Question: Two resistors, Two capacitors (Ch 19 #50)

Ch 19 #50 part b

I am not sure how to distinguish whether a capacitor is charging or discharging from this problem. Sometimes, I can't tell through which resistor will a capacitor will discharge...is it through the one in parallel or in series?

Let's just look at simplifying what we have when the switch is open, then closed.

When the switch is open there are two resistors in series which are in parallel with two capacitors in series. The voltage across the set of resistors will be the same as the voltage across the set of capacitors (24 V in this case). At point "a" there will be some voltage/potential between 0 and 24 V due to the voltage dropped (IR) across one of the two resistors. Likewise at point "b" there will be some voltage/potential between 0 and 24 V as there is some voltage dropped (Q/C) across one of the two capacitors. (Note that to determine this you will have to use the equivalent resistances and capacitances of each branch of the circuit to determine the net current and charge, respectively.)

Now, when the switch is closed, charge will flow from a to b or vice versa because the potential at a and b were not equal initially. Once the switch is closed this potential difference cannot be maintained. Without working through to see which is greater and thus which direction the charge will flow I can't say whether the capacitors are charging or discharging. However, I can tell you if the potential at "a" is greater than at "b" the current (charge) will flow toward the capacitors and thus the capacitors will charge (partially). Conversely if the potential at "b" is greater than at "a" the current (charge) will flow away from the capacitors and the capacitors will discharge (partially).

Good luck. This is a tricky problem.

Correction: Example Ch 16 #61

Sorry, there is an error in the example posted on the main page: Electrostatics / Ch. 16 #61

In the first part, I forgot to square the velocity, so instead of 5.4e-9 m, I should have gotten a distance of 0.115 m. The error is also carried forward into part b, instead of 1e-12 s, I should have gotten 2e-8 s for the time.

I will try to fix the original page soon.

WebTA survey

To get some feedback on the usefulness and usage of the webTA project I have developed a short survey. Paper copies should have been (will be?) available in class at some point, however, if you missed your chance to have your say and are interested in providing feedback I will give you another chance here.

You may download a PDF version of the 2006 webTA project survey HERE [35 kB PDF].

Print it out, fill it out, and return it to my mailbox (Sarah Burke, in the Grad students section) in the mailroom of the Rutherford Physics building.

(Location of the mailroom: Enter Rutherford through main doors, turn right and go through a big brown door. At the end of the hall is a staff lounge. The mailroom is on the right just before this. It's a room with a photocopier and a bunch of labelled mailboxes. Mine's on the left hand side near the beginning as they are more or less alphabetical.)

Of course any creative ways you think of to send it back electronically are good too... sorry I don't have a way to make a form-fillable pdf. :)

End of term looming...

Just a quick warning in case you were not already aware. The webTA site will not be maintained after the end of classes. I will do my best to answer as many of the questions you ask as possible until then.

Question: Stable and unstable equilibria

How do we tell the stable and unstable equilibrium mathematically? cuz using the definition in the text to determine what equilibrium i'm dealing with is ambiguous and obscure...

This is a very good question which is actually very general. Let's start with some definitions...

Equilibrium is a state of a system in which the variables which describe the system are not changing (note that a system can be in a dynamic equilibrium where things might be moving or changing, but some variable(s) which describe the system as a whole is(are) constant). One example you are all familiar with is a mechanical system in equilibium where positions of objects are not changing (ie. no net forces acting).

In a Stable equilibrium if a small perturbation away from equilibrium is applied, the system will return itself to the equilibrium state. A good example of this is a pendulum hanging straight down. If you nudge the pendulum slightly, it will experience a force back towards the equilibrium position. It may oscillate around the equilibrium position for a bit, but it will return to its equilibrium position.

In an Unstable equilibrium if a small perturbation away from equilibrium is applied, the system will move farther away from its equilibrium state. A good example of this is a pencil balanced on it's end. If you nudge the pencil slightly, it will experience a force moving it away from equilibrium. It will simply fall to lying flat on a surface.

Ok, now that we know more about equilibria it will be easier to determine what "kind" we have. Strictly speaking, mathematically we determine whether a mechanical equilibrium is stable or unstable by looking at the second derivative of the energy with respect to the coordinate of interest. If the equilibrium is at a minimum (second derivative is positive) the system is in a stable equilibrium. If the equilibrium is at a maximum (second derivative is negative) the system is in an unstable equilibrium.

However, there is a simpler way to quickly test whether an equilibrium is stable or unstable. If we think about the definitions, each involves the response to a small perturbation. So let's "apply" a perturbation. If we say change the position slightly, is there a net force? in what direction? does the perturbation result in a force which drives the system back toward equilibrium (stable) or away from equilibrium (unstable)?

If you can't easily picture the situation in your head (as with the pencil and the pendulum), what this means in practical terms is that you re-calculate the forces at some position near, but not at equilibrium and determine whether they are driving the system back towards equilibrium (again, this means stable) or away from equilibrium (this means unstable).

Hope that clarifies things!

Question: pulling apart capacitor plates

A parallel-plate capacitor with plate area 2.0 cm^2, air-gap .5mm, is connected to a 12V battery, if the air-gap is pulled to .75mm, how much work is required?

calculated Q=4.2E-11 C, V(d=0.75)=18V, delta PE=Q*delta V = 2.52E-10 J, but the answer is 1.3E-10 J, which could be a half of my answer. PE/2? why? Would you please explain this to me? Thanks

Ok, so the thing you are changing when you pull a parallel plate capacitor apart is the capacitance (it's ability to store energy).

The capacitance depends on distance like:

C=εo*(A/d)

so clearly the capacitance will decrease if the distance between the plates is increased. Let's consider the capacitance before (C), and after (C'):

C=εo*(A/d)
C'=εo*(A/d')

Ok. So now let's consider how this will effect the energy stored in the capacitor. This can be written in 3 different ways (by substituting Q=CV into the first):

PE=(1/2)QV=(1/2)CV²=(1/2)Q²/C

Which of these should we use? Well, since the plates are connected to a 12 V battery, the voltage will be the same before and after moving the plates. Note that the charge may change so we can't use the expressions with Q in them because we don't know how Q is changing (ok, yes, we could figure it out, but it would be equivalent to using the second expression, so let's not bother).

All right, so let's put it together:

W=-ΔPE=-((1/2)C'V²-(1/2)CV²)
W=-((1/2)εo*(A/d')V²-(1/2)εo*(A/d)V²)
W=-εo*A*V²/2((1/d')-(1/d))

Plugging in the numbers given in the equation above for the work done, I get: 8.49e-11J (using 12V). You also write 18V in your question, with that number I get an answer: 1.91e-10J. Sorry to muddy the waters with more answers... the method should be correct. (maybe you can recheck which numbers the question uses?)

If you haven't already, I'd suggest taking a look at the similar example (though with constant charge not voltage) posted on the main page: Ch. 17, #50.

Hope that helps. Feel free to ask questions by posting comments.

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Fair Game! Topic open: everything

At this point in the year, everything is fair game for you, so any question you have is fair game for me. Feel free to ask me anything you have covered in the course (or even things from last semester that might come up).

On that note, a piece of wisdom from my 3rd year mechanics professor:

You don't have problems, you have exercises! ~Dr. Melvin Calkin

Question: difference between RC, RL, LC

How do you tell the difference between RC, RL, LC circuits?

Each of these circuits, and what they qualitatively do in DC and AC circuits...

RC - a resistor and capacitor in series. Exhibits charging behaviour with characterisitic time constant with DC voltage source. Acts as a high pass filter (allows high frequency currents, but not low frequency currents) in AC circuits.
RL - a resistor and inductor in series. Acts as a short with a DC voltage source, but smooths out rapid variations in current. Acts as a low pass filter (allows low frequency currents, but not high frequency currents) in AC circuits.
LC (and RLC) - an inductor and capacitor (and resistor) in series. If initially charged, has oscillitory behaviour (damped if also has a resistor). Has resonant behaviour with AC driving voltage (damped if also has a resistor).

Question: LC in parallel

How to calculate the resistance of a capacitor and an inductor connected in parallel?
(-XcXL/(XL-Xc))^2 ?

This is a little bit trickier that considering elements in series because the inverse of the impedances have to be added. I'm pretty sure it is beyond the scope of this course to need to do this calculation, but I will put it here anyway. I'm also not sure how you would arrive at the answer without using the appropriate complex represenations of the impedance, so I will use them but try to be careful to explain.

So, for elements in parallel:

1/Z_tot=1/Z_C+ 1/Z_L+ 1/Z_R ...[1]

We will consider just an LC circuit (with ideal components), so R=0. Now, in complex notation Z_Cand Z_L are:

Z_C=1/iωC ...[2]
Z_L=iωL ...[3]

So substituting [2,3] into [1]:

1/Z_tot=iωC+1/iωL

or, using the fact that 1/i=-i:

1/Z_tot=iωC-i/ωL
1/Z_tot=i(ωC-1/ωL) ...[4]

so, we can now replace 1/ωC with X_C and ωL with X_L:

1/Z_tot=i(1/X_C-1/X_L) ...[5]

Now we can simplify the part in the brackets, giving us:

1/Z_tot=i(X_L-X_C)/X_CX_L ...[6]

or taking the inverse of the fraction on both sides (and once again using 1/i=-i):

Z_tot=-iX_CX_L/(X_L-X_C) ...[7]

Now, if we had also had a resistor in parallel in the circuit (R), we would end up at equation [7] with both real and imaginary parts. To find X_tot we would have to find the magnitude of the resultant phasor. Since Z_tot is purely imaginary, we need only to find the magnitude of this number, which we can express as:

X_tot=-X_CX_L/(X_L-X_C) ...[8]

Which is pretty much what I think you suggested. Note that if we hadn't used the imaginary numbers we would not have been able to get the right signs and would have come up with a different answer.

Looking back at eqn. [7], let's think about what this represents... In the imaginary plane Z_tot would be a phasor pointing along the y-axis. For values of X_L and X_C which make Z_tot positive, the impedance could be considered "inductive", and inversely if Z_tot were negative, the impedance could be considered "capacitive".

Let's consider two extreme cases...

If the frequency is high, X_L is very large, and X_C is very small (let's say negligible). In the numerator of Z_tot the frequency cancels out, so this is a constant. If we look at the denomenator, X_L >> X_C, so the denomenator is approximately X_L. Since X_L is large, Z_tot will be small and negative (capacitive). In fact if you substitute in equations [2,3] ignoring X_C in the denominator, you will recover Z_C. This is consistant with the inductor providing a break in the circuit, and the capacitor providing a short.

Similarly, if the frequency is very low, X_C becomes very large, and X_L very small. In this case, the result is small and positive (inductive), and indeed again, if you work it out you will recover Z_L as Z_tot. This is consistant with the capacitor providing a break in the circuit, and the inductor providing a short.

To summarize:

ω→∞ Z_tot→Z_C and current flows through the capacitor
ω→0 Z_tot→Z_L and current flows through the inductor

These limiting cases can be worked out simply as a thought experiment (consider which branch of the circuit has a small resistance and which has a large resistance) and are often more insightful than getting through the algebra.

Hope this helps.

Topic Open: RC, RL, LC, LRC circuits

The following topic is now open for questions: RC, RL, LC, LRC circuits (and any AC circuits)

To pose a question, please post a comment to this post by clicking on the comments link below.

Question: Impedance of LR circuit with "real" inductor

Just had a question about LR circuits. Just say there was a circuit consisting of a resistor and inductor. And just say the inductor had a resistance of its own. How do you find the total impedance of the circuit. This was similar to one of the CAPA questions. My friend told me to find the impedance of just the resistor and inductor and then add the resistance of the inductor to Z. I am confused on why you do this? Why couldn't you just add the resistance of the inductor and resistor first, then use the formula Z = sqrt(R^2 + XL ^2), R being the resistance of the inductor and resistor combined?

Actually, you are right! You can (and should) just add the resistance of the inductor to the resistance of the resistor then find the impedance including the inductive term.

Here's how it works. A real inductor will have some inherent resistance due to the fact that it is a (usually fairly large) coil of wire. To analyze a circuit, we would replace a "real" inductor with and "ideal" inductor and a resistor (to represent the pure resistance of the inductor). So, in an LR circuit you would go from having 1 resistor and 1 "real" inductor in series to having 2 resistors in series and one "ideal" inductor also in series. You would then find the net resistance of the circuit, and this would be "R" in the formula you quote above.

From this you would find the magnitude of the inductance from Z=sqrt(R²+X_L²)
(this formula comes from the fact that R and X_L are 90^o out of phase, so the magnitude is the hypotenuse of the two "phasors" see: Basics of AC circuits/RLC circuit example). Since R and X_L do not add linearly, I do not think you should get the right answer if you go about it in the opposite order (except maybe in some special circumstance).

EDIT: A similar question was also asked regarding an LRC circuit. In this case, the methodology is the same. The inductor with a resistance is replaced by a resistor and an inductor in series. The resistance of the whole cicuit is calculated, and then the magnitude of the impedance is determined. In the case of an LRC circuit, Z is given by:

Z=sqrt(R²+X_L²+X_C²)

Otherwise, the whole problem is the same.

Hope this helps!

Question: Solenoid in a coil

If a solenoid is placed inside a coil, how does the current flowing through it affect said coil? How would you find the inductance value for the coil?

That's a very good question. Qualitatively, the solenoid (with some current running through it) establishes a magnetic field parallel to the axis of the solenoid. If a coil is then placed on the outside of it, the coil will "sense" the magnetic field established by the solenoid. If the current in the solenoid is time-varying (changing in time) then the magnetic field established by it will also change with time. The coil will then "sense" this changing magnetic field and respond to oppose the change (Lenz's law).



Quantitatively, we will have to look at what the magnitude of the magnetic field produced by the solenoid is, and how that may depend on time. The magnetic field of a solenoid is given by:

B_sol=μ₀I[N/l]

Thus, if the current, I is a function of time, the magnetic field will have the same time dependence. If, say, the time dependence is linear (eg. we ramp up the current to the solenoid at a constant rate), then we can use the slope as ΔB/Δt, which is related to the EMF:

EMF=-ΔΦ_B/Δt=-A_{perpendicular}ΔB/Δt

Chapter 21, Question #80 is an example of this, and might be good to look at.

I hope this is helpful and answers your question.

Topic open: Charges and currents in magnetic fields

The following topic is now open for questions: Charges and currents in magnetic fields.

To pose a question, please post a comment to this post by clicking on the comments link below.

Topic open: Magnetic fields

The following topic is now open for questions: Magnetic fields.

To pose a question, please post a comment to this post by clicking on the comments link below.