Conversation started Feb 17, 2020 at 10:51.
Guru Vishnu
Guru Vishnu
Feb 17, 2020 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
John Rennie
@GuruVishnu hi
Guru Vishnu
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
John Rennie
Do you want to go through the calculation?
Guru Vishnu
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
John Rennie
The starting point is to calculate the current as a function of time. Did you do that?
Guru Vishnu
Guru Vishnu
Feb 17, 2020 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
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
Guru Vishnu
@JohnRennie Ok sir. Understood.
John Rennie
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
Guru Vishnu
Yes sir. And I see you're getting close to the correct book answer.
John Rennie
John Rennie
Yes, you're going to end up with $t = RC\ln2$
Guru Vishnu
Guru Vishnu
Feb 17, 2020 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
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
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
John Rennie
The substitution only works at time zero.
Guru Vishnu
Guru Vishnu
@JohnRennie Yes sir. Now understood this completely. Thank you sir :-)
John Rennie
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
Guru Vishnu
Feb 17, 2020 11:09
@JohnRennie Yes sir. May I know how were you so careful on the first try itself?
John Rennie
John Rennie
@GuruVishnu I'm not sure what you're asking ...
Guru Vishnu
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
John Rennie
The question asks about:
> the power delivered by the battery
Guru Vishnu
Guru Vishnu
Yes sir.
John Rennie
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
Guru Vishnu
Feb 17, 2020 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
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
Guru Vishnu
@JohnRennie Ok sir. I see we can bring the resistor $R$ in the RC circuit into the battery :)
John Rennie
John Rennie
So the power delivered is $P = EI - I^2R$.
Guru Vishnu
Guru Vishnu
@JohnRennie Ok sir. Thank you :-)
 
Conversation ended Feb 17, 2020 at 11:16.