Series parallel capactitive circuit

[freshbox]

Hi guys, I want to ask if I can use the voltage divider rule to solve for part (ii).


Thanks!

 

[gneill]

Hi guys, I want to ask if I can use the voltage divider rule to solve for part (ii).


Thanks!

You could; do you know the voltage divider rule for capacitors?

Alternatively, if you've already worked out the total capacitance and the charge on the total capacitance, then take note that C1 and C2 are in series in that total... so what charge must be on them?

 

[freshbox]

charge is the same, sorry let me sort out something first

 

[freshbox]

Total capcitance of two capacitors in series:

Ct=C1xC2/C1+C2 is also the same as 1/Ct=1/C1+1/C2

Is the any difference between them? How come there are 2 type of ways to find the total C. I have always use the 2nd one "1/Ct" to find the total C. Is there any special for the first one?

 

[gneill]

Total capcitance of two capacitors in series:

Ct=C1xC2/C1+C2 is also the same as 1/Ct=1/C1+1/C2

Is the any difference between them? How come there are 2 type of ways to find the total C.

There's no difference. Simplify algebraically the one on the right and you'll see.

 

[freshbox]

Ok, thank you. Can I ask if the above diagram capacitors are being replaced by inductor, can I still use the voltage divider rule to solve?Thanks.

 

[gneill]

Ok, thank you. Can I ask if the above diagram capacitors are being replaced by inductor, can I still use the voltage divider rule to solve?


Thanks.

There would be a problem replacing all the capacitors with inductors if the power supply remains 25V DC. Do you see why?

 

[freshbox]

No, why?

Does the voltage divider rule for inductor exist? Because I can't find the formula in my book but I did some googling and found a formula for it.

 

[gneill]

No, why?

The DC resistance of an ideal inductor is zero. While capacitors behave as open circuits and can charge up to present a potential difference, inductors behave as short circuits and just conduct more and more current. I = (V/L)t.

Does the voltage divider rule for inductor exist? Because I can't find the formula in my book but I did some googling and found a formula for it.

It would be more useful for AC circuits where inductors have an impedance (sort of like an "AC resistance"). Voltage divider rules can be applied to any components with impedance.

In a DC circuit, the eventual voltage across all inductors after any transients die away (some time after power is applied) is zero. It's hard to have a voltage divider mean anything useful if the potentials involved are all zero

 

[freshbox]

Is the VDR for capacitor the same as Inductor?

 

[gneill]

Is the VDR for capacitor the same as Inductor?

No, the divider rule for inductors looks like the one for resistors. But again, you would probably only use it for AC circuits (or during transient analysis for DC circuits).

 

[freshbox]

Ok. I can't get the answer for part (iii.)

My working:

V3=CtE/C5 = 2.42x10^-6x25/14=4.321uC

Wc=1/2CV²
=1/2x8x10^-6x(4.321x10^-6)²
=7.468x10^-17J

Ans is:0.733J*I combined C3 and C4 together to form C5. From there, i tried to find the voltage across the new C5 reason being voltage for C3 and C4 is the same in parallel. My new V3 is also equal to the V across C3 and C4.

After getting the new voltage "C5", I sub the values inside the formula including the 8uf (C4) but i cannot get the answer.

 

[gneill]

Ok. I can't get the answer for part (iii.)

My working:

V3=CtE/C5 = 2.42x10^-6x25/14=4.321uC

Shouldn't the result be in volts if it's a voltage? Where are the units for your C5?

Also, you may be propagating roundoff errors from intermediate results. Don't round intermediate values that are to be used in further calculations -- keep a few extra digits in all intermediate values to prevent previous roundings from creeping into your significant digits in the final values.

Wc=1/2CV²
=1/2x8x10^-6x(4.321x10^-6)²
=7.468x10^-17J

Ans is:0.733J

That provided answer looks suspiciously large. 0.733J is a lot of energy for a small capacitor to hold... check the given units.

*I combined C3 and C4 together to form C5. From there, i tried to find the voltage across the new C5 reason being voltage for C3 and C4 is the same in parallel. My new V3 is also equal to the V across C3 and C4.

After getting the new voltage "C5", I sub the values inside the formula including the 8uf (C4) but i cannot get the answer.

Your overall method looks okay. Check the items I pointed out.

 

[freshbox]

7.468x10^-17J is my answer

0.733J is the book answer.

So if ignore the rounding off errors am I'm right?

 

[gneill]

7.468x10^-17J is my answer

0.733J is the book answer.

So if ignore the rounding off errors am I'm right?

No, your result has an incorrect power of 10. Check the items I pointed out. Your C5 is not "14"; it needs appropriate units.

 

[freshbox]

My bad. 74.6uJ. I'm sorry can you help me take a look at another question...

My working:

250x10^-3=500²x1800x4∏x10^-7xA / 200x10^-3

50000x10^-6=500²x1800x4∏x10^-7xA

A=50000x10<sup>-6[/SUP / 5002x1800x4∏x10-7

A=0.000008841x101

A=0.00008841m2

A=0.08841mm2 My ans

Book ans is:88.4mm2</sup>

 

[gneill]

My bad. 74.6uJ. I'm sorry can you help me take a look at another question...

Yes, that's closer to what I found: 74.32 μJ.

My working:

250x10^-3=500²x1800x4∏x10^-7xA / 200x10^-3

50000x10^-6=500²x1800x4∏x10^-7xA

A=50000x10<sup>-6[/SUP / 5002x1800x4∏x10-7

A=0.000008841x101</sup>

^{Your powers of ten seem to have gone awry in that last step. You may find it easier to "normalize" the big (or small) numbers to the form n.nnn x 10n before continuing. That will make slipping decimal places more difficult.}

 

[freshbox]

law of indices division = minus

-6/-7 = -6 - -7 = 1

This what I'm thinking..

 

[gneill]

law of indices division = minus

-6/-7 = -6 - -7 = 1

This what I'm thinking..

Okay, sorry, my eyes slipped over the x10¹ tacked onto the end of the decimal number. You really should present the figures in normalized form.

It would appear that your conversion from m² to mm² is incorrect. How many square mm are there in a square m?

 

[freshbox]

Sorry, going back to post #12. Is there something wrong with my concept+answer?

I missed out C1 earlier on and i use VDR on C2 and C34 instead. I thought VDR only apply for 2 capacitors and how can i use it if i have 3 capacitors at that time?

 

[freshbox]

So if what i said in post #20 is correct, redoing everything...Since Ct=2.42uF
Q=VC
Q=25x2.42=60.5uC
Qt=Q1=Q2 (Series Circuit)

V2=8.64v

Finding Voltage across C1, naming it as V1
Q=VC
60.5uC=V x 5uF
V1=12.1V

V2+V1=8.64+12.1
= 20.74VEt-20.74=4.26V
4.26V is the voltage across C3 and C4Subbing all the values into the energy stored in C formula, the answer i got is 72.59uJ. So can I say post #12 working+concept = wrong while this is the correct one.

Please help me take a look, thank you.

PS: Just curious, is part (ii) still doable using the VDR method?

 

[gneill]

Sorry, going back to post #12. Is there something wrong with my concept+answer?

I missed out C1 earlier on and i use VDR on C2 and C34 instead. I thought VDR only apply for 2 capacitors and how can i use it if i have 3 capacitors at that time?

Combine two of the capacitors leaving you with two, then apply the voltage divider rule.

But really, it's probably easier to find the total capacitance first and determine the charge; the same charge will be on each of the capacitors in series, so you can find the voltage on each one quite easily.

 

[gneill]

So if what i said in post #20 is correct, redoing everything...

I notice that you are rounding intermediate results as you go along and not keeping enough guard digits to prevent rounding errors from entering your later calculated values. You apply two different (but correct) methods to find a given value and end up with different numbers in the significant figures. This is rounding/truncation errors biting you.

In red below I've shown the same values that you've calculated but without successive roundings being applied along the way, and with an extra guard digit.

Since Ct=2.42uF (2.414 μF)
Q=VC
Q=25x2.42=60.5uC (60.345 μC)
Qt=Q1=Q2 (Series Circuit)

V2=8.64v (8.621 V)

Finding Voltage across C1, naming it as V1
Q=VC
60.5uC=V x 5uF
V1=12.1V (12.069 V)

V2+V1=8.64+12.1
= 20.74V (20.690 V)


Et-20.74=4.26V (4.310 V)
4.26V is the voltage across C3 and C4


Subbing all the values into the energy stored in C formula, the answer i got is 72.59uJ (74.316 μJ then rounded to 74.32 μJ for "final answer"). So can I say post #12 working+concept = wrong while this is the correct one.

Please help me take a look, thank you.

The method in post #12 was fine, but you made some errors in applying it, such as using '14' for the value of C5 when it should have had units of microfarads included.

PS: Just curious, is part (ii) still doable using the VDR method?

Sure, just combine C1 and C5 into a single capacitor (call it C15) and then use the voltage divider rule.

 

[freshbox]

Thank you gneill for your explanation so far. So can I say that post #12/23 question has a couple of ways to solve it?

I would like to ask you a question relating to the picture below:

"currents and voltages have reached their final values" -> Does it have another meaning as current and voltage = @ max value. ?
Thank you.

 

[gneill]

Thank you gneill for your explanation so far. So can I say that post #12/23 question has a couple of ways to solve it?

Sorry, I don't follow your question, but it is true that most problems can be solved via several approaches.

I would like to ask you a question relating to the picture below:

"currents and voltages have reached their final values" -> Does it have another meaning as current and voltage = @ max value. ?

It means that the currents and voltages settle into a steady state after which no further changes in their values take place unless the circuit is disturbed in some fashion (like a switch being thrown).

When switches open or close, sudden changes in potentials or currents can cause transient events like big voltage spikes across inductors, so minimum or maximum magnitudes can occur during the transient state, and then settle down to some constant "end state" over time.

For the most part, for RL or RC circuits you'll see exponential curves that either decay from a starting value to a minimum, or rise from some starting value to a maximum "plateau".

Circuits containing both L and C components can be trickier as there will be oscillations that decay over time, so just saying "minimum" or "maximum" value won't work cleanly.

Anyways, "Assume that currents and voltages have reached their final values" means that the circuit has been left undisturbed for long enough that any transient conditions have died away and operation has reached a steady state (all values remain constant as time goes on).

 

[freshbox]

Is 5 time constant = steady state/final values? Because I know that at 5 time constant, for capacitor: v is max/i is 0. Inductor: v is 0/I is max. So for this case, is Vc=35v (max value) ?

 

[gneill]

Is 5 time constant = steady state/final values? Because I know that at 5 time constant, for capacitor: v is max/i is 0. Inductor: v is 0/I is max. So for this case, is Vc=35v (max value) ?

5 time constants is a "rule of thumb" (engineering approximation) for when transients are assumed to have died out and first order circuits have settled into their steady state. Here you have both inductors and capacitors, so the circuit is not first order.

But you don't need to worry about that here. After sufficient time, any circuit with only passive components (no active components like amplifiers) powered by DC will eventually settle into a steady state where inductors look like short circuits and capacitors like open circuits. That's your "for capacitor: v is max/i is 0. Inductor: v is 0/I is max" condition. So to solve this problem replace inductors with wires and remove the capacitor, then solve for the potentials and currents.

What does the circuit look like when you've done that? Are the capacitor connection points going to have a 35V potential difference? Look closely at what components remain after suppressing the inductors and capacitor.

 

[freshbox]

Do you mean that the 5 time constants concept is only applicable when the circuit only have 1 inductor or 1 capacitor by itself. So for this question, because there is a mixture of inductor and capactor, the 5 time constants concept cannot be apply here?


Thanks.

 

[gneill]

Do you mean that the 5 time constants concept is only applicable when the circuit only have 1 inductor or 1 capacitor by itself. So for this question, because there is a mixture of inductor and capactor, the 5 time constants concept cannot be apply here?


Thanks.

It applies when there is only one TYPE of energy storing device (inductor or capacitor) in a circuit. There can be multiple capacitors or multiple inductors, but not a mix of both.

When there are both types of device present in a single circuit their interaction can produce oscillations and longer time constants which are not trivial to deal with.

 

[freshbox]

So the question says that it has reached its final value, does it mean that it is at 5 time constant?

 

[gneill]

So the question says that it has reached its final value, does it mean that it is at 5 time constant?

The number of time constants does not matter if enough time is allowed to pass. Assume that 10¹⁰⁰time constants have passed...

 

[freshbox]

Do you mean that it can happen at any time? Why does it not matter? Is it because this circuit has 2 type of device?

Thanks.

 

[gneill]

Do you mean that it can happen at any time? Why does it not matter? Is it because this circuit has 2 type of device?

The problem statement says to assume that the currents and voltages have reached their final values. There are no constraints placed on how long you are allowed to wait to ensure that this happens. So just assume that enough time has passed regardless of how long it might be (and you do not care about the specific amount of time!); once a circuit reaches steady state it will no longer change no matter how much more time passes.

Any DC circuit that contains resistance will eventually reach steady state because the energy driving any transients must eventually by dissipated by Ohmic losses. This is true even for RLC circuits containing both L and C components, even if the time constants involved may be tricky to work out. No matter how long it might take for 'ringing' or 'oscillations' to damp out, they must eventually damp out as resistance bleeds away their power, so one just assumes that a sufficiently long time has been allowed for this to occur.

You don't need to know precisely how long this process takes; just assume it has been long enough and the circuit is at steady state.