Discharge of a capacitor and current in an RC circuit

[greg_rack]

Homework Statement

When the switch is open, the capacitor is charged and there isn't any current flowing through the resistors.
In light of this, the potential between the resistors is $16V$.
As soon as the switch gets closed, this potential varies immediately and the capacitor starts to discharge.
Determine the currents in $R_{1}$ and $R_{2}$:
-as soon as the switch gets closed;
-when the capacitor is empty.

DIAGRAM ATTACHED BELOW$\downarrow$

Relevant Equations

$q=CV$
$q(t)=q_{0}e^{-\frac{t}{\tau}}$
$I=\frac{V}{R}$

I'm having a few troubles understanding a few things about this circuit...
Firstly, what does "In light of this, the potential between the resistors is $16V$" exactly mean? If current isn't flowing, how could there be a potential between resistors?
Secondly, how does current flow with the switch closed? There is still a generator attached to the circuit... how could a capacitor discharge if attached to a generator?

 

[berkeman]

"In light of this, the potential between the resistors is "

It's confusing wording, but they just mean that the voltage all along the top rail (including the resistors) is 16V. So all 3 nodes are at 16V (top left, top middle, top right).

Secondly, how does current flow with the switch closed?

When the switch is closed, it shorts the middle top node to ground, forming two separate circuits. The left circuit is the voltage source and resistor, with a steady current flowing I=V/R, and the right circuit is the capacitor in series with the right resistor, which discharges the capacitor with the time constant RC. Does that help?

 

[etotheipi]

You have my sympathy, it would be clearer if they wrote "the potential of the node between the two resistors is 16V".

@berkeman gave you a very brilliant answer! To derive the results he stated, you should apply Kirchhoff's voltage law to the left loop, then the right loop, individually.

 

[greg_rack]

It's confusing wording, but they just mean that the voltage all along the top rail (including the resistors) is 16V. So all 3 nodes are at 16V (top left, top middle, top right).

But, if current isn't flowing(right?), can we speak of voltage?

When the switch is closed, it shorts the middle top node to ground, forming two separate circuits. The left circuit is the voltage source and resistor, with a steady current flowing I=V/R, and the right circuit is the capacitor in series with the right resistor, which discharges the capacitor with the time constant RC. Does that help?

That does definitely help!
But now, how could I determine the current flowing through the second resistor as soon as the switch gets closed? If the right mesh is now separate from the first, in order to know this current I should know the voltage of the capacitor, given by $V_{in}-I_{1}R_{1}-I_{2}R_{2}$, and then apply Kirchhoff... right?

 

[greg_rack]

You have my sympathy, it would be clearer if they wrote "the potential of the node between the two resistors is 16V".

@berkeman gave you a very brilliant answer! To derive the results he stated, you should apply Kirchhoff's voltage law to the left loop, then the right loop, individually.

But how could I apply Kirchhoff's law without knowing the operating voltage of the capacitor?

 

[berkeman]

But, if current isn't flowing(right?), can we speak of voltage?

When you hold a 9V battery and look at the two terminals, there is no current flowing between them, but there certainly is 9V between them, right?

So when the voltage source in the circuit is first turned on with the switch open, a current will flow through the two resistors to gradually charge up the capacitor to 16V. Once the cap is fully charged, the whole top 3 nodes are all at 16V and current stops flowing.

If the right mesh is now separate from the first, in order to know this current I should know the voltage of the capacitor

The capacitor is initially charged to 16V and stays there until the switch is closed. What is the equation for V(t) for the capacitor as it is discharged by the right resistor R?

 

[greg_rack]

When you hold a 9V battery and look at the two terminals, there is no current flowing between them, but there certainly is 9V between them, right?

So when the voltage source in the circuit is first turned on with the switch open, a current will flow through the two resistors to gradually charge up the capacitor to 16V. Once the cap is fully charged, the whole top 3 nodes are all at 16V and current stops flowing.

Ok, great! But why does the capacitor charge to 16V and not to $V_{in}-I_{1}R_{1}-I_{2}R_{2}$? Resistors cause a drop in potential... so won't the cap get a lower voltage than 16V?

 

[berkeman]

Think how an RC circuit behaves. You get an exponential rise or fall in the cap voltage (depending on if you are charging or discharging it)...

 

[greg_rack]

Think how an RC circuit behaves. You get an exponential rise or fall in the cap voltage (depending on if you are charging or discharging it)...

Yup, you're right... and cap voltage increases until reaching the EMF of the battery!

 

[greg_rack]

Yup, you're right... and cap voltage increases until reaching the EMF of the battery!

But I have a doubt: I remember having solved an exercise by calculating the drop in tension caused by a resistor... in order to find the maximum charge in a capacitor:

here, I calculated the current flowing(2A) when the capacitor reached its maximum charge, and then calculated the drop in tension caused by the first resistor(4V) for calculating the max q=C(12V-4V).
Why, here, the maximum charge isn't q=C*EMF? @berkeman

 

[berkeman]

here, I calculated the current flowing(2A) when the capacitor reached its maximum charge, and then calculated the drop in tension caused by the first resistor(4V) for calculating the max q=C(12V-4V).
Why, here, the maximum charge isn't q=C*EMF? @berkeman

I'm not sure I understand your question or your confusion. That voltage divider in steady state puts 8V across the capacitor, so the charge is as you wrote q=C*8V.

 

[greg_rack]

I'm not sure I understand your question or your confusion. That voltage divider in steady state puts 8V across the capacitor, so the charge is as you wrote q=C*8V.

Right, and assuming the voltage divider wasn't there, what would the maximum charge of the capacitor have been like?

 

[berkeman]

You mean no resistor in parallel with the capacitor? Then the cap would charge all the way up to the power supply voltage (with an RC time constant exponential charging waveform) and stay there. So the q=CV calculation just uses the full power supply voltage.

 

[greg_rack]

You mean no resistor in parallel with the capacitor? Then the cap would charge all the way up to the power supply voltage (with an RC time constant exponential charging waveform) and stay there. So the q=CV calculation just uses the full power supply voltage.

Ok, I'm finally there!
Circuits are messing me up
Thank you so much for your patience!