Calculating Capacitor Load: Using Capacity and Formula from Capacitor Layout Appendix

Question

Hello, how to calculate the load ( ) on a capacitor with a capacity ? (capacitor layout in the appendix) I am trying to use the formula here: I substitute the data from the drawing for this formula. however, this is not a valid result. Thank you in advance for your help :)

ryszard1955 · Accepted Answer

Buddy, you are supposed to calculate the load in the C2 capacitor, and not in all three. In order to calculate the C2 charge, you need to find what will be the voltage on the C2 and C3 systems, the total capacity of which is their sum, i.e. 2 uF. As you can see, since C1 is also 2uF, the voltage across this circuit will be equal to half the voltage, i.e. 2V.
So according to your formula, Q2 is going to be 2x1.5 = 3

kryststgpom · Answer

You substituted the supply voltage, not the voltage, on C2. It should be Q2 = U2xC2. The formulas for parallel and series connection of capacitors can be used.

12pawel · Answer

After connecting C2 and C3, Ci are connected to the capacitors C1 (2uF) and C23 (2uF). The supply voltage is U = U1 + U23 and since they both have 2uF, the voltage is distributed in half, so 2V. So you have 2V on the replacement capacitor. You break the capacitor back into C2 and C3 and that they are in parallel, so the voltage is the same, i.e. 2V
Such a trivial task. I advise you to go to a humanist because you will not be an engineer. Because if an engineer cannot, he knows where to look.

quaka456 · Answer

Thanks for the answers.
I would like to make sure that I understand it correctly.
If I have a layout like the picture and want to calculate $$Q_{4}$$ and $$U = 9V$$ this:
$$U = U_{12} + U_{34} + U_{5}$$ that is $$U_{34} = 3V$$ and I'm using the formula now $$ Q = UC $$ that is $$ Q_{4}= 3 \cdot 2 = 6 $$ .
Do I understand it correctly ?

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12pawel · Answer

The C4 capacitor will be 1.57V so the charge is 1.57 * 2

You cannot divide 9V evenly across 3 capacitors because they have different capacitance.
You calculate the equivalent capacity of the entire system, then you calculate the total charge, and from the charge you calculate the individual voltages on the capacitors.

bodzio507 · Answer

You have to define everything exactly, because everything can be calculated from the dependence Q = CU appropriately transformed.

We do this type of task by folding the circuit into one capacitor and then returning it to the full circuit.

Artur k. · Answer

You have to remember the basic rules for connecting capacitors - in a parallel connection, we have the same voltages on each capacitor, the charges accumulated in them are different (according to the known formula). On the other hand, in a series connection it is the other way round - the charges accumulated in the capacitors are the same, but the voltages on the capacitors are different (of course also according to the known formula).

Only this is needed to properly solve these tasks.

jesion40 · Answer

This is probably the answer the test was about. However, in fact, without additional assumptions, it is impossible to answer this question. I do not mean the trivial assumption that capacitors do not have leakage. Much more questionable is the assumption that their charges were zero before they were connected and connected to this system.

Peter134 · Answer

In such calculations, it is assumed that the system is perfect.

Artur k. · Answer

And why is that?

Calculating Capacitor Load: Using Capacity and Formula from Capacitor Layout Appendix

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