capacitance

Okay so I just found out that the charge of each capacitor would be different, so how would I work them out because the voltage for each is also different.

Reply 2

Original post by rm2

Okay so I just found out that the charge of each capacitor would be different, so how would I work them out because the voltage for each is also different.

Your description of the circuit is a bit ambiguous. If you can post a scan of the circuit diagram, that would be a great start to helping you arrive at the correct answers.

A couple things for you to think about are:

a) the wired together isolated plates of series/parallel connected combinations of capacitor, must all be at the same voltage potential.

b) no charge can be added to or taken away from isolated plates. (no current path exists because the central plates are isolated.

c) Total charge will be shared in the ratio of individual capacitances.

(edited 9 years ago)

Reply 3

image post

So does a) mean that C1 and C23 have the same voltage?

Reply 4

Original post by rm2

image post

So does a) mean that C1 and C23 have the same voltage?

That makes more sense, thanks. :smile:

Your original calculations are correct, including the 40V developed across C1 (120uF).

Energy stored:

E = \frac{1}{2}CV^2

Use the calculated voltage dropped across each capacitor and their individual capacitance values to find the energy stored in each capacitor.

(edited 9 years ago)

Reply 5

Thanks. Okay so using the same voltage formula, I got

V2 = 160V (is this right? It's higher than the battery supply)
V3 = 96V

Then using those voltage values

E = ½CV²

E1 = ½ x (120x10^-6) x 40² = 0.096J
E2 = 0.384J
E3 = 0.2304J

Energy stored in the circuit is E1 + E2 + E3 = 0.7104J

Going back to the question, the order it is given is weird because I've already worked out V1. Is this just bad question structure or some sort of trick question?

Also, I don't get why the same charge value is used for all 3 (to work out voltage) because they have different capacitance.

Reply 6

Original post by rm2

Thanks. Okay so using the same voltage formula, I got

V2 = 160V (is this right? It's higher than the battery supply)
V3 = 96V

?????

Total equivalent capacitance = 48x10^-6F = 48uF

Total charge stored Q = CV = 48x10^-6 x 100 = 4.8x10^-3C

The voltage developed across C1 is:

V = Q/C = 4.8x10^-3 / 120x10^-6 = 40V

The voltage developed across the parallel pair can be calculated in more than one way:

a) C2 and C3 are in parallel. i.e. The voltage developed across this pair must be identical:

C2 in parallel with C3 = C2 + C3 = 80uF

V_C2 = V_C3

= Q/C_parallel

= 4.8x10^-3 / 80x10^-6

= 60V

b) using Kirchoff's Voltage law, the sum of the votage drops around a circuit must equal the supply.

This means V_S = V_C1 + V_C2¦¦3 ( ¦¦ = parallel )

V_C2¦¦C3 = V_s - V_C1 = 100 - 40 = 60V

Original post by rm2

Then using those voltage values

E = ½CV²

E1 = ½ x (120x10^-6) x 40² = 0.096J
E2 = 0.384J
E3 = 0.2304J

Energy stored in the circuit is E1 + E2 + E3 = 0.7104J

The total energy stored is calculated using the equivalent combined capacitance value of 48uF and the 100V supply voltage.

E = \frac{1}{2}CV^2 = \frac{48\mathrm x10^{-6} \mathrm x 100^2}{2} = 2.4\mathrm x10^{-3}J

Energy stored in each individual capacitor is calculated using the voltages previously calculated. i.e. 40V across the 120uF and 60V across the 30uF and 50uF capacitors.

E1 is correct, the other two (E2 and E3) are not.

Redo the calculations (including E = E1 + E2 + E3) and post your answers here and I can check them for you.

Original post by rm2

Going back to the question, the order it is given is weird because I've already worked out V1. Is this just bad question structure or some sort of trick question?

No tricks, it's a reasonable set of questions and you really have to think about what is going on at the electron level.

Original post by rm2

Also, I don't get why the same charge value is used for all 3 (to work out voltage) because they have different capacitance.

Think about how a capacitor is constructed (conducting plates seperated by an insulating gap) and stores energy in the electric field between the capaitor plates.

This e-field is created by the repulsion between electrons on either side of the plates by voltage pressure from the supply. The atoms of the plate are fixed in position and cannot move (protons and neutrons), but electrons are free to migrate and therefore bringing the plates close together creates a repulsive force between the electrons which depletes electrons from the surfaces closest together.

But the plates have a finite size (surface area) and so can only store energy in proportion to the number of electrons available within that surface area.

Connecting capacitors together:

Parallel capacitors increase the available plate area able to store charge from the same voltage source. Hence the total capacitance is the sum of the individual parallel capacitor values.

Series capacitors have isolated plates (plates linked together between the capacitors). Thus the available charge (electrons) already on the isolated plates is fixed (trapped) and therefore has to be shared between the connected plates. i.e. charge (current) cannot flow across the insulation gap. The repulsive charge force acts over the distance of the insulation gap. This is why the same charge value is used for all three capacitors - they all share the same charge because of those isolated plates.

i.e. the repulsive forces must be balanced out and shared equally between those capacitors with isolated plates sharing the same conduction path (wired together). Hence the effective capacitance of series connected capacitors is a function of the smallest plate area. In the case of this question, that would be created by the two 30uF and 50uF (80uF) capacitors in parallel which has a combined plate area smaller than the plate area of the 120uF capacitor.

You may now want to read through the question again and see how this description applies to your answers.

(edited 9 years ago)

Reply 7

Thanks, those paragraphs were detailed and helped a lot.
Okay I get why they did the question like that. Before, I worked out the individual energy of the capacitors and added them together but to do that I needed the individual voltages. But since I can just use 0.5QV and I already have the total charge (48x10^-4C) and voltage (100V) I don't need the individual voltages.

E = 0.5 × (48×10^-4) × 100^2 = 0.24J

I got the same by adding the energy of C1 and C23.

So now that I have the total capacitance, charge, total energy and V1 I think the question is complete.

... So there's this other question which seems to use the same things I did here and is of a lower level (doesn't look like it though).

Question:

Capacitors of 3, 6 and 12 uF are connected in parallel across a 100V DC supply. State the voltage across each capacitor, the charge on and the energy stored in each capacitor. Theres no diagram this time, and theres a part 2 I'll get to later which involves almost the same thing but in series.

I have found the total capacitance is 3 + 6 + 12 = 21uF

Charge of the capacitors is Q=CV so Q = (21×10^-6) × 100 = 21×10^-4 C

Since all capacitors are in parallel would the voltage of each capacitor be 33.3V?

Reply 8

Original post by rm2

... So there's this other question which seems to use the same things I did here and is of a lower level (doesn't look like it though).

Question:

Capacitors of 3, 6 and 12 uF are connected in parallel across a 100V DC supply. State the voltage across each capacitor, the charge on and the energy stored in each capacitor. Theres no diagram this time, and theres a part 2 I'll get to later which involves almost the same thing but in series.

I have found the total capacitance is 3 + 6 + 12 = 21uF

Total capacitance is correct. Parallel capacitors sum.

Original post by rm2

Charge of the capacitors is Q=CV so Q = (21×10^-6) × 100 = 21×10^-4 C

Total charge is correct.

Original post by rm2

Since all capacitors are in parallel would the voltage of each capacitor be 33.3V?

No. They are all 100V.

All common points connected together in a circuit MUST be at the same voltage potential measured w.r.t. the supply.

Since the capacitors are in parallel, all three must have the same potential across them. i.e. they are all connected to the same 100V dc supply in all three cases.

Think about it logically: where would the voltage drops (energy loss) occur to make the potentials different? A: nowhere.

Reply 9

Ignore this post. See below.

(edited 9 years ago)

Reply 10

I got that the pd across C1 must be 40V and across C2 and C3 (since they are in parallel) is 60V, I originally got 23.1 and 76.9V which wasn't consistent with the total energy stored.

Reply 11

So using E = ½CV²

E1 = 0.015J
E2 = 0.03J
E3 = 0.06J

Charge for all 3 capacitors = 21×10-⁴C

Voltage for all 3 capacitors = 100V

And that is the first part done.

Second part says the same capacitors are connected in series across the 100V supply. Calculate the charge on, and the energy stored in each capacitor.

Total capacitance = (1/3 + 1/6 + 1/12)-¹ =1.71uF

Charge = CV = (1.71×10^-6) × 100 = 1.71×10-⁴C

V1 = Q/C = (1.71×10-⁴) ÷ 3×10^-6 = 57.14V

V2 = 28.57V

V3 = 14.29V

E = ½QV

E1 = ½ × (1.71×10-⁴) × 57.14 = 4.9×10-³J

E2 = 2.45×10-³J

E3 = 1.22×10-³J

Correct, hopefully?

Reply 12

Original post by rm2

So using E = ½CV²

E1 = 0.015J
E2 = 0.03J
E3 = 0.06J

Charge for all 3 capacitors = 21×10-⁴C

Voltage for all 3 capacitors = 100V

And that is the first part done.

All correct.

Original post by rm2

Second part says the same capacitors are connected in series across the 100V supply. Calculate the charge on, and the energy stored in each capacitor.

Total capacitance = (1/3 + 1/6 + 1/12)-¹ =1.71uF

Correct.

Original post by rm2

Charge = CV = (1.71×10^-6) × 100 = 1.71×10-⁴C

V1 = Q/C = (1.71×10-⁴) ÷ 3×10^-6 = 57.14V

V2 = 28.57V

V3 = 14.29V

Correct.

As a quick check, sum V1, V2 and V3 and they should equal the supply voltage of 100V.

Original post by rm2

E = ½QV

E1 = ½ × (1.71×10-⁴) × 57.14 = 4.9×10-³J

E2 = 2.45×10-³J

E3 = 1.22×10-³J

Correct, hopefully?

I have to be away for a while. Will look at the last bit shortly.

Reply 13

Original post by uberteknik

Connecting capacitors together:

Parallel capacitors increase the available plate area able to store charge from the same voltage source. Hence the total capacitance is the sum of the individual parallel capacitor values.

Series capacitors have isolated plates (plates linked together between the capacitors). Thus the available charge (electrons) already on the isolated plates is fixed (trapped) and therefore has to be shared between the connected plates. i.e. charge (current) cannot flow across the insulation gap. The repulsive charge force acts over the distance of the insulation gap. This is why the same charge value is used for all three capacitors - they all share the same charge because of those isolated plates.

Can I just ask a question here. Is this the same as saying that the number of electrons that flow onto the negative plate = number of electrons that flow off the positive plate and then those that 'came off' now flow onto the negative plate of the next capacitor and thus the charge stored on both are equal?

Reply 14

Original post by rm2

Back again!

Yes, all correct now.

More importantly, have you understood how and why parallel and series capacitors work to arrive at the maths descriptions?

Reply 15

Original post by Protoxylic

As an analogy, you can think of it like that to help you remember but it's not an accurate description of the physical process:

In reality it is the 'ripple' of the repulsive electric force (electrons are force carriers) throughout the conductor that is doing the moving.

Think of the electrons in a conductor like the links of a bicycle chain. Move one end a small amount and all of the links in the chain also move. This is how pedals move the road wheels.

In a similar way, the power source ripples charge force to the first capacitor in the series connection. That voltage pressure generates a repulsive force across the capacitor gap and the electrons on the opposite plate ripple the same charge force to the next plate, and so on and so forth.

The voltage potential developed across the capacitor plates is a function of the plate size. Bigger plates = more charge and hence more force available across the capacitor insulating gap and vice versa.

So the smallest plate size in the series chain (smallest capacitor) will dictate (limit) to all of the other capacitors how much force is allowed to ripple through the chain.

That's because the smallest capacitor can only store a maximum of charge proportional to its capacitance and the supply voltage.

i.e. Q = CV

And once that smallest capacitor saturates, the limit of the force available to push on all other electrons in the series path is reached.

The voltage developed across the other capacitors V = Q/C is then dictated by that limited charge available to the smallest capacitor.

It does not matter where in the chain that smallest capacitor is because the electron repulsion acts in both directions.

Hence ALL capacitors must have equal charge and because of that, the voltage developed across the other capacitors is dictated by that charge and the different plate sizes. i.e. the larger capacitors have spare capacity.

Does that make sense?

(edited 9 years ago)

Reply 16

Original post by uberteknik

So if you had a 1uF cap in series with a 100uF cap, the charge stored will be limited to the amount of charge the 1uF cap can store? Isn't the effective Capacitance 101uF so why the charge stored limited to the amount an effective 101uF capacitor can store. Isn't the positive plate of the 1uF connected to the negative plate of the 100uF (if connected in series in this order) and this arrangement is disconnected from the whole circuit?

Reply 17

Original post by Protoxylic

The effective capacitance of the series connected capacitors is:

C_{eff} = (\frac{1}{C_1} + \frac{1}{C_2})^{-1}

C_{eff} = (\frac{1}{1 \mathrm x10^{-6}} + \frac{1}{100 \mathrm x10^{-6}})^{-1} = 0.99 \mathrm x 10^{-6}F = 0.99 \mu F

Capacitances for parallel connected capacitors sum. I.e. 1 + 100 = 101.

(edited 9 years ago)

Reply 18

Original post by uberteknik

The effective capacitance of the series connected capacitors is:

C_{eff} = (\frac{1}{C_1} + \frac{1}{C_2})^{-1}

C_{eff} = (\frac{1}{1 \mathrm x10^{-6}} + \frac{1}{100 \mathrm x10^{-6}})^{-1} = 0.99 \mathrm x 10^{-6}F = 0.99 \mu F

Schoolboy error, I understand now cheers.

Reply 19