ = I₂R₂
And the potential difference across R₃ is ΔV₃ = (I₁ - I₂)R₃
07:32
Then do i have sum them together?
@JohnRennie But when its used the formula to calculate for example 2 resistors in parallel
@mo-_- Now, I think it's just asking for the voltages separately.
In fact the voltages across R₂ and R₃ will be the same because they are connected in parallel.
Oh okay
Thank you :-)
 
3 hours later…
10:22
@JohnRennie Hi
Hi :-)
I had a few theory based doubts on circular motion.
OK ... ?
Like I was doing this question. While calculating the centrifugal force of a rotating body, do we take the rel angular velocity (the angular velcoity of the body wrt the rotating frame)?
I think I would have to see the question as it isn't clear to me what you are asking.
10:25
And the distance? Will it be the distance of the body from the axis of rotation? I don't understand centrifugal force well in such cases.
okay, sure
This was the question. I'm having trouble with the first statement.
It is a tricky question.
The way I would do this is to say that in the rest frame of the cylinder Abhi has an acceleration or rω².
But John is not in an inertial frame. He is in an accelerating frame so in his frame there will be fictitious forces.
OK so far?
So in John's frame the force on Abhi is the real force mrω² plus another fictitious force mrω² due to John being in a non-inertial frame.
So in John's frame the total force on Abhi is 2mrω²
But the question doesn't make clear whether it means you to count fictitious forces or not ...
The answer is option (C)..
They say statement 1 is true
In that case they meant you to not count the fictitious force and only count the real force.
But the question did not make that clear.
10:33
Okay.. I didn't understand this part. We have to find out the centrifugal force, right? Could you please explain once again?
About the real and fictitious forces
Basically it's a terrible question. If that was in the JEE I bet it got appealed.
Yes...
Thank you
You're welcome :-)
10:58
@JohnRennie Hi
Calculate the magnetic induction field at point $O$ marked in the figure, generated by the current $I$ flowing through the circuit, with $a$ and $b$ being the given distances. Furthermore, place a small magnetic needle at $O$ with a magnetic dipole moment $m$, oriented perpendicularly to the magnetic field $B$ produced by the circuit.
Write the torque of the forces acting on it and the work done in rotating it by $90^\circ$.
$(Data: I = 2 \, \text{mA}; a = 1 \, \text{cm}; b = 2 \, \text{cm}; m = 3 \, \text{Am}; \frac{\mu_0}{4\pi} = 10^{-7} \, \frac{\text{Tm}}{\text{A}})$
Can we discuss this?

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