Artificial Capacitor

Electricity and Magnetism Level 3

A conventional capacitor's charge and the voltage across it are related by:

$\Delta V = \frac{q}{C_{conv}}$

Here symbols have their usual meanings. Now imagine an artificial capacitor, the behaviour of which is governed by the relation:

$\Delta V = \frac{q^2}{C_{art}}$

Now, consider this artificial capacitor of capacitance $C_{art}=4 V/C^2$ discharging through a $3\Omega$ resistor. The capacitor initially has $20 C$ of charge. How much energy does the resistor dissipate in the first 5 seconds? Let this value be $E_1$ . Carry out the same calculation for the case where a conventional capacitor $C_{conv}=4F$ is used. Let the value of energy dissipated be $E_2$ . Enter your answer as $\boxed{E_1/E_2}$

How can this result be interpreted physically?

Inspiration - 1

Inspiration - 2

The answer is 23.55.

1 solution

Steven Chase
Aug 24, 2019

For an ordinary capacitor (note that a passive convention is being used, with current flowing into the capacitor):

$Q = C V \\ \dot{Q} = I = C \dot{V} \\ \dot{V } = \frac{I}{C}$

For the artificial capacitor:

$Q^2 = C V \\ 2 Q \, \dot{Q} = 2Q \, I = C \dot{V} \\ \dot{V} = \frac{2 Q I}{C}$

Simulation code is attached. It appears that the artificial capacitor effectively yields a shorter time constant than the ordinary capacitor, as seen from the graph of the charges $Q_1$ and $Q_2$ over time.

import math

R = 3.0   # Resistance

C1 = 4.0  # Capacitance and alt capacitance values
C2 = 4.0

dt = 10.0**(-5.0)  # Time step

##############################################################

Q1 = 20.0   # Initial charges
Q2 = 20.0

V1 = (Q1**2.0)/C1  # Initial voltages
V2 = Q2/C2

V1d = 0.0 # Initial voltage derivatives
V2d = 0.0

E1 = 0.0  # Initial resistor energies
E2 = 0.0

t = 0.0

count = 0

print "t Q1 Q2 I1 I2 P1 P2 E1 E2"

while t <= 5.0:

    V1 = V1 + V1d * dt   # Euler integration - voltage
    V2 = V2 + V2d * dt

    I1 = -V1/R           # Currents, convention has them flowing into caps
    I2 = -V2/R

    Q1 = Q1 + I1 * dt    # Update charge values
    Q2 = Q2 + I2 * dt

    V1d = 2.0*Q1*I1/C1  # Calculate voltage rates of change
    V2d = I2/C2

    P1 = (V1**2.0)/R   # Resistor power
    P2 = (V2**2.0)/R

    E1 = E1 + P1 * dt   # Resistor energy
    E2 = E2 + P2 * dt

    t = t + dt
    count = count + 1

    if count % 1000 == 0:
        print t,Q1,Q2,I1,I2,P1,P2,E1,E2

##############################################################

print ""
print ""

print (E1/E2)
# 23.5535606998   # ratio of E1 to E2 at t = 5

Thanks for the solution. As for the physical interpretation, I was thinking more on the lines of something analogous to least-action. The actual configuration of the capacitor corresponds to one where energy dissipated is minimum. Does this line of thought make sense to you?

Karan Chatrath - 1 year, 9 months ago

Let's see. Recalling the resistor case, the formal statement is essentially:

"Currents distribute in a resistor network in such a way as to minimize the power dissipated, subject to Ohm's Law constraints"

What would the analogous statement be for the capacitor?

Steven Chase - 1 year, 9 months ago

The statement in mind is: " Given a potential difference across the plates of a capacitor, the charge acquired by it varies with potential difference such that all energy exchanges in a circuit (or with other electrical elements) are minimal."

Specific to this case, I assert that the energy dissipated by the capacitor is least given its realistic configuration, as compared to any artificial one.

The next logical step is to test said assertion.

Karan Chatrath - 1 year, 9 months ago

I'm also kind of interested in another notion of an alternate capacitor, in which there are two derivatives in the differential equation, instead of just one.

Steven Chase - 1 year, 9 months ago

I have solved your recent problem based on this notion. Interesting idea

Karan Chatrath - 1 year, 9 months ago

The problem can be solved analytically also. For the artificial capacitor, current $I_{art}$ is obtained as a function of time by solving the equation $R\dfrac{dq}{dt}+$ $\dfrac{q^2}{C_{art}}=0$ . Then integrating the element of energy dissipated $dE_1=$ $\dfrac{(q_0C_{art})^2R^3}{(RC_{art}-q_0t)^4}dt$ over the given time interval, we get $E_1$ .

A Former Brilliant Member - 1 year, 9 months ago

Yes, this is the recommended approach in order to obtain a closed-form solution. Thank you. Although, even I took a numerical approach, so am unfamiliar with the final expression.

Karan Chatrath - 1 year, 9 months ago

How do you reach the conclusion that the conventional capacitor yields the minimum power dissipation in the circuit? Do you have any analytical proof? For example, you can assume a Charge-Voltage relation as $Q=f(V)$ . Then use the definition of $C$ as $C=\dfrac{Q}{V}$ and proceed.

A Former Brilliant Member - 1 year, 9 months ago

@A Former Brilliant Member – My statement is more of an assertion than a conclusion. I am thinking of a mathematical framework where I can prove this. My thought points towards the calculus of variations but I am yet to pen down a proof.

I was solving the following problem:

https://brilliant.org/problems/discharging-capacitor/

And I thought whether any other capacitor configuration, that is not the real one, would dissipate more energy or not? And so I got the inspiration for this problem. Turns out that the result of this problem corroborates my assertion.

Karan Chatrath - 1 year, 9 months ago

@Karan Chatrath – Think about my suggestion.

A Former Brilliant Member - 1 year, 9 months ago

@A Former Brilliant Member – Will keep in mind. Thank you

Karan Chatrath - 1 year, 9 months ago

@Karan Chatrath – The functional is the integral of voltage, and charge which is a function of voltage over time which is a parameter. For it's extrimum, Euler's equation yields change must be directly proportional to voltage. (Partial derivative of charge with respect to voltage is constant). So the conventional capacitor yields an extrimum of dissipated energy. From your example, it is asserted that this extrimum is the minimum indeed.

A Former Brilliant Member - 1 year, 9 months ago

@A Former Brilliant Member – Thanks for the comment. As for the choice of functional, I didn't get what the time integral of voltage represents physically.

Karan Chatrath - 1 year, 9 months ago

@Karan Chatrath – When voltage depends on time, the time integral of voltage, like any other variable, gives the action of voltage over time interval given.

A Former Brilliant Member - 1 year, 9 months ago

@A Former Brilliant Member – Thanks for all your inputs. I will dwell on your suggestions and reply to this thread.

Karan Chatrath - 1 year, 9 months ago

Artificial Capacitor

The answer is 23.55.

1 solution

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