🔹RC Circuit with Series Capacitors — One Pre-Charged, One Uncharged

[Uplakshya]

Homework Statement

Circuit Description:
• A battery of emf is connected in series with:
– an uncharged capacitor of capacitance 2C
– a second capacitor of capacitance , C which is already charged to CE (its left plate is negative, right plate positive)
– a resistor of resistance .

The order (clockwise) is:
Battery (+ terminal) → Capacitor 2C → Capacitor C (left plate negative, right plate positive) → Resistor R → Battery (– terminal)

Initial Conditions (at )

1. Capacitor C has voltage E, charge .CE
Negative charge on left plate


3. No current has yet flown through the resistor.



Task:

1. Write the correct Kirchhoff-voltage equation around the loop, carefully assigning signs to each term.


2. Derive the differential equation relating (the charge that flows into the capacitor) and .


3. Solve for and explicitly, applying the proper initial condition.
Im an 12 student so use understandable language
"My overall priority is that to find a trick a logical one to generalise this question and find time equation without integration or differentiation". But the trip should be logical and that can be explained"

Relevant Equations

KVL
E -q/(2C) + (CE - q)/C -iR = 0

My solution

 

[PhDeezNutz]

What is the meaning of CE? (Capacitance times electric field?)

I’m not sure that has units of charge as the problem statement indicates.

 

[Steve4Physics]

What is the meaning of CE? (Capacitance times electric field?)

I’m not sure that has units of charge as the problem statement indicates.

I think the OP intends 'E' to be the cell's emf.

'CE' is the initial charge of the pre-charged capacitor. The initial charges on the capacitor's plates are -CE and +CE. When some charge, q, is transferred, the charges on the plates become (q - CE) and (q + CE). This ties-up with the diagram and working.

There are various problems with the question. These include:

- it looks like the start of first equation in the attachment has been lost; the equation should probably start "E - ..."

- the OP has incorrectly used "i" and $\frac {dq}{dt}$ (which are the same) in the second equation.

There are various other minor errors, the working is sometimes unclear and some of the handwriting is difficult hard-on-the-eye tor read.

Answering a question which has a number of mistakes/problems is messy. That's probably why there are so few replies. (@Uplakshya may wish to note this if they are still following.)

Minor edits.

 

[Uplakshya]

I think the OP intends 'E' to be the cell's emf.

'CE' is the initial charge of the pre-charged capacitor. The initial charges on the capacitor's plates are -CE and +CE. When some charge, q, is transferred, the charges on the plates become (q - CE) and (q + CE). This ties-up with the diagram and working.

There are various problems with the question. These include:

- it looks like the start of first equation in the attachment has been lost; the equation should probably start "E - ..."

- the OP has incorrectly used "i" and $\frac {dq}{dt}$ (which are the same) in the second equation.

There are various other minor errors, the working is sometimes unclear and some of the handwriting is difficult hard-on-the-eye tor read.

Answering a question which has a number of mistakes/problems is messy. That's probably why there are so few replies. (@Uplakshya may wish to note this if they are still following.)

Minor edits.

Got it!
Assumptions:

E is the battery’s emf.

CE is the capacitor’s initial charge: one plate has +CE, the other –CE.

q(t) is the extra charge transferred after t = 0.

1. New plate charges:

Positive plate: +CE → +CE + q

Negative plate: –CE → –CE + q
And a plate of capacitance 2C has charge q now

2. Voltage across the capacitor:
V = q / C

3. Current:
i = dq/dt

4. KVL around the loop:
E – (q / C) – iR = 0
→ E – (q / C) – R(dq/dt) = 0

I mixed I and DQ/dt as to final integration


"Also don't see my solution see the post I posted there you will get the whole scenario"

f you want a shorter or even simpler version!

 

[Uplakshya]

What is the meaning of CE? (Capacitance times electric field?)

I’m not sure that has units of charge as the problem statement indicates.

Don't worry about units all are in si units
And CE means capacitance times potential difference
May be this question is not for you man

 

[DaveE]

May be this question is not for you man

Which is why most of us didn't respond.
Do you have a question? You may need to make it more obvious for volunteers like us.

 

[berkeman]

May be this question is not for you man

Watch the attitude please. And please do a better job of defining your terms and your question. Thank you.

 

[PhDeezNutz]

I think a partial answer to OP’s question “without integration or differentiation” is using Laplace transforms. I will give it some more thought as to how I would incorporate initial conditions.

 

[DaveE]

"My overall priority is that to find a trick a logical one to generalise this question and find time equation without integration or differentiation". But the trip should be logical and that can be explained"

Sorry, this circuit requires either integration or differentiation to solve. It's a mistake to avoid this with tricks.
However, once you've solved circuits like this 20, 30, 50... times, there are powerful tricks from both memorizing the solution of simple canonical variants and linear system theory. So, I'm too lazy to do it, but here are some comments:

1) Real EEs (as opposed to many, many physics texts) will always name each component. They each need names, polarities, and sometimes values, ICs, etc. So, I'll do it for you: R1=R, C1=C, C2=2C, Vs=E, Vc1(0)=E, Vc2(0)=0. IRL clear communication is really important, especially for engineers.

2) Two capacitors in series is odd and can be modelled as one. C'=(2/3)C, Vc'(0)=E. Now you have a simple RC circuit. If you care about the values for each capacitor, that is just a voltage divider. For every volt that leaves C2, 2 two volts are put onto C1 because they share the same current (charge is conserved).

3) The dynamic response of a linear system does not change with different initial conditions or input(s). Every time dependent term in your solution will be proportional to $e^{\frac{-t}{RC'}}$.

4) You can identify what each parameter of interest is at t=0 and at t=∞. They will transition from one to the other according to the dynamic solution above. Note this isn't very helpful with complex networks where the dynamics aren't simple.

5) You can use superposition to breakup the problem into the sum of simpler solutions, like the natural response plus the driven response. But, for this circuit that's probably not necessary.

6) Yes @PhDeezNutz is correct. Analog EEs always do these solutions with Laplace transforms, it's easier. But it's still calculus, don't pretend you're doing algebra (like I do, LOL).

 

[DaveE]

https://math.arizona.edu/~system/archives/pwaters/IPSA math 2013/Math1 HW/LaplaceNotes1.pdf

 

[PhDeezNutz]

I think the answer for the charge across the second capacitor that was initially charged to $C\epsilon$ should be

$\frac{1}{3} C \epsilon \left(e^{-\frac{3t}{2CR}} + 2 \right)$

Which i was able to get through Laplace Transforms. This answer at the very least supports the initial condition that at time $t=0$ the charge on the second capacitor (of capacitance $C$) is indeed $C \epsilon$

However it doesn’t corroborate that the initial current through the resistor is 0.

 

[DaveE]

However it doesn’t corroborate that the initial current through the resistor is 0.

There isn't really an "initial" current in the resistor. It's value will instantly change when the switch closes. The solution in this format only applies to after the switch has closed. Anyway, initial conditions in these linear systems are the capacitor voltages and inductor currents only; components that store energy and "remember" their history.

The solution in this format only applies to after the switch has closed. Initial conditions in these linear systems are the capacitor voltages and inductor currents.