Discharging a capacitor - how long does it take

[jsmith613]

Homework Statement

A capacitor stores 1.09 C of charge.
The capacitor has a p.d of 16V and a capacitance of 68 000 µF.
The maximum leakage current is 0.003 µA per µF per V therefore for this capacitor the maximum leakage current = 3260 µA.
How long does it take for the capacitor to discharge by leaking?

Homework Equations

The Attempt at a Solution

So Q=It
therefore t = Q/I = 334 seconds
BUT apparently this is the time constant, why?

(now I realize my method is wrong because discharging current will change but I have no idea how to do this Q so just thought I would use Q=It)

 

[NascentOxygen]

At any time t, the current flowing from the capacitor is given by the differential equation:
1/C * dV_c(t)/dt = ...

 

[jsmith613]

At any time t, the current flowing from the capacitor is given by the differential equation:
1/C * dV_c(t)/dt = ...

how does this help me fidn the time constant and then (5RC) time to discharge?

 

[NascentOxygen]

The dots ... indicate where you fill in using the data you are given.

 

[jsmith613]

The dots ... indicate where you fill in using the data you are given.

but I don't have dV_c(t)/dt

 

[NascentOxygen]

At any time t, the current flowing from the capacitor is given by the differential equation:
1/C * dV_c(t)/dt = ...

Expressing it in symbols rather than words:
I_c(t) = 1/C * dV_c(t)/dt = ...

 

[jsmith613]

Expressing it in symbols rather than words:
I_c(t) = 1/C * dV_c(t)/dt = ...

sorry...I am very much confused...

What i still don't see is why the t I found using Q/I was time constant...I realize it is not discharge time but cannot see why it is time constant...

 

[NascentOxygen]

You are told what the capacitor's leakage current is, here:

the leakage current is 0.003 µA per µF per V

 

[jsmith613]

You are told what the capacitor's leakage current is, here:

yes so leakage current changes BECAUSE p.d will change...but this also means dV/dt changes...

 

[NascentOxygen]

yes so leakage current changes BECAUSE p.d will change...but this also means dV/dt changes...

It means V changes. So you express leakage current in terms of V, and this leads to a differential equation to solve.

 

[gneill]

sorry...I am very much confused...

What i still don't see is why the t I found using Q/I was time constant...I realize it is not discharge time but cannot see why it is time constant...

You may gather some insight though unit analysis. Take a look at the units of the leakage specification. If capacitance remains constant, how does the leakage current vary with the capacitor voltage? What do the units remind you of?

 

[jsmith613]

It means V changes. So you express leakage current in terms of V, and this leads to a differential equation to solve.

this is A2 physics...we don't need to solve differential equations...note the question was only worth 2 marks so clearly this cannot be the solution

This is what the MS wrote:

Estimate of time for capacitor to discharge with reasoning:
Q/I₀ = 334 s [This is time constant] (1) - why??
Q=1.09

Numerical example such as: for less than 0.7% remaining t = 5τ =
1670 s OR well-reasoned estimate showing t >> 300 s (1)
(I know this bit BUT NOT the first bit??)

 

[jsmith613]

You may gather some insight though unit analysis. Take a look at the units of the leakage specification. If capacitance remains constant, how does the leakage current vary with the capacitor voltage? What do the units remind you of?

Leakge: uA/uF/V
If capacitance is constant then we get uA/V
this = 1/R

 

[NascentOxygen]

Avoiding the D.E. makes it too easy! :tongue:

 

[jsmith613]

Avoiding the D.E. makes it too easy! :tongue:

yes but that's all I need (no D.E but working as per mark scheme)...I am still struggling to get why the mark scheme is correct

 

[gneill]

yes but that's all I need (no D.E but working as per mark scheme)...I am still struggling to get why the mark scheme is correct

Given the voltage V on capacitor C, what's the value of the charge Q on the capacitor?

 

[jsmith613]

given the voltage v on capacitor c, what's the value of the charge q on the capacitor?

q=cv
=v*680000*10^-6

 

[gneill]

q=cv
=v*680000*10^-6

Yes, but the numerical value is not important for the moment. If you divide both sides by I what do you get?

 

[jsmith613]

Yes, but the numerical value is not important for the moment. If you divide both sides by I what do you get?

Q/I = RC...

ok...so the next question is why must I use the initial current...I presume this is because I am also using the initial charge, right?

 

[gneill]

Q/I = RC...

ok...so the next question is why must I use the initial current...I presume this is because I am also using the initial charge, right?

You can use any set of Q,I, and V which hold at a given instant; It just so happens that you know those particular values. And the R=V/I is constant... that's the leakage resistance you worked out earlier.

 

[jsmith613]

You can use any set of Q,I, and V which hold at a given instant; It just so happens that you know those particular values. And the R=V/I is constant... that's the leakage resistance you worked out earlier.

you are amazing...thanks so much for explaining this to me so well (and walking me through rather than telling me straight out...it means I won't make the same mistake in the exam) :)

 

[gneill]

Glad to help. Good luck in your exams

 

[jsmith613]

Glad to help. Good luck in your exams

sorry for brining this up again...but I was looking at the charge/discharge curve and realized something...I just wanted to check:

would I be correct in thinking that this method would only hold true for a DISCHARGING capacitor...for a charging capacitor charge increases and current decreases...thus initial Q/I = small and final Q/I = large

 

[jsmith613]

silly me

I should be looking at the Q and V curve

both either increase or decrease at the same rate...therefore it should work for bothh...i think...really not sure though

 

[gneill]

silly me

I should be looking at the Q and V curve

both either increase or decrease at the same rate...therefore it should work for bothh...i think...really not sure though

Yup, should work for both. The initial slopes of the Q vs t or V vs t curves are finite so long as the resistance in the discharge path is nonzero and finite.

 

[jsmith613]

Yup, should work for both. The initial slopes of the Q vs t or V vs t curves are finite so long as the resistance in the discharge path is nonzero and finite.

actually this could ONLY work for discharging as they are talking about leakge current...how can a charging capacitor leak charge...

but...actually I cannot think of an example...I was trying to make an example to prove the situation for charging and discharging...maybe you can?
if not I will just take your word for it and hope nothing too hard comes up :)

 

[jsmith613]

Yup, should work for both. The initial slopes of the Q vs t or V vs t curves are finite so long as the resistance in the discharge path is nonzero and finite.

also is not R the resistance of the resistor...its V is NOT the same as the V of the capacitor...how then does this work for a charging circuit?

 

[jsmith613]

Yup, should work for both. The initial slopes of the Q vs t or V vs t curves are finite so long as the resistance in the discharge path is nonzero and finite.

I realize the comment was foollish...it is discharging to the surroundingg...the resistance to disharging is R=V/I...silly me

just to check though, for a discharging capacitor, is the voltage across the resistor the same as that cross the capacitor BUT the for a charging capacitor the sum of the p.d's across the capacitor and resistor = source (i.e: they are not the same at all points)??

 

[gneill]

For this problem the circuit resistance happens to be provided by a leakage resistance of the capacitor, but it's not a requirement; The resistance could be given as an external component (or even an external resistance in parallel with a leakage resistance).

The point to take from this problem is that the initial slope of the Q vs t, V vs t, or I vs t curves for charging or discharging is determined by the time constant of the circuit.

If this problem had been about a charging capacitor rather than a discharging one they couldn't have given you the initial charge on the capacitor; They would have had to provide some other set of parameters that would allow you to determine the time constant.

 

[jsmith613]

the circuit resistance

this is something new I learned from this...I always assumed resistance cam from an extneral source...but it can come from anywhere...this is definitely a great insight into the physics of capacitors :)

 

[gneill]

just to check though, for a discharging capacitor, is the voltage across the resistor the same as that cross the capacitor BUT the for a charging capacitor the sum of the p.d's across the capacitor and resistor = source (i.e: they are not the same at all points)??

For the typical charging capacitor circuit, the sum of the resistor and capacitor p.d.'s is constant (equal to the voltage source).

Of course, for any complete "tour" around a given circuit the sum of all potential changes must be zero according to Kirchhoff.

 

[jsmith613]

For the typical charging capacitor circuit, the sum of the resistor and capacitor p.d.'s is constant (equal to the voltage source).

yes...thanks :)

Of course, for any complete "tour" around a given circuit the sum of all potential changes must be zero according to Kirchhoff.

for this I just learn sum of p.d across componenets = emf of source (assuming wire have no resistance / no internal resistance)