JR0004 : HCV-32-Ex-83

Guru Vishnu

Feb 18, 2020 05:54

JR0004 - HCV-32-Ex-83

Question:
A capacitor of capacitance $C$ is given a charge $Q$. At $t=0$, it is connected to an uncharged capacitor of equal capacitance through a resistance $R$. Find the charge on the second capacitor as a function of time.

@JohnRennie: Hi sir. Good morning :-)

John Rennie

@GuruVishnu hi :-)

Guru Vishnu

@JohnRennie Are you free now sir? I need some help in the above question.

John Rennie

You need to write an expression for $dQ/dt$.

Suppose the initial change on the capacitor is $Q_0$, then the voltage is $V_0 = Q_0/C$. Yes?

Guru Vishnu

@JohnRennie But here there are two capacitors involved. One is charging and the other is discharging. In an ordinary circuit with only one capacitor I know the expressions for the charge as a function of time for both charging and discharging. But here I don't understand how to relate the effect one capacitor has on the other, sir.

@JohnRennie Yes sir.

John Rennie

Now suppose a charge $Q$ has flowed off the first capacitor onto the second, so the charge on the first capacitor is $Q_0-Q$ and the charge on the second capacitor is $Q$.

Guru Vishnu

Feb 18, 2020 06:02

@JohnRennie Yes sir.

John Rennie

Then the voltage on the first capacitor is $(Q_0-Q)/C$ and the voltage on the second capacitor is $Q/C$.

Can you see where I'm going with this?

Guru Vishnu

@JohnRennie Yes sir. But the presence of resistor on only one branch causes some lack of symmetry. Does it have any role in the final answer?

John Rennie

The potential difference between the two capacitors is $\Delta V = V_1 - V_2 = (Q_0-Q)/C - Q/C = (Q_0 - 2Q)/C$.

And this potential difference is across the resistor, so the current is $dQ/dt = \Delta V/R = (Q_0-2Q)/(RC)$

Guru Vishnu

@JohnRennie: Ok sir.

Is the next step is to apply Kirchhoff's voltage/loop law?

John Rennie

No. You've now got a diffarential equation:

$$ \frac{dQ}{dt} = \frac{Q_0-2Q}{RC} $$

Just solve it for $Q(t)$

Guru Vishnu

Feb 18, 2020 06:10

@JohnRennie Ok sir. I'll do it and ask if I have any doubts. Thank you.

John Rennie

@GuruVishnu are you happy you understand what I did and how I got this equation?

You'll see a lot of questions like this so you need to be sure you understand the approach for solving them.

Guru Vishnu

@JohnRennie I'm happy but still the unsymmetrical condition gives some trouble.

The presence of resistor on only one branch would differentially cause drag in one wire more than the other connecting the plates of the two capacitors.

Did I make it clear how the system is unsymmetrical sir? Or may I draw a diagram?

John Rennie

That's the circuit.

Guru Vishnu

Exactly yes sir.

John Rennie

The first capacitor starts with a charge $Q_0$ and charge is going to flow round the circuit until both capacitors end up with a charge $Q_0/2$.

Guru Vishnu

Feb 18, 2020 06:15

Yes sir.

John Rennie

I don't see what the problem is ...

Guru Vishnu

@JohnRennie Resistor retards the flow of electrons. Am I right, sir?

John Rennie

I wouldn't put it that way. The current through a resistor is I = V/R. That's all you need to know.

Guru Vishnu

@JohnRennie Ok sir. Is it ok to assume two resistances of R/2 each in two branches satisfying the symmetric condition?

John Rennie

Why would you do that?

Guru Vishnu

Feb 18, 2020 06:18

@JohnRennie Just for my satisfaction to see the system as symmetric :-) Also to know whether is it ok to swap the order of capacitors and resistors in a closed loop.

But the second reason is the more important one, sir.

John Rennie

If you do that you make the calculation more complicated because then you have two resistors each with a potential difference across them.

Guru Vishnu

@JohnRennie I agree sir. But will the final result vary on this fact? Because, as per our previous discussions a zero resistance wire causes a lot of trouble logically.

So I thought it would be better to assume some resistance in both wires. That's why I thought of sharing the total resistance with the other branch.

John Rennie

The wire connecting the bottom plates of the capacitors just sets them to be the same potential, so the potential difference we calculated is then between the top plates of the capacitors i.e. across the resistor.

Guru Vishnu

@JohnRennie Yes sir. But in reality, same amount of current must flow through the wires in order to maintain the same amount of charges on the opposite plates of the capacitor and this is why I'm having some trouble understanding the unequal resistance distribution.

John Rennie

Current is like a fluid - like water flowing in a pipe. The current has to be the same everywhere so if you restrict the flow at one point you reduce the flow everywhere.

Guru Vishnu

Feb 18, 2020 06:30

Fine sir. Doesn't this call for same resistance on both branches?
Using your analogy:
Speed of flow is constant only if the area of cross section is uniform at constant volume flow rate. But it varies drastically if the area of cross section is different at different regions. And the un-symmetrical condition is similar to that of here, sir.

John Rennie

Not speed of flow, volumetric flow rate. The current is like a volumetric flow rate.

i.e. number of electrons per second passing any point in the circuit.

Guru Vishnu

@JohnRennie Ok sir. Now understood. So irrespective of the resistances of the two branches, the effect on one branch is equally transmitted to the other one. Sounds similar to Pascal's law in fluids :-)

John Rennie

> Sounds similar to Pascal's law in fluids

Yes.

Circuits are remarkably similar to fluids.

Hydraulic analogy

The electronic–hydraulic analogy (derisively referred to as the drain-pipe theory by Oliver Lodge) is the most widely used analogy for "electron fluid" in a metal conductor. Since electric current is invisible and the processes at play in electronics are often difficult to demonstrate, the various electronic components are represented by hydraulic equivalents. Electricity (as well as heat) was originally understood to be a kind of fluid, and the names of certain electric quantities (such as current) are derived from hydraulic equivalents. As with all analogies, it demands an intuitive a...

Guru Vishnu

@JohnRennie Cool :-)

Guru Vishnu

Feb 18, 2020 07:09

Thank you sir.

Guru Vishnu

Feb 18, 2020 07:22

And solving the differential equation, I obtained the expression:
$$Q(t)=\frac Q 2 (1-e^{-2t/RC})$$

End of #JR0004 - HCV-32-Ex-83

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