# Capacitor Charging: Voltmeter, Charge & Voltage Calculations

• Biker
In summary, The voltmeter reading is 5 volts when the switch K is open and C1 is not charged. When you close K calculate the following: 1- Voltmeter reading 2- Charge of each capacitor 3- Voltage of each capacitor.
Biker

## Homework Statement

In the following diagram, The voltmeter reading is 5 volts when the switch K is open and C1 is not charged.
When you close K calculate the following:
2- Charge of each capacitor
3- Voltage of each capacitor

## Homework Equations

Usual electronics equation

## The Attempt at a Solution

Well, I was looking through my friend's private tutor papers and I found this question. My solution is different that his so I thought I could bring it here so someone can check it and see if I am missing something.

C1 is not charged thus it doesn't contribute to the voltage of the voltmeter (assuming actual current flowing or assuming it is negligible) so the voltage reading is purely C2's voltage

Now instead of assuming Q as transferred charged, you could choose a much simpler way.

C1 and C3 (it didnt say if C3 is charged so I assumed it is not) are in series then the resultant is parallel with C2
So
##\frac{Q_t}{C_{eq}} = V_{after~opening~ the switch} ##
you get ## V = 4.25 volts ##
The Charge for any of C1 and C3 capacitors is the voltage times the equivalent capacitance of C1 and C3 which is in series
## V ~ ( \frac{1}{C_1} + \frac{1}{C_3} ) = Q_{3 ~ or ~ 1} ##
## Q = 9 uC ##
the charge for C2 is simply
## Q_2 = C_2 V ##
## Q_2 = 51 uC ##

Now the first requirement needs the voltmeter reading which is:
##V = V_{C_2} - V_{C_1} ##
## V = 4.25 - Q/C_1 ##
## V = 4.25 - 2.25 ##
## V = 2 ## which he found to be the same
the rest is straight forward.

For the charges and voltages, he got it different because he assumed that C2 and C1 are connected in series which is wrong because they definitely don't have the same charge

Your method of solution looks correct to me. Your answers agree with what I got by first solving for the charge q that leaves C2 when the switch is closed. Your solution gets to the answer quicker!

Biker
TSny said:
Your method of solution looks correct to me. Your answers agree with what I got by first solving for the charge q that leaves C2 when the switch is closed. Your solution gets to the answer quicker!
Thank you TSny :).
I usually use that method to derive if I could treat those capacitor is a parallel or series way. Sometimes it can get complicated so I just use the Q method :D

Anyway, Thank you again. Much appreciated

Biker said:
C1 is not charged thus it doesn't contribute to the voltage of the voltmeter (assuming actual current flowing or assuming it is negligible) so the voltage reading is purely C2's voltage
Biker said:
(it didnt say if C3 is charged so I assumed it is not)
I would have thought that either the diagram is labelled wrongly or the given statement that C1 starts uncharged is a typo; should say C3. A voltmeter works by allowing a trickle current, so the two in series must both be charged.
On that basis, the two in series have combined capacitance of 3uF, and including the one in parallel we get 7.5uF. The final voltage is therefore (3/7.5)5=2. I think that was your argument, but I did not find it very clear.

I worried about the effect of the voltmeter, too. I ended up assuming that the voltmeter had no influence. I interpreted it as just a way to say that the total potential difference in going from the left side of C2 to the right side of C1 is 5 V when C1 is uncharged and the switch is open. That's a weird way to just say that initially C2 has 5 V while the other two caps are uncharged.

haruspex said:
I would have thought that either the diagram is labelled wrongly or the given statement that C1 starts uncharged is a typo; should say C3. A voltmeter works by allowing a trickle current, so the two in series must both be charged.
On that basis, the two in series have combined capacitance of 3uF, and including the one in parallel we get 7.5uF. The final voltage is therefore (3/7.5)5=2. I think that was your argument, but I did not find it very clear.
Doesn't the ideal voltmeter have infinite resistance? Which means a negligible amount of charges gets transferred from C2 to C1

I don't know about C1 being charged or no but I clearly remember that the question statement said C1 is not charged but It didnt mention C3Even if we assume C1 is charged with certain charge Q and C3 isn't, we can't be sure that ## Q_2 = Q_1 ## so that we can take it as series unless the question states that they perhaps were connected to the same voltage source initially or providing that they have the same charge

Last edited:
Biker said:
Doesn't the ideal voltmeter have infinite resistance? Which means a negligible amount of charges gets transferred from C2 to C1
No real voltmeter has infinite resistance, just very high, which means the current will be very small. How much charge is transferred depends on how long the voltmeter is connected. That said, I think we have to assume no significant charge is leaking through, otherwise it would eventually be reading 0. The two capacitors would have equal and opposite voltages.
Biker said:
we can't be sure that Q2=Q1
I don't think it matters. All we need to know is that the initial voltage reading is 5V. How the charge is arranged on those two capacitors is unimportant.
I suggest the question should be saying that the third capacitor starts with no charge; it's a misprint in either the text or the diagram.

haruspex said:
No real voltmeter has infinite resistance, just very high, which means the current will be very small. How much charge is transferred depends on how long the voltmeter is connected. That said, I think we have to assume no significant charge is leaking through
Yea I get that no real voltmeter has infinite resistance that is why I said ideal. Also it would be a bit difficult to calculate at least for my level.

haruspex said:
I don't think it matters. All we need to know is that the initial voltage reading is 5V. How the charge is arranged on those two capacitors is unimportant.
I suggest the question should be saying that the third capacitor starts with no charge; it's a misprint in either the text or the diagram.

Yes, I derived that just few secs ago. Assuming Q charge transfer method or Using the method you gave me above they both yield in the same variables.

But just a question, We can't find the B and C part of the question because we have no knowledge of Q1 and Q2 right?

Can anyone confirm the last question in the comment above?

Biker said:
But just a question, We can't find the B and C part of the question because we have no knowledge of Q1 and Q2 right?
The question does say that C1 is initially uncharged. So that's not an issue. The only possible impediment to finding all the steady-state charges is that no mention was made of an initial charge for C3. But you would be justified in assuming it to be initially uncharged and proceeding; Just state your assumption as part of your answer. Alternatively, assume some unknown initial charge and give the answers as expressions (symbolically). That would be more work, and probably not what the question author intended.

Biker

## 1. What is a capacitor?

A capacitor is an electronic component that stores energy in an electric field. It is made up of two conductive plates separated by an insulating material.

## 2. How does a capacitor charge?

A capacitor charges when a voltage is applied across its plates. Electrons from one plate are attracted to the other, creating an electric field. As more electrons accumulate, the voltage across the capacitor increases until it reaches the same voltage as the source.

## 3. How do you measure the voltage of a charging capacitor?

The voltage of a charging capacitor can be measured using a voltmeter. The voltmeter should be connected in parallel to the capacitor, meaning one probe on each of the capacitor's plates.

## 4. How do you calculate the charge of a capacitor?

The charge of a capacitor can be calculated using the equation Q = C x V, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor.

## 5. What is the relationship between voltage and capacitance in a charging capacitor?

The voltage and capacitance of a charging capacitor are directly proportional. This means that as the voltage increases, so does the capacitance, and vice versa. This relationship is described by the equation C = Q/V, where C is the capacitance, Q is the charge, and V is the voltage.

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