1:51 AM
@MadhuchhandaMandal uh sorry, you're right. Shielding doesn't effect potential. Sorry for my earlier message. if you do the calculation again, there should be five terms in your expression (charge-A, charge-B, A-B, self energy A, self energy B). The answer will again be the same as you expect :)

@JohnRennie Are Difference in Potential Energy and Potential Difference, the same thing?

3 hours later…
5:14 AM
@RaviPrakash the potential difference is the potential energy per unit charge. That is, if I move a charge of 1 coulomb from point A to point B then the energy needed to do this is equal to the potential difference.

hi

Morning

Morning
what is the reason for space expansion ?

@Akash.B Have a look at:
0

Why does General Relativity with no dark energy, no cosmological constant, predict that the universe should expand? All the matter in the the universe causes collapse of spacetime, not expansion. So, where does the "expansion" come from? An intuitive answer is appreciated.

can you explain it in more simple way?

5:28 AM
That is about as simple as it gets I'm afraid.
The real explanation involves pages of equations.

okay leave it then
what are forces that can act on light?

5:45 AM
hmm
@JohnRennie sir

6:17 AM
@JohnRennie Hi

Ok ... ?

If we are given $u$ and $v$ , i.e. the object and image distance respectively , how do I calculate the error in the focal length ?
$u$ and $v$ are given along with errors .

The percentage error in 1/x is the same as the percentage error in x

@JohnRennie ok

You're presumaby using $$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$$

6:20 AM
yeah

call the percentage errors in $u$, $v$ and $f$ $X_u$, $X_v$ and $X_f$.

@JohnRennie okay

The percentage errors in the reciprocals are the same.

@JohnRennie yes

So we need the error for two terms that are added. For this we need the absolute errors $\sigma_u$, etc, not the percentage errors

6:23 AM
@JohnRennie yeah
hmm , I got it . Thanks :)

So the steps are:
1. start with the absolute error for $u$ and $v$, convert it to a percentage.
2. convert this percentage to an absolute error in 1/u and 1/v
3. add the squares to get the absolute error in 1/f
4. convert to percentage in 1/f
5. convert to absolute error in f
All a bit tedious, but straightforward

yeah , thanks once again :)
@JohnRennie do you have time for another question ?

@Tanuj Ask it and let's see :-)

@JohnRennie Thanks a lot
@JohnRennie I see it's a zener diode basically , but what is exactly the principle that has to be used to solve problems like this ?

A zener diode has a breakdown voltage. Above this voltage it conducts.
This means there is a maximum voltage that can be applied across it. If you you try to put more voltage across thezener diode it just conducts and pulls the voltage down again.

6:37 AM
@JohnRennie yea

If you had zero resistance for $R_s$ then as soon as the voltage increased above the breakdown voltage you'd get a huge current and burn out the diode.
So you need some resistance in series with the voltage source to limit the current.

How do I figure out the breakdown voltage here , is it $V_L$ ?

The breakdown voltage is $V_L$ because that's the voltage across the diode.

@JohnRennie cool

@Tanuj they've drawn everything in the circuit itself :P answer is (1)?

6:40 AM
@GaurangTandon yea , but I don't get what to do exactly , so..

@Tanuj The way I'd approach this is to calculate $R_L$, which is just $V_L/I_L$

yea

Then you want $$\frac{R_L}{R_S + R_L} V = V_L$$
Oh, hang on ...
You want the voltage drop across $R_S$ to be given by $$V_S = V - V_L$$
And $V_S = IR = (nI_L + I_L)R_S$

@JohnRennie okay , because above this it won't work

@Tanuj yes, if you make $R_S$ too big then it drops so much voltage you can't get your required voltage across the zener diode.

6:45 AM
@JohnRennie I didn't get this step. How did you write I ?

The current through the diode is written on the diagram as $nI_L$, and the current through the load is $I_L$. Just add the two currents.

why ?

@Tanuj huh?

@JohnRennie why to add them ?

The current flowing through $R_S$ divides in two when it meets the junction between the diode and the load

6:47 AM
Oh no yea that's fine
lol that was embarassing

:-)

@JohnRennie Thanks , I got it :)
@JohnRennie I'm wondering what if the current through the diode wasn't given , could we somehow find it out ourselves ?

@JohnRennie Aren't they same?

@Tanuj Yes, because the voltage drop across $R_S$ has to be $V - V_L$. So you can calcukate the current through $R_S$.

@RaviPrakash Potential at center of a solid uniformly charged sphere is $1.5$ times the potential that at its surface.

6:52 AM
And the current through $R_L$ is just $V_L/R_L$

@JohnRennie yea
@JohnRennie cool

@RaviPrakash the potential energy when you move a charge $Q$ through a potential difference $V$ is $QV$.

so $$\dfrac{V-V_L}{R_S}-\dfrac{V_L}{R_L}$$

@RaviPrakash So the energy and the voltage are not the same
@Tanuj yes

@JohnRennie awesome ! Thanks once again :)

6:55 AM
@RaviPrakash that's why the voltage is the potential energy per unit charge, because if we write $U = QV$ then when $Q=1$ we get $U=V$.

7:19 AM
@JohnRennie quick question
How many minutes in $\left(\dfrac{1}{60}\right)^{°}$

@Tanuj One sixtieth of what?

@JohnRennie edited

Well, how many minutes are there in one degree?

1/60 ?

Perhaps you need to step back and clarify what you are asking. Are you asking how many minutes it takes the minute hand on a clock to sweep through one degree?

7:22 AM
@JohnRennie hmm I'm just thinking of an analog watch
$360^{°}$ means $1$ minute , right ?

Or the second hand?

@JohnRennie nope , but that is what I was what I was thinking to get a relation between a degree and a minute

I suspect you're hunting a snark here. There is no relation between a minute of time and a minute of angle.

Okay , so that's where I went wrong

A minute of arc, arcminute (arcmin), arc minute, or minute arc is a unit of angular measurement equal to 1/60 of one degree. Since one degree is 1/360 of a turn (or complete rotation), one minute of arc is 1/21600 of a turn. A minute of arc is π/10800 of a radian. A second of arc, arcsecond (arcsec), or arc second is 1/60 of an arcminute, 1/3600 of a degree, 1/1296000 of a turn, and π/648000 (about 1/206265) of a radian. These units originated in Babylonian astronomy as sexagesimal subdivisions of the degree; they are used in fields that involve very small angles, such as astronomy, optometry,...

7:28 AM
how many minutes in a degree then ?

See the Wikipedia article I linked

60

Yes

same confusion .

Indeed :-)

7:32 AM
Thanks. I'm glad I got it out of my head , with your help of course !
:)

1 hour later…
8:51 AM
@GaurangTandon Ya! Now looks perfect!!
@GaurangTandon They matched. I checked that yesterday. Actually I forgot the Shell-Shell Pair in Calculation , so the answer was not matching !!!

9:26 AM
hi

10:05 AM
Hello

2 hours later…
12:04 PM

1:01 PM
hi

1:18 PM
I am getting really confused in this question, we can find the total circuit impedance by $\overline{Z}= \overline{R}+\overline{X_L}+\overline{X_C}$
But what about potential difference across AB? Shouldn't it be $-\overline{i}\cdot(\overline{X_L}+\overline{X_C})$?
where $\overline{i}=\displaystyle\frac{\overline{e}}{\overline{Z}}$?

2:15 PM
I think you've to take the cosine component of the net V vector for the magnitude
and for the phase, subtract as many degrees as you took the cosine of
this I do by looking at the phasor diagram
is the answer e=6cos(100*pi*t - 53deg)?

2:33 PM
@GaurangTandon Yes, but this is at any general time t

@Rick yes, but so what?

So did you get the impedance between A and C as 6Ω (-90deg) ?
They asked the value of the pd at a particular time when it's half of e
@GaurangTandon In general, if I have an RLC circuit connected in series to an AC emf source, is $\overline{e} - \overline{i} \cdot(\overline{R}+\overline{X_L}+\overline{X_C}) = \overline{0}$

@Rick oh! then just solve for it am lazy :P
@Rick yep, right!
@Rick i don't know, never saw that vector equation before :/

Hmm..how do you solve AC questions then?

just use the phasor diagram

2:47 PM
hi
can I ask you a question?
hmm
hello anyone?

@Akash.B you don't need permission to ask questions here. This room was created to handle questions.
In any case, always ask any questions you have. Someone or the other at some stage will probably answer them

@GaurangTandon Okay I will try stop asking such type of question

@GaurangTandon That's basically KVL applied across the load.

what are the forces that can influence light?

2:51 PM
@Rick Z is 10 ohms. For Potential difference, just multiply I and The reactance across the inductor and capacitor.
Always use phasor diagrams.

@Sid hmm

@Akash.B i meant, don't stop asking at all. Just post, someone will answer if they know :)
@Sid oh, then what's the scalar product for?

okay then

@GaurangTandon What Scalar product?
@Akash.B if you mean to ask whether gravity is experienced by light, then, yes it is

@Sid the scalar product of current vector with impedances' sum

2:54 PM
@GaurangTandon the site is blocking me from asking further question

@GaurangTandon oh I hate that room

@GaurangTandon Isn't that what Rick wrote? Basically apply KVL.

@Sid Ok, also: I know that phasors are used to simplify the calculations, I also get why current in a capacitor leads the voltage by 90deg, and lags for an inductor (we can see that by differentiating/integrating), but how do we know it will solve differential equations too?

Be very careful about the complex terms. People tend to make mistakes there.

2:55 PM
@Sid what are the other forces other than gravity?

@Rick differential equations? I don't quite understand what you are asking here..

hmm

ok 1 min..

can I know what he is asking?
hmm
Rick?
hello
anyone respond

Suppose I have an RL circuit connected in series to an AC emf like this:

3:00 PM
okay then
@Rick go on

@Rick Okay. Basically $X_C=0$

then, we would say $\overline{Z} = \sqrt{R^2 + L^2 \omega^2}\angle\tan^{-1}(\omega L/R)$
or $\overline{i} = \displaystyle\frac{e_0}{\sqrt{R^2 + L^2 \omega^2}}\angle\tan^{-1}(-\omega L/R)$
but without using phasors, we would have written $+e_0\sin(\omega t)-iR - L\frac{\text{d}i}{\text{d}t} = 0$ and try to solve that differential equation, right?

Yeah.

How do we know we're solving that differential equation by playing around with those phasors?
By the expression I got for $\overline{i}$, $i(t) = \displaystyle\frac{e_0}{\sqrt{R^2 + L^2 \omega^2}} \sin(\omega t - \tan^{-1}(\omega L/R))$

@Rick In a phasor, all you are doing is vector calculations. You are imagining Voltages, Currents and Impedences to be vectors and then just performing basic arithmetic on them.
@Rick this is correct, yeah.

3:13 PM
Is it the solution to the differential equation $+e_0\sin(\omega t)-iR - L\frac{\text{d}i}{\text{d}t} = 0$?
I don't know how to solve such differential equations
@Sid Yep, I was asking by doing all that how do we know that it's solving the differential equation

Probably JR can explain better on this.
@Rick you don't need to solve these equations. You have to just use Ohm's Law and/or KVL and KCL...

I applied KVL for the loop and got that
Oh no..I looked up the solution to that equation and it's slightly different youtu.be/aiW3Qg_UVoE?t=541 According to his answer, the phasor answer is correct if c=0