Problem Solving Strategies

sammy gerbil

00:00

But A also looks correct. Either no current flows through inductor, in which case PD across it is zero. Or current does flow but it gradually becomes steady, therefore PD becomes zero.

However, there could be an oscillation in the RC loop, resulting in a periodic current through the inductor. So the PD across inductor is not necessarily zero,

Anonymous

03:33

@sammygerbil I didn't understand how you plotted the graphs. I understood the equations.

Anonymous

Ah, understood!

Anonymous

Without the graphs.

sammy gerbil

@IceInkberry I used a spreadsheet. I started with a column of values for $\beta$. Then I calculated in turn $\sin\beta, \cos\beta$ and then using the geometrical constraint equation $\tan\alpha$.

Next I calculated $\alpha, \sin\alpha$ followed by $\alpha+\beta$ and $\sin(\alpha+\beta)$.

Anonymous

From the last constraint equation, $$\dfrac{\sin{\beta}}{\tan{\alpha}} + \cos{\beta} = \dfrac{H}{r} = \text{constant}$$

Anonymous

After, simplifying that, we get $$\dfrac{\sin{\alpha + \beta}}{\sin{\alpha}} = \text{constant}$$ Also, $$\dfrac{R}{W} = \dfrac{\sin{\alpha}}{\sin{\alpha + \beta}}$$

Anonymous

03:42

@sammygerbil That would have been tedious.

Anonymous

Here, $$H = \text{Height of base from pulley}$$ and $$r = \text{Radius of the hemisphere}$$

sammy gerbil

@IceInkberry Yes. So $R$ is constant. I didn't follow it through.

Anonymous

You were just a step away from that!

sammy gerbil

@IceInkberry Yes I was.

I would like to know if there is a simpler way of seeing why $R$ is constant.

Anonymous

@sammygerbil If I find out any, I would share it.

Anonymous

03:46

It was a nice question. I never dealt with triangular forces on which sine law could be applied. I learnt something new. Thankyou!

Loop Back

04:00

@JohnRennie r u there?

John Rennie

04:14

@LoopBack morning :-)

Loop Back

@JohnRennie do you anything about joule expansion (free expansion) of ideal gas

@JohnRennie morning

John Rennie

@LoopBack yes ... ?

Loop Back

@JohnRennie why doesn't the temperature decreases when an ideal gas undergoes free expansion. As the ideal gas does work in increasing it's volume , it's temperature should decrease

John Rennie

A gas does work when it pushes on something. In a free expansion the gas is expanding into a vacuum so it isn't pushing on anything.

Loop Back

@JohnRennie in questions during an adiabatic process gas pushes against a piston which is generally taken massless, so why is such process adiabatic and not free expansion.

@JohnRennie a massless piston is equivalent to opening value of a small container so that the gas may fill in the larger container.

John Rennie

04:28

If the piston really was massless and not connected to anything then the expansion would be free. In practice we assume the piston is connected to something e.g. a train engine, and the work is transferred through the piston to the train (or whatever).

Loop Back

@JohnRennie thanks. It really helped.

John Rennie

@LoopBack there is another way to think about this if you're interested ...

Loop Back

@JohnRennie always interested

John Rennie

A gas is made up of atoms or molecules all whizzing around in random directions at high speed. The pressure of the gas is due to these molecules hitting the walls of the container and bouncing off. You probably already know this. Yes?

Loop Back

@JohnRennie Yes

John Rennie

04:35

For an ideal gas the temperature is related to the average kinetic energy of the atoms by:

$$ \tfrac{1}{2}mv^2 = \tfrac{3}{2}kT $$

Again you probably know this.

Loop Back

Actually average translation kinetic energy

John Rennie

Yes, though for an ideal gas there is no other form of energy.

Loop Back

Really?

John Rennie

Anyhow, suppose an atom of our gas hits the piston. Some of the atom's energy is transferred to the piston so the speed of the piston is increased and the speed of the atom is decreased. This is what we mean by doing work at an atomic/molecular scale. OK so far?

Loop Back

While solving numericals we even consider diatomic gases as ideal

Ok

John Rennie

04:39

And since the velocity of the gas atoms is decreased that means the temperature of the gas is decreased. That's why the temperature falls when the gas does work. Basically kinetic energy of the gas atoms is converted to kinetic energy of the piston.

Loop Back

Ok

John Rennie

But in a Joule expansion the gas atoms are just travelling into a vacuum so they don't collide with anything. They just keep moving at a constant speed. Since their kinetic energy doesn't change the temperature doesn't change.

Loop Back

Okay

This explanation clears all my doubt about free expansion

Thanks once again

Abcd

06:30

@JohnRennie Morn

John Rennie

@Abcd morning :-)

Abcd

@JohnRennie Question Q and S.

John Rennie

@Abcd I'm just doing something. I'll be ten minutes.

Abcd

i'll wait

John Rennie

06:46

@Abcd back

Abcd

@JohnRennie please c

John Rennie

For Q isn't it just a matter of applying a voltage to an inductor and capacitor in series and calculating the current?

Likewise for S

Abcd

@JohnRennie not getting the answer

John Rennie

Should I attempt the calculation?

Abcd

@JohnRennie there is really no calculation...

just analysis right?

in inductive circuit current is pi/2 phase behind

in capacitve it is pi/2 phase ahead

John Rennie

06:52

The way I would approach this is using complex impedance. $Z_L = i\omega L$ and $Zc = 1/(i\omega C)$. Since they are in series they just add, so we get:

$$ V = IZ = e^{i\omega t}\left( i\omega L + \frac{1}{i\omega C}\right) $$

Multiply this out and the ratio of the imaginary to real part gives the phase.

A bit of quick multiplication gives:

$$ V = i (\cos\omega t + i\sin\omega t)\left( \omega L-\frac{1}{\omega C} \right) $$

Hmm, isn't that always going to give $\pm 90º$ ... ?

The sign just depends on the relative magnitude of L and C

That would be (4) then

Abcd

@JohnRennie got it silly mistake

@JohnRennie More questions

John Rennie

Yes?

Abcd

@JohnRennie In question 1, wont the resistance get shorted?

John Rennie

Yes, in position 2 it's just an LC series circuit, so the resonance condition is $\omega L = 1/(\omega C)$

Abcd

@JohnRennie Hmm did question 1 correctly then

@JohnRennie Please help with questions 2 and 3

John Rennie

07:08

@Abcd what was the problem. Did your answer not agree with the answer key?

Abcd

@JohnRennie I meant "OK, I have solved it correctly then and got the right answer! Was just verifying my method with you"

John Rennie

So for 2 $\omega = 100$, $L = 0.01$ and $C=0.01$ ?

Abcd

@JohnRennie ???

John Rennie

@Abcd I'm not sure I understand what 2 is saying about the values of the frequency, indcutance and capacitance. Is it saying use the values you got in q1?

No wait, it's using different values of L and C as the choices.

Abcd

@JohnRennie just omega of first

John Rennie

07:16

Ah wait, I think it means that the circuit is assumed to be resonant without the resistor so we have the condition $\omega^2 = 1/(LC)$

Abcd

@JohnRennie nooo

John Rennie

So the value of $\omega$ is different for each of the four options.

Abcd

@JohnRennie It is just saying use omega of question 1. Nothing else.

@JohnRennie nooo

John Rennie

Ok, we can proceed on that basis and see what happens.

The voltage across the capacitor is $e^{i\omega t}Z_C/(Z_L + Z_C + R)$ and the voltage across the resistor $e^{i\omega t}R/(Z_L + Z_C + R)$

Abcd

@JohnRennie ohh please listen

John Rennie

07:20

OK

Abcd

just call the deominator $|Z|$

And voltage is $V_o Z_c/|Z|$ and so on...

no need to bring e in the story.

John Rennie

In that case we just get:

$$ \frac{2Z_c}{|Z|} = \frac{R}{|Z|} $$

I'm not wholly convinced by this ...

Abcd

@JohnRennie you only told me this formula :(

John Rennie

@Abcd the question says $V_C = 0.5V_R$. I'm just taking this and substituting in your expression for $V$.

Maybe that's correct and it does give the correct answer ...

Abcd

Aug 16 at 6:34, by John Rennie

Likewise the voltage across the inductor is $V_0 Z_L/Z_T = V_0 (120j)/11$

John Rennie

07:26

Yes, but you wrote $|Z|$ not $Z$.

Abcd

@JohnRennie in $Z_t$ you have used $|Z|$ only

John Rennie

@Abcd in that problem it turned out that $Z$ was real ...

but that was just the way that particular problem worked out.

Abcd

@JohnRennie can you please tell me how that formula is obtained?

$V_R = \dfrac{V_o Z_R}{Z_T}$

How is this derived?

John Rennie

@Abcd if you use complex impedance everything works just like regular resistance. That's what makes it so cool. If you have three resistors in series the voltage across $R_1$ would be the usual expression $V_0 R_1/(R_1 + R_2 + R_3)$.

For an inductor, capacitor and resistor in series you just replace $R$ by $Z$ and the formula is exactly the same.

Abcd

@JohnRennie oh wow ok

@JohnRennie Now how to do 2?

John Rennie

07:34

The voltage across the capacitor and resistor wont be in phase, so I'm guessing the question means the peak voltage across the capacitor is half the peak voltage across the resistor.

So we calculate the peak voltages using the complex impedances and compare them.

Ah, it's 8:30. I have to work for about 20 minutes.

Abcd

@JohnRennie didnt get, please let me know when you are back

John Rennie

08:05

@Abcd back.

Abcd

@JohnRennie 2

John Rennie

Is the answer A?

Abcd

Yes

John Rennie

Ok, the working is a bit messy so let's go through this step by step.

We have three circuit elements with complex impedances $Z_L = i\omega L$, $Z_C = 1/(i2\omega C)$ and $Z_R = R$. $Z_C$ has a factor of two because there are two capacitors in parallel so the capacitance is $2C$. OK so far?

Abcd

@JohnRennie yes

John Rennie

08:13

Let's start with the capacitor. We need to work out $Z_C/(Z_L + Z_C + R)$ and this is:

$$ \frac{\frac{1}{i2\omega C}}{i\omega L + \frac{1}{i2\omega C} + R} $$

To simplify this I'm going to multiply the top and bottom by $i2\omega C$ to get:

$$ \frac{1}{-2\omega^2 LC +1 + i 2\omega RC} $$

Abcd

@JohnRennie is this true for any instant?

@LoopBack Please replace the images now

Loop Back

Images?

John Rennie

@Abcd The complex impedance is not dependent on time. The full equation would be $V = IZ$ and $V$ and $I$ are functions of time because they are oscillating, but I'm leaving them out for now.

Abcd

@LoopBack screenshots of your meta question.

@JohnRennie kk

John Rennie

Now let's do $Z_R/(Z_L + Z_C + Z_R)$. Using the same approach we get:

Loop Back

08:19

Done

John Rennie

$$ \frac{R}{i\omega L + \frac{1}{i2\omega C} + R} $$

And again I'm going to multiply by $i2\omega C$ to get rid of the fraction. This gives:

$$ \frac{i 2\omega RC}{-2\omega^2 LC +1 + i 2\omega RC} $$

We would need to multiply these by the current $I_0 e^{i\omega t}$ to get the voltages but the current is the same through both elements and we're going to be taking the ratio of these two so we can ignore the current.

Is all this still OK? I appreciate it all seems a bit involved.

Abcd

@JohnRennie yes

John Rennie

Note that the two equations we got are 90º out of phase because the first one has 1 on the top and the second has $i$ on the top. What we need to do is take the modulus of the two equations and look at their ratio. We want the modulus because that's what a voltmeter measures when you connect it across a circuit element.

Abcd

I ll be back in 15 minutes. its urgent.

John Rennie

OK, I'll go off and surf Facebook for a bit.

$$ 2 \left| \frac{1}{-2\omega^2 LC +1 + i 2\omega RC} \right| = \left| \frac{i 2\omega RC}{-2\omega^2 LC +1 + i 2\omega RC} \right| $$

$$ 2 \left| \frac{1}{-2\omega^2 LC +1 + i 2\omega RC} \right| = 2\omega RC \left| \frac{i}{-2\omega^2 LC +1 + i 2\omega RC} \right| $$

Abcd

09:02

@JohnRennie back\

@JohnRennie ??

John Rennie

@Abcd hi

The question says the voltage across the capacitor is half the voltage across the resistor. Yes?

Abcd

yup

@JohnRennie will that be true at every instant of time

John Rennie

@Abcd no, and that's the key point.

Abcd

@JohnRennie $:($

John Rennie

The voltage ratio could only be constant if the two voltages were in phase. Yes?

Abcd

09:06

yes

John Rennie

And in general the voltage across a capacitor and resistor in series won't be in phase. So the question must mean the modulus of the voltages are in the ratio 1:2.

i.e. if you connect a voltmeter across the capacitor then the resistor what you're going to measure is the modulus of the voltage not it's instanataneous value.

Abcd

@JohnRennie you will take modulus of such complex thing O_O ??

John Rennie

@Abcd it turns out to be really straightforward

Abcd

@JohnRennie how?

ohh

cancel the denomintors s ??

John Rennie

Look at my first equation:

32 mins ago, by John Rennie

$$ 2 \left| \frac{1}{-2\omega^2 LC +1 + i 2\omega RC} \right| = \left| \frac{i 2\omega RC}{-2\omega^2 LC +1 + i 2\omega RC} \right| $$

Abcd

09:09

@JohnRennie SO the answer is $\omega = 1/RC$ ??

John Rennie

@Abcd BOOM! :-)

Abcd

@JohnRennie got it thanks.

@JohnRennie will you be here after 3.25 hours

I have lots of doubts piled up

John Rennie

I usually go out around 1 p.m. UK time, which is in 2 hours and 50 minutes, so probably not.

Abcd

Okay I will ask @sammygerbil then.

John Rennie

I will be around later, say 5 p.m. UK time, or failing that I'll be back tomorrow.

Abcd

09:11

@JohnRennie no body uses facebook now AFAIK

user64829

@JohnRennie Morning :-)

John Rennie

@Abcd Sammy is the afternoon guy while I'm the morning guy :-)

4

@Abcd Facebook is allegedly becoming less popular with teenagers, but more popular with us old timers. All my (old timer) friends use Facebook.

@user64829 morning :-)

Abcd

@JohnRennie Oh I see.

user64829

John Rennie

@user64829 Q247?

user64829

09:14

Torque= gR(m2-km1)=Iα right?

@JohnRennie Yes

Will we get $a(tangential acceleration) = 2g(m2-km1)/m$ ?

John Rennie

@user64829 no

user64829

Umm..where did I go wrong then?

John Rennie

The torque on the pulley is the difference in the tension in the strings and the tension in the bottom part of the string is not $m_2g$. If it was that would mean $m_2$ wasn't accelerating because the tension would exactly balance the gravitational force.

user64829

Oh yeah, damn

Hold on, lemme try it again

user64829

09:36

Will the acceleration affect tension in the left part of string?

I guess no since a and g are perpendicular to each other

Am I right?

John Rennie

@user64829 yes

The acceleration of $m_1$ is given by $m_1a_1 = T_1 - m_1 g \mu$ i.e. the tension minus the frictional force.

And $a_1 = a_2$ since they are connected by a string.

I need to work for a while now. Back later ...

user64829

Sure, Thank you, got it!

user64829

09:51

@JohnRennie Are you still there?

user64829

10:04

I found excess pressure inside. It is $10^5 pascals$

Idk what to do next, tried to use P^(2-γ).T^(γ-1)=constant but it didn't work out

Loop Back

10:33

@user64829 is B the answer

And if, it is then the there's a printing mistake in the question. The expansion in volume should be 0.1m and not 1m. If you take 1m as expansion then the answer will be 14300K.

@user64829 option B means, option 2 or 800K

Abcd

11:55

@sammygerbil Please let me know when you are free

Abcd

12:35

@IceInkberry Please see ...

in option A I am getting mu_o i/ a

user64829

@LoopBack Yes it's B, I found something was wrong in printing, since it was a previous JEE problem I googled it and got the right value

But how do you do it

Anonymous

@Abcd I don't understand some things in the question.

Abcd

@IceInkberry Just check option A at least

Anonymous

@Abcd Is the magnetic field in the middle?

Abcd

@IceInkberry hmm

Anonymous

12:46

So, I'll have to integrate it from each smaller strip of dx length until b/2 length. So much why

Anonymous

Sorry, but I was a little irritated. I have forgotten some things. Need to revise!

Abcd

@IceInkberry use ampere's law?

Anonymous

@Abcd No, I give up. Sorry.

Anonymous

I don't even remember how to use that law.

Abcd

@IceInkberry Okay

@IceInkberry haha happened with me a few days ago XD

Can totally relate

Anonymous

12:53

This gives me hope! I can get on track with revision!

sammy gerbil

13:13

@Abcd I am available.

Abcd

@sammygerbil Thanks I was waiting for you desperately!!

Loop Back

@user64829 since the gas is heated slowly it's pressure will be infinitesimally greater than outside pressure. The gas will expand infinitesimally slowly and at any instant it's pressure is equal to the external pressure. When equilibrium is reached pressure of the gas $$P = P_o+\dfrac{kx}{A}$$, where $P_o$ is atmospheric pressure.

Abcd

40 mins ago, by Abcd

@sammygerbil Please seee^^^

Loop Back

@user64829 its new volume $V=V_i +Ax$

sammy gerbil

@Abcd That problem looks familiar. I have seen it before. What part are you having difficulty with?

Loop Back

13:17

@user64829 now you can use $$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$ to get your answer

Abcd

@sammygerbil For part a I am getting a in denominator not b!!!

@sammygerbil Should I show you my attempt for part a

sammy gerbil

@Abcd Yes please.

Found it : physics.stackexchange.com/questions/424708/…

Abcd

@sammygerbil See I am getting l in denominator!!

sammy gerbil

@Abcd Your Amperian Loop is perpendicular to the plane of the page, yes? While the conducting strips are in the plane of the page?

user64829

@LoopBack Why can't we use p^(1-gamma)T^(gamma) = constant? We will not have to deal with volume at all

Abcd

13:28

@sammygerbil Hmm so B.dl = 0 $:($

@sammygerbil So ampere's law is not valid here? Why??

sammy gerbil

@Abcd You also mention charge density. I am not sure what you mean by this because the charge is moving as current. Maybe you mean current density.

Abcd

@sammygerbil miswrote it. Yes i mean current density

sammy gerbil

Ampere's Law is valid but the magnetic field around the loop is not constant.

Abcd

@sammygerbil so how to do part a??

sammy gerbil

@Abcd There might be a better way but you could consider the current in each strip to be made of an infinite number of very thin wires. Label them 1, 2, 3 etc from the top of each conducting strip downwards.

Then every pair of currents with the same number forms a loop of dimensions $\ell \times (a+b)$.

The magnetic flux density $B$ inside the loop is uniform because $a \ll \ell$. (Wait : I should be comparing $b$ with $\ell$.)

Back to drawing board. ...

The diagram is misleading because the width of the strips $b$ is much greater than the gap - ie $b \gg a$. But the diagram gives the impression that $a \gt b$.

@Abcd ok my 2nd attempt is to use the same idea of the conducting strips being divided into an infinite number of wires. But this time label the wires 1, 2, 3 etc from the gap. That is, the two wires closest to the gap are #1, those 2nd closest are #2. etc.

... Sorry I am thinking as I write and finding problems with my approach.

@Abcd Possibly A is incorrect because the magnetic field between the strips (ie in the gap) is not going to be uniform. It will have different values in the horizontal direction and also in the vertical direction (even though the gap is very small).

Loop Back

13:52

@user64829 that equation is specifically for adiabatic process. Adiabatic process is a process where neither heat enters or leaves the system. Here the gas is being heated

sammy gerbil

(I presume this is a multiple-choice question. We have to select all options which are correct, and there may be more than one correct option.)

So I think a calculation is not necessary for A.

Abcd

@sammygerbil Answer is ABD

@sammygerbil what is that?

sammy gerbil

@Abcd Don't you mean "Why is that?"

Abcd

@sammygerbil No i am asking you "what is a drawing board, why are you going there?"

sammy gerbil

@Abcd dictionary.cambridge.org/dictionary/english/…

Abcd

13:56

@sammygerbil oh lol

sammy gerbil

@Abcd My answer on the main site (linked above) addresses option D.

Abcd

@sammygerbil any idea about a b

@sammygerbil I have more questions ...

sammy gerbil

@Abcd ok we can come back to this question later.

Abcd

@sammygerbil please C^^

sammy gerbil

@Abcd What do you think?

Abcd

14:09

@sammygerbil option A only should be correct $:($

sammy gerbil

@Abcd Why do you think A is correct?

Abcd

@sammygerbil Sending diagram

@sammygerbil Sorry not a. But do you think my diagram below is correct:

sammy gerbil

@Abcd No I think your diagram is incorrect.

Can you explain why you have drawn the EMFs as in your diagram?

ie Why are they in the same direction, pushing current clockwise in each of the lower 2 loops?

user64829

@LoopBack Ah right. I was confused since they mentioned everything is thermally insulated.

sammy gerbil

@Abcd EMF on right is correct. The flux through that loop is decreasing, therefore a current is induced which will increase flux. The induced current must create a magnetic field pointing downwards, therefore it must be clockwise as shown.

Abcd

14:21

@sammygerbil ????

Anonymous

~~I won't confuse~~ Edited

sammy gerbil

@Abcd Flux through loop on left is increasing. Induced current must oppose this by creating a magnetic field which points upwards. The current in this loop must flow anti-clockwise. So the EMF should point in the opposite direction.

Abcd

@sammygerbil ohhhhh no wrong analysis on my part

positive charges on ends

and negative charge at centre

@IceInkberry other way round^

Anonymous

@Abcd Ah yes, sorry. I rotated it anti clockwise.

sammy gerbil

@Abcd That cannot be the explanation. If you got the symbols wrong way round then both should be reversed. But I think only the EMF on the left should be reversed.

Anonymous

14:25

@Abcd But, that doesn't change the magnitude of current. You'll get the direction wrong.

Abcd

@sammygerbil no

sammy gerbil

@Abcd So my conclusion it that the conventional current flows upwards in the resistor in both loops. They don't cancel. So option A is incorrect.

Do you agree?

Abcd

@sammygerbil thats the same as what I said after correction

sammy gerbil

@Abcd So you agree that option A is false?

Abcd

@sammygerbil yes

sammy gerbil

14:35

@Abcd Good. What about option B?

Abcd

So we have:

$\Delta V = B\omega R^2/ 2$

For the two batteries

Now potential differences in parallel.

But same potential differences so resistor can choose any

Thus current should be:

$I = \dfrac{\Delta V}{R}$

$\implies I = \dfrac{B\omega R^2}{2r}$

@sammygerbil $:($

Anonymous

(Remember that the rod also has some resistance)

Abcd

DARN IT

Retrying

$I = \dfrac{B\omega R^2 }{2(r+ r/2)}$

sammy gerbil

@Abcd That is the current for each loop, I think.

Abcd

@sammygerbil $:( ^\infty$

sammy gerbil

14:42

So option B is also false because the denominator should be $3r$ not $5r$. Total current through stationary resistor is $I=2\omega a^2/3r$.

Abcd

56 secs ago, by Abcd

@sammygerbil $:( ^\infty$

@sammygerbil Answer is BD

sammy gerbil

@Abcd Maybe there is a mis-print : somebody wrote 3 by hand and it looked like 5.

Anonymous

I got option (B)

Abcd

3 mins ago, by sammy gerbil

So option B is also false because the denominator should be $3r$ not $5r$. Total current through stationary resistor is $I=2\omega a^2/3r$.

Anonymous

The resistors are in parallel.

Abcd

14:45

@sammygerbil I am not even getting a 2 at top

@IceInkberry please send

Anonymous

r/2, r/2 and r are in parallel.

Anonymous

I meant that current will flow from B to A*

Abcd

@IceInkberry they have asked current through r

Anonymous

@Abcd I am so dumb :D

Anonymous

14:51

Sorry.

Abcd

@IceInkberry r u being sarcastic. I am very poor at understanding sarcasm

Anonymous

No, I am not being sarcastic.

Anonymous

I didn't read it.

Abcd

@sammygerbil Please help

Anonymous

@Abcd But, I don't think this is correct.

sammy gerbil

14:56

@IceInkberry What do you think is the current in the fixed resistor?

Anonymous

@sammygerbil $$V = i × R$$ $$B\omega a^2/2 = i × r$$ $$i = B\omega a^2/2r$$

Abcd

@sammygerbil @IceInkberry I am done and have got option B as the answer :D

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