@GuruVishnu I'll only be a couple of minutes while I wash up the cereal bowl and boil the kettle. Post the question now and I'll have a look as soon as the coffee is made.
Dielectric constant of the slab between plates of a capacitor is $18$ and its resistivity is $4\pi\times10^3~\Omega\rm m$. Then time constant of this capacitor when directly connected to a battery will be:
(a) $2~\mu\rm s$ (b) $3~\mu\rm s$
(c) $1~\mu\rm s$ (d) $9~\mu\rm s$
Book's solution and my confusion:
The author used $\tau=RC$ where $R$ is the resistance of the dielectric and $C$ is the capacitance of the capacitor with the dielectric and found the answer to be $2~\mu\rm s$. I don't understand how he can assume the resistance of this circuit to be that of the dielectric. First of all, I don't even see how a capacitor with a conducting dielectric is an RC circuit.
In a RC circuit, the resistor and the capacitor are in series, but here they are in parallel. I know this and have also learnt this from this Q&A.
So, could you please explain how to approach this question? I think the solution provided by the book is incorrect. If not, can you tell why is it correct despite the circuit is a parallel combination of a resistor and a capacitor instead of a series combination of resistor and a capacitor?
That's an interesting question, though basically a simple one.
We effectively have a capacitor and resistor in series, though seeing that is not obvious.
Hmm, I wonder what exactly the question means i.e. does charge flow between the plates of the capacitor through the dielectric? If so then yes I agree with you that it would be a capacitor and resistor in parallel.
Yes sir. That's what I interpreted from a conducting dielectric. Actually this is a second part of a two-part question. And I hope this is what is meant by the author.
But if the dielectric is not electrically connected to the plates of the capacitor then the resistance of the dielectric affects the rate at which the charge separation develops so it behaves like a series resistor.
Ok sir. Then I think there is no significance of "resistance of the dielectric" as electrons can't jump from the metal plates to the dielectric and vice versa.
Or in other words, when there's no contact between the dielectric and the plates, the capacitor acts as an ideal capacitor i.e., non conducting at steady state.
Ok sir. This is the entire section from the beginning:
Passage VII [Q. Nos. 15-16] In RC circuit, answer the following questions:
Question 15:
During charging of a RC circuit let $t_1$ and $i_1$ be the time constant and initial charging current when the capacitor is assumed to be filled by a perfect insulator and $t_2$ and $i_2$ be the corresponding values when it is assumed imperfect. Then
(a) $t_1<t_2$ (b) $i_1>i_2$
(c) both (a) and (b) are correct (d) both (a) and (b) are wrong
Question 16:
Dielectric constant of the slab between plates of a capacitor is $18$ and its resistivity is $4\pi\times10^3~\Omega\rm m$. Then time constant of this capacitor when directly connected to a battery will be:
(a) $2~\mu\rm s$ (b) $3~\mu\rm s$
(c) $1~\mu\rm s$ (d) $9~\mu\rm s$
That's it!
@JohnRennie: Do you want to see how I approached the question 15? But it would deviate us from the question 16...
I agree with you. It looks to me like a parallel circuit so the capacitor would charge to the battery voltage immediately (apart from the internal resistance of the battery).
The top resistor is the whatever resistance there is in the circuit. It assumes the capacitor isn't connected directly to a perfect voltage source. It could be the internal resistance of the PSU or a resistor deliberately placed in the circuit.
I haven't solved the equation myself, but your solution looks plausible. The dimensions all look fine and it has the right behaviour at $t = 0$ and $t = \infty$.
So yes the time constant would be $C\frac{R_1R_2}{R_1 + R_2}$
When $R_2 \to \infty$ this becomes $CR_1$ as it should.
Fine sir. I understood so far. The final result obtained is the time constant of classic RC circuit. I think we also need to consider the limit $R_1\to0$ for the sake of this question. But it seems to approach $0$ instead of $2~\mu\rm s$. Am I missing anything else sir?
Any time you want to understand what happens with a zero resistance you need to start with a small but non-zero resistance and see what happens in the limit of $R \to 0$.
It doesn't matter whether I attain the answer in the proper approach as I think the method used by the author is incorrect. But I had a good time discussing this with you. Thank you sir :-)
During charging of a RC circuit let $t_1$ and $i_1$ be the time constant and initial charging current when the capacitor is assumed to be filled by a perfect insulator and $t_2$ and $i_2$ be the corresponding values when it is assumed imperfect. Then
(a) $t_1<t_2$ (b) $i_1>i_2$ (c) both (a) and (b) are correct (d) both (a) and (b) are wrong
My approach:
I used the fluid dynamics analogy to solve this. If charging of a capacitor can be considered as filling a beaker with water by pouring it, we can take the capacitor with non-conducting dielectric as a good beaker with no holes and a conducting dielectric as a beaker with holes in the bottom. Intuitively, filling the beaker with holes would take more time and constant pouring compared to a good beaker. And hence option (a) $t_1<t_2$ is correct.
As per this Q&A on Electrical Engineering StackExchange, an ideal capacitor acts as a conductor at $t=0$. And a capacitor with a conducting dielectric would act as a resistor. And hence option (b) $i_1>i_2$ is also correct.
Hence the option that needs to be chosen is (c). And this is the correct answer as per my book.
Does this approach sound good sir? Or are there any better explanations for this?
And if we differentiate the expression for $Q(t)$ to get $I(t)$ then we find $I(0) = V/R_1$, which is exactly what I would expect. So $I(0)$ is independent of $R_2$.
@JohnRennie No sir. If you meant the question 15, the answer according to the book is (c); But it might be a printing mistake too as sometimes I myself spot mistakes in the key.
Hm. I didn't notice this as I approached this using the fluid dynamics analogy. Or in other words, I found the equation only after typing my theoretical method. But if you feel this is the correct answer then the book might be incorrect. Sometimes it's easy to find the mistake. But this time this doesn't seem obvious to me.
If the question is too long please comment to shorten it atleast , I really need an answer.
I have just started learning about RC circuits and these circuits are confusing me.My questions-1)What will be the current in diagram A just at the moment the switch is closed , The capacitor and resistor ...
It looks like a big thread and I need some time to fully go though it. Do I need to read that now, or is it ok to see that when you're not available, sir?
Ok sir. I understood the answer for 16 is wrong. For 15, I'll think about it for some time and ask if I find any issues. Mainly I need to re-analyse my intuitive reason and why it failed here.
Shall we continue our discussion after some time sir? I'm going to have my lunch.
Ok sir. Now is it right to say that a capacitor with a conducting dielectric causes a simultaneous discharging cycle while it's charging? Or in other words, in steady state, the current in the DC circuit is non zero.
Fine. So when we have a conducting dielectric in a capacitor, the time constant is larger than the case when there's no dielectric at all. Then how can we tell the opposite is true while choosing option (d) for question 15 sir?
The presence of the conducting dielectric means the capacitor won't charge to as high a voltage. In our circuit the final voltage will be $V_c = V R_2/(R_1 + R_2)$ instead of $V$.
So while it's true that the current going onto the capacitor is decreased, it doesn't have to charge as far as it did when $R_2$ wasn't present.
Ok sir. I see it not possible to simply say that the time needed to attain saturation/steady-state is larger than the normal non-conducting dielectric case.
But I think we can see the validity of option (b) without these details.
Yes. It is far from intuitively obvious to me what would happen.
@GuruVishnu At time zero the charge on the capacitor is zero and therefore the voltage on the capacitor is zero, and this is true with or without the parallel resistor. Yes?
Yes sir. In case of a non-conducting dielectric, the capacitor would act as a conducting wire at t=0 and I think when the dielectric conducts, there would be some non-zero resistance on converting the capacitor resistor system to entirely resistor system.
So at time zero the voltage across $R_1$ is $V$ in both cases, and therefore the current is $I(0) = V/R_1$ in both cases. That is, the initial current is identical in both cases.
At time $t=0$ the voltage across $R_2$ is zero because it is in parallel with a capacitor that has a voltage across it of zero. So at time zero (but only at time zero) we can ignore the parallel resistor.
For any time $t > 0$ we do need to take it into account. To see what happens we need to take our expression for $Q(t)$ and differentiate it.
Oh, wait, we can do it without differentiating ...
@JohnRennie Ok. Now how can there be some non-zero current in one part of the circuit and a mandatory zero current at the other part? Isn't this a violation of Kirchhoff's law?
As per this Q&A on Electrical Engineering StackExchange, an ideal capacitor acts as a conductor at $t=0$. And a capacitor with a conducting dielectric would act as a resistor. And hence option (b) $i_1>i_2$ is also correct.
I used the fluid dynamics analogy to solve this. If charging of a capacitor can be considered as filling a beaker with water by pouring it, we can take the capacitor with non-conducting dielectric as a good beaker with no holes and a conducting dielectric as a beaker with holes in the bottom. Intuitively, filling the beaker with holes would take more time and constant pouring compared to a good beaker. And hence option (a) $t_1<t_2$ is correct.
Now we can't say the above happens for sure as we consider beakers of two different volumes with holes in one of them.
So the time constants $t_1$ and $t_2$ have no fixed relationship.
Or with only the details in the question, we can't come to a proper conclusion.
@JohnRennie: Could you tell whether I understood this properly, sir?
This would be the hydraulic analogy (diagram incoming):
On the left we have an infinite reservoir so water flowing out of it doesn't change its level. That means the pressure at the left side of the pipe is a constant $P = \rho g h$.
The water flows along the pipe into the beaker, so the pressure at the right end of the pipe is $P' = \rho g x$, where $x$ is the height of the water in the beaker (I forgot to draw this on the diagram).
It's somewhat similar to pouring water in a beaker with hole as I suggested earlier. This seems slightly better as we'd maintain almost constant flow rate due to the reservoir.
If you assume the flow rate in the pipes is proportional to the pressure difference between the ends then this should have the same equations as the capacitor/resistor setup.
Yes. If there is no parallel resistor/outflow then the height in the beaker eventually reaches the same height as the reservoir.
But when we add the parallel resistor/outflow the equilibrium height in the beaker is when the rate of flow in from the reservoir equals the rate of flow out through the outflow. That equilibrium height is going to lower than the height in the reservoir.
I just wanted to see where my analogy failed. Now I understood the main reason it failed was, in the second case, the equilibrium height is lower than the reservoir level. And this made me to conclude the time constant for the second case is greater than the first case.
@JohnRennie One small doubt, in the case of capacitor with conducting dielectric, how do we account for the increased capacitance of the capacitor? Do we increase the cross sectional area of the second container to the right? This seems, it would further increase the difference caused by the hole and the bottom and the decreased equilibrium limit.
Ok sir. Can you please give an alternate method to the following question? :
When both $\lambda_1$ and $\lambda_2$ are positive or negative, the vertical field component cancels each other and only the field component along the $x$ axis exists. I considered the extreme case of $\theta=90$ and some arbitrary acute angle. As there are two opposing horizontal fields, the direction might either point towards the positive $x$ axis or the negative $x$ axis.
I think using the formula for field due to a semi-infinite wire would consume some time. Are there any other ways by which we can easily get the solution sir?
If you consider the dashed line at some angle to the line charge then there is a point near the end where the field lines are normal to the dashed line and the component along the dashed line is zero. This is the middle of the three points I've drawn. This point always exists no matter how long or short the line charge is.
And ether side of this point the component along the dashed line changes direction.
Ok sir. I understood this point. And it's the relative position of this point wrt. the line charge which determines whether it's finally towards the positive or negative x axis.
So depending on where you put the point P the component along the dashed line can be in either direction i.e. to positive x or negative x.
@GuruVishnu Yes
The question doesn't say where P is, so we're forced to conclude it could be anywhere and therefore the component along the x axis could be in either direction.
Yes sir. From a similar approach (different from this one) I concluded for A and B the answer is T. But I think this would be bit complicated when both line charges are of different nature.
The problem is telling for sure whether it's entirely +y or -y or variable like the horizontal component.
I'm just asking, since it's easier to deal with like charges (like "+ and +" or "- and -" ), is it possible to assume the negative charge distribution as a different positive charge distribution which gives the same effect? I think this would simplify the problem.
The reason that equal charges can have a component in either direction on the x axis is that you have equal charge either side of point P. Then the net x component depends on how much charge we have either side and how far away that charge is.
If the charges are opposite then all the positive charge is on one side and all the negative charge is on the other. e.g. all positive charge is at $y > 0$ and all negative at $y < 0$ or the opposite way around.
You never have the same charge on both sides so the field can never change direction as you move P.
I need to go I'm afraid. We'll need to continue this tomorrow.
@JohnRennie No problem sir. I understood this method completely. Thank you for showing the better approach. I was just drawing a diagram to show my first method.
