The voltages across capacitors in series

[Josielle Abdilla]

Homework Statement

In a circuit, 2 capacitors of 8 microFarads and 6 microFarads are fully charge. These are also connected to 20kohms resistor. What is the charge across each resistor. The emf of cell is 6V

Relevant Equations

Q=CV
Q1 =Q2 (in series)

I found total capacitance and inserted the total capacitance and emf of cell in equation CV =Q. However I know that there is a resistor connected so that this accounts for lost volts

 

[Doc Al]

What is the charge across each resistor.

I assume the question asks for the charge on each capacitor, not the resistor.

However I know that there is a resistor connected so that this accounts for lost volts

Once the capacitors are fully charged, what's the voltage across the resistor?

 

[DaveE]

The voltage across the resistor is V=IR. What is the current in the circuit when the capacitors are fully charged?

 

[Josielle Abdilla]

The current is not known and hence I can not find the voltage across the resistor

 

[Doc Al]

The current is not known and hence I can not find the voltage across the resistor

I know it. And so should you!

 

[Josielle Abdilla]

I assume the question asks for the charge on each capacitor, not the resistor.


Once the capacitors are fully charged, what's the voltage across the resistor?

I understand that the question is asking for charge of each capacitor but in order to do so the voltage of each capacitor must be found. I couldn't find the voltage across each capacitor by the potential divider rule due to unknown Voltage of the resistor connected in series

 

[Josielle Abdilla]

I assume the question asks for the charge on each capacitor, not the resistor.


Once the capacitors are fully charged, what's the voltage across the resistor?

I understand that the question is asking for charge of each capacitor but in order to do so the voltage of each capacitor must be found. I couldn't find the voltage across each capacitor by the potential divider rule due to unknown Voltage of the resistor connected in sereis

I know it. And so should you!

How can I calculate the current then?

 

[Josielle Abdilla]

Should I calculate the current using V = IR by inserting emf of battery and the resostance of resistor. Does however capacitor offer reactance?

 

[Josielle Abdilla]

Can you send me a picture of the worked out answer because I am confused

 

[jbriggs444]

How can I calculate the current then?

If a capacitor is fully charged, is it getting charged further?

 

[Josielle Abdilla]

The capacitor is fully charged

 

[DaveE]

Can you send me a picture of the worked out answer because I am confused

I don't think you'll get that here. We can help you figure it out. Providing answers isn't the same as teaching.

Have you studied Kirchhoff's Voltage Law (KVL) yet? How can you relate the voltage on the capacitors and battery to the voltage on the resistor?

 

[Doc Al]

The capacitor is fully charged

Once the capacitors are fully charged, what happens to the current? Does it keep flowing? (Where can it go?)

Once you realize what we're hinting at, it will all make sense.

 

[Josielle Abdilla]

Current is 0 so does this mean that since no current passes through resistor, the total voltage is equal to the sum of pd across each capacitor?

 

[Doc Al]

Current is 0 so does this mean that since no current passes through resistor, the total voltage is equal to the sum of pd across each capacitor?

Exactly!

When the capacitors are uncharged and the switch is thrown, current flows until the capacitors are fully charged.

 

[Josielle Abdilla]

Thanks a lot! Hence, I can use the potential divider rule to solve i.e.
Vtotal/V(of capacitor) = Ctotal/C(of capacitor)

 

[DaveE]

Yes, good work!

 

[ehild]

Yes, good work!

No. The voltage across series capacitors is inversely proportional to the capacitance.

 

[Josielle Abdilla]

Yes, good work!

Thanks a lot. This conversation has been fruitful after all

 

[epenguin]

It might have been more or sooner fruitful if you had made it possible to read the question.
I have the impression there is a key one there not answered yet because we can't see what it is.

 

[Josielle Abdilla]

I am sorry to bother you again but in an LC circuit the current does no drop to 0 but varies sinusoidally. The capacitor is recharged with a different polarization. Why is this so?

 

[Doc Al]

I am sorry to bother you again but in an LC circuit the current does no drop to 0 but varies sinusoidally. The capacitor is recharged with a different polarization. Why is this so?

I assume you realize that the polarity of the capacitors depends on the polarity of the imposed voltage. So, if the latter changes -- as it does in an AC circuit, which is the situation here -- then the former must change as well.

Things are a bit more involved than with the DC circuit that started this thread, so you'll need to do some reading.

 

[DaveE]

This is a little hard to explain well if you haven't studied simple differential equations. But basically the current (or voltage, you can analyse it either way) in the components is out of phase with the other. Inductors will store current just like capacitors store voltage. So the inductor keeps the current flowing after the capacitor has discharged which transfers the energy stored in the inductor into the capacitor as the reverse polarity voltage. The process will then repeat in reverse with the energy in the capacitor causing the inductor current to increase in the opposite polarity.

This circuit is exactly analogous to a mass bouncing on a spring. For example, hang a mass from the ceiling with a spring. Notice the resting positions of the mass and spring. Now pull down on the mass and let it go. The mass will oscillate up and down with energy transferred cyclically between the mass and spring in the same way energy is transferred between the inductor and capacitor in your circuit.

I know this isn't that clear they way I described it. Look at this video for an example. This would also be a great way to use a circuit simulation, like LTSpice (which is good and free) to see how the voltages and currents interact.