Particle Accelerator

Guru Vishnu

10:51

Question: How many time constants will elapse before the power delivered by the battery drops to half of its maximum value in an RC circuit?

What did I do?:

Found an expression for power delivered by the battery as per our previous discussion as - $P=i^2R$.

Substituted the expression for $i$ the current.

John Rennie

@GuruVishnu hi

Guru Vishnu

@JohnRennie Hi sir :-)

But finally I got the answer to be $\ln 2/2$ instead of $\ln 2$; The problem is entirely simple, and my answer varies from the correct one by a factor of $2$.

So, who is incorrect - HC Verma or Guru Vishnu?

John Rennie

Do you want to go through the calculation?

Guru Vishnu

@JohnRennie I think yes. I'm unable to think of other ways to know the correct answer. Is it possible sir?

I just described the method I used, above.

John Rennie

The starting point is to calculate the current as a function of time. Did you do that?

Guru Vishnu

10:58

Yes sir. $i=Q_0/RC \times (1-e^{-t/RC})$

Where $Q_0$ is the charge the capacitor would carry after some years.

John Rennie

You can simplify this since $V = Q_0/C$ and $I_0 = V/R$, where $I_0$ is the initial current.

So you get $I(t) = I_0 ( 1 - e^{-t/RC} )$

Guru Vishnu

@JohnRennie Ok sir. Understood.

John Rennie

And the power being supplied by the battery is $P = I(t)V$

So the power falls to half the initial value when the current falls to $I_0/2$ i.e. when $1-e^{-t/RC} = 0.5$

Guru Vishnu

Yes sir. And I see you're getting close to the correct book answer.

John Rennie

Yes, you're going to end up with $t = RC\ln2$

Guru Vishnu

11:03

But what if we use $P=i^2R$ instead of $P=iV$; we must get the same result right? But I tried a lot of times and got $t=\ln 2/2$.

John Rennie

No, that's giving you the power dissipated in the resistor R not the power being supplied by the battery.

You're ignoring the power being converted into the electric field in the capacitor.

Guru Vishnu

@JohnRennie Ok sir. I can see where I went wrong. But could you explain why a simple substitution of $V=iR$ makes the situation different. I'm unable to understand this point.

John Rennie

The substitution only works at time zero.

Guru Vishnu

@JohnRennie Yes sir. Now understood this completely. Thank you sir :-)

John Rennie

At time zero the voltage across the capacitor is zero because no charge has flowed onto it, and that means the whole voltage $V$ is dropped across the resistor. So the current at time zero is $I_0 = V/R$.

@GuruVishnu cool :-)

Guru Vishnu

11:09

@JohnRennie Yes sir. May I know how were you so careful on the first try itself?

John Rennie

@GuruVishnu I'm not sure what you're asking ...

Guru Vishnu

@JohnRennie How were you so cautious on using $P=iV$ instead of $P=i^2R$ (like me)?

It didn't strike my mind even after a lot of attempts.

I thought of various other scenarios.

John Rennie

The question asks about:

> the power delivered by the battery

Guru Vishnu

Yes sir.

John Rennie

So you need to write down the power delivered by the battery, and this is the battery voltage times the current through the battery. This is true regardless of what the battery is connected to.

Guru Vishnu

11:13

@JohnRennie Ok sir. Is this true only for ideal batteries or can we extrapolate this to batteries with non-zero internal resistance?

I know the potential must drop with current. But how is this related to power, sir?

It seems, both the $i$ and $V$ terms vary.

John Rennie

If the battery EMF is $E$ then the power produced by the battery is always $P = EI$. But for real batteries some of this power is dissipated in the battery due to the internal resistance $R$.

Guru Vishnu

@JohnRennie Ok sir. I see we can bring the resistor $R$ in the RC circuit into the battery :)

John Rennie

So the power delivered is $P = EI - I^2R$.

Guru Vishnu

@JohnRennie Ok sir. Thank you :-)

May I ask another doubt?

John Rennie

@GuruVishnu Yes.

Guru Vishnu

11:19

@JohnRennie In a problem, I saw internal resistance of a battery varies with the fact whether it's charged or discharged. I asked this question on the main site. And I got some answers indicating the conductivity of batteries vary depending upon the electrolytic composition. If it's discharged it's resistance is higher because it contains less number of charge carries. If this is so, will it not alter the potential difference of the battery? If so does a 1.5V battery would become a 0.5V battery

after usage. Then it seems the fundamental definition of battery fails - "a battery maintains a constant potential difference in a circuit"

Question on the main site: physics.stackexchange.com/q/530846/238167

John Rennie

A real battery isn't an ideal battery.

Luckily the JEE will only ask you questions about ideal batteries not real ones :-)

Guru Vishnu

@JohnRennie Does a battery die because of loss in potential difference or is that because of its inability to give away current?

John Rennie

Both.

Do you know the Nernst equation?

Guru Vishnu

@JohnRennie Yes sir. I learnt it in chemistry. But now it seems in my brain "Error 404 not found"; I forgot it :-(

John Rennie

Nernst equation

In electrochemistry, the Nernst equation is an equation that relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation. It was named after Walther Nernst, a German physical chemist who formulated the equation. == Expression == The Nernst equation is derived from the standard changes in the Gibbs free energy associated with an electrochemical transformation. For any electrochemical reducti...

The Nernst equation tells you how the EMF of a cell varies with the concentrations of the reagents.

As a battery runs down the concentrations of the reagents in the battery changes, so the battery EMF also changes according to the Nernst equation.

Guru Vishnu

11:24

@JohnRennie Ok sir. Now understood your "Both" statement :-)

@JohnRennie Yes sir :-) I just wanted to be clear so that it might help in my college or beyond.

John Rennie

In general there's no easy way to say how the internal resistance changes because it depends on the design of the battery.

Guru Vishnu

@JohnRennie Fine sir. That was my second question about :-)

I just wanted to see a graph which shows the internal resistance variation with the energy in a battery.

John Rennie

I'm sure such graphs exist if you Google for them. But they would need to be measured experimentally.

Guru Vishnu

Do you know any such graphs sir?

@JohnRennie I tried sir. But didn't get one.

John Rennie

I don't know of any such graphs. You could always do the experiment :-)

Guru Vishnu

11:27

I got only straight lines.

:-) That's why there wasn't anyone on the main site to answer that part.
Yes sir. But I don't have a potentiometer and also some patience to eat away a fully charged battery.

Guru Vishnu

> Car batteries have to be almost 'flat' (no energy left) before they won't turn the starter-motor, and an appreciable internal resistance would prevent them from doing so. This paragraph, though, is merely speculative.

Sir, if possible could you explain the above paragraph? I asked the answerer but still didn't get it.

John Rennie

There is some data on internal resistance here

> Figure 6: Typical internal resistance readings of a lead acid wheelchair battery. The battery was discharged from full charge to 10.50V. The readings were taken at open circuit voltage (OCV).

Guru Vishnu

Yes sir. Thank you for sharing the website. It seems the internal resistance rises exponentially and I guess there is a minimum potential below which a battery is of no use. So even a battery completely dead - say our mobile phone's battery has some potential difference - say 2V if it's initially 100V. Am I right, sir?

John Rennie

Yes

Guru Vishnu

@JohnRennie Thank you sir :-)

John Rennie

11:40

The voltage is going to fall in a roughly exponential fashion, so in principle it never quite falls to zero.

Guru Vishnu

@JohnRennie Thank you again sir.

John Rennie

:-)

Transcript for

♦ Particle Accelerator