Problem Solving Strategies

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 :)

Ravi Prakash

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

John Rennie

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.

Akash. B

hi

John Rennie

Morning

Akash. B

Morning

what is the reason for space expansion ?

John Rennie

@Akash.B Have a look at:

0

Q: Expansion from General Relativity

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.

Akash. B

can you explain it in more simple way?

John Rennie

5:28 AM

That is about as simple as it gets I'm afraid.

The real explanation involves pages of equations.

Akash. B

okay leave it then

what are forces that can act on light?

Akash. B

5:45 AM

hmm

@JohnRennie sir

Tanuj

6:17 AM

@JohnRennie Hi

John Rennie

Ok ... ?

Tanuj

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 .

John Rennie

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

Tanuj

@JohnRennie ok

John Rennie

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

Tanuj

6:20 AM

yeah

John Rennie

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

Tanuj

@JohnRennie okay

John Rennie

The percentage errors in the reciprocals are the same.

Tanuj

@JohnRennie yes

John Rennie

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

Tanuj

6:23 AM

@JohnRennie yeah

hmm , I got it . Thanks :)

John Rennie

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

Tanuj

yeah , thanks once again :)

@JohnRennie do you have time for another question ?

John Rennie

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

Tanuj

@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 ?

John Rennie

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.

Tanuj

6:37 AM

@JohnRennie yea

John Rennie

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.

Tanuj

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

John Rennie

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

Tanuj

@JohnRennie cool

Gaurang Tandon

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

Tanuj

6:40 AM

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

John Rennie

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

Tanuj

yea

John Rennie

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 $V_S = IR = (nI_L + I_L)R_S$

Tanuj

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

John Rennie

@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.

Tanuj

6:45 AM

@JohnRennie I didn't get this step. How did you write I ?

John Rennie

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.

Tanuj

why ?

John Rennie

@Tanuj huh?

Tanuj

@JohnRennie why to add them ?

John Rennie

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

Tanuj

6:47 AM

Oh no yea that's fine

lol that was embarassing

John Rennie

:-)

Tanuj

@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 ?

Ravi Prakash

@JohnRennie Aren't they same?

John Rennie

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

Tanuj

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

John Rennie

6:52 AM

And the current through $R_L$ is just $V_L/R_L$

Tanuj

@JohnRennie yea

@JohnRennie cool

John Rennie

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

Tanuj

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

John Rennie

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

@Tanuj yes

Tanuj

@JohnRennie awesome ! Thanks once again :)

John Rennie

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 $U=V$.

Tanuj

7:19 AM

@JohnRennie quick question

How many minutes in $\left(\dfrac{1}{60}\right)^{°}$

John Rennie

@Tanuj One sixtieth of what?

Tanuj

@JohnRennie edited

John Rennie

Well, how many minutes are there in one degree?

Tanuj

1/60 ?

John Rennie

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?

Tanuj

7:22 AM

@JohnRennie hmm I'm just thinking of an analog watch

$360^{°}$ means $1$ minute , right ?

John Rennie

Or the second hand?

Tanuj

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

John Rennie

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

Tanuj

Okay , so that's where I went wrong

John Rennie

Minute and second of arc

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,...

Tanuj

7:28 AM

how many minutes in a degree then ?

John Rennie

See the Wikipedia article I linked

Tanuj

60

John Rennie

Yes

Tanuj

mathforum.org/library/drmath/view/65468.html

same confusion .

John Rennie

Indeed :-)

Tanuj

7:32 AM

Thanks. I'm glad I got it out of my head , with your help of course !

:)

Madhuchhanda Mandal

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 !!!

Akash. B

9:26 AM

hi

Madhuchhanda Mandal

10:05 AM

Hello

Gaurang Tandon

12:04 PM

@MadhuchhandaMandal glad to help, you're welcome

Akash. B

1:01 PM

hi

Rick

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 $\overline{i}=\displaystyle\frac{\overline{e}}{\overline{Z}}$?

Gaurang Tandon

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

let me calculate the answer

is the answer e=6cos(100*pi*t - 53deg)?

Rick

2:33 PM

@GaurangTandon Yes, but this is at any general time t

Gaurang Tandon

@Rick yes, but so what?

Rick

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 $\overline{e} - \overline{i} \cdot(\overline{R}+\overline{X_L}+\overline{X_C}) = \overline{0}$

Gaurang Tandon

@Rick oh! then just solve for it am lazy :P

@Rick yep, right!

@Rick i don't know, never saw that vector equation before :/

Rick

Hmm..how do you solve AC questions then?

Gaurang Tandon

just use the phasor diagram

Akash. B

2:47 PM

hi

can I ask you a question?

hmm

hello anyone?

Gaurang Tandon

your questions are too conceptual, sorry but I can't help you out in them

Sid

@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

Akash. B

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

Sid

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

Akash. B

what are the forces that can influence light?

Sid

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.

Akash. B

@Sid hmm

Gaurang Tandon

@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?

Akash. B

okay then

Sid

@GaurangTandon What Scalar product?

@Akash.B if you mean to ask whether gravity is experienced by light, then, yes it is

Gaurang Tandon

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

Akash. B

2:54 PM

@GaurangTandon the site is blocking me from asking further question

Gaurang Tandon

@Akash.B please post this query on the h bar chatroom instead

Akash. B

@GaurangTandon oh I hate that room

Sid

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

Rick

@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?

Sid

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

Akash. B

2:55 PM

@Sid what are the other forces other than gravity?

Sid

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

Akash. B

hmm

Rick

ok 1 min..

Akash. B

can I know what he is asking?

hmm

Rick?

hello

anyone respond

Rick

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

Akash. B

3:00 PM

okay then

@Rick go on

Sid

@Rick Okay. Basically $X_C=0$

Rick

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?

Sid

Yeah.

Rick

How do we know we're solving that differential equation by playing around with those phasors?

"playing around" = add/subtract/multiply/divide

By the expression I got for $\overline{i}$, $i(t) = \displaystyle\frac{e_0}{\sqrt{R^2 + L^2 \omega^2}} \sin(\omega t - \tan^{-1}(\omega L/R))$

Sid

@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.

Rick

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

Sid

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...

Rick

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

Transcript for

Problem Solving Strategies