Calculating Charge and Potential in a Series RC Circuit

In summary: In the first case the switch has been closed for a long time and there is a path that current can follow. It passes through R2. That's when it dissipates the power that's given in the problem statement. You should be able to determine several things given that information, including the voltage drops across R2 and R1 and the voltage of the battery. But first you need to be clear on what path the current takes in the...In the first case, the current would flow through R2 and then dissipate the power.In the second case, the switch has been opened for a short time and no current would flow because the left side of the switch is connected to the ground.
  • #1
terryds
392
13

Homework Statement


rc.jpg
An electric circuit is shown as above.
The switch S is closed for a long time and the current flowing is constant.
Given that C1 = 3 μF, C2 = 6 μF, R1 = 4 kΩ, R2 = 7 kΩ
If the power at R2 resistor is 2,4 W

a) Determine the charge at C1
b) When the switch is opened, after some miliseconds, how much does charge at C2 change?

Homework Equations


KVL, KCL

The Attempt at a Solution



I still have no idea whether the current at right loop is zero when the switch is opened. Please help..
 
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  • #2
terryds said:
I still have no idea whether the current at right loop is zero when the switch is opened. Please help..
Why? What analyses have you done? What are the initial conditions and final conditions for C2?
 
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  • #3
gneill said:
Why? What analyses have you done? What are the initial conditions and final conditions for C2?

When the switch is closed, the current flow through all the components, C1, R2, R1, and C2.
After the switch is opened, the current doesn't flow through C2 (because the left junction from C2 is a dead-end). The current just flows through C1 and R2.
Is it right?

Sorry for replying too slow... I've got busy lately
 
  • #4
terryds said:
When the switch is closed, the current flow through all the components, C1, R2, R1, and C2.
The problem states that the switch has been closed for a long time. Will current be flowing through C1 and C2 after a long time? What does the phrase, "after a long time" imply for DC circuits that have capacitors in them?
After the switch is opened, the current doesn't flow through C2 (because the left junction from C2 is a dead-end). The current just flows through C1 and R2.
Is it right?
No. When the switch is open, what's the difference between the C1 branch and the C2 branch (other than component values)? Is there any difference?
 
  • #5
gneill said:
The problem states that the switch has been closed for a long time. Will current be flowing through C1 and C2 after a long time? What does the phrase, "after a long time" imply for DC circuits that have capacitors in them?

No. When the switch is open, what's the difference between the C1 branch and the C2 branch (other than component values)? Is there any difference?

Sorry, I missed a thing.
After a long time means that the capacitor has fully discharged.

After the switch is open, the voltage between C1 equals C2, right? (assume that the time is after a long time after opening the switch)
But, still, I cannot determine the voltage at the capacitors. But, I think I can get the voltage at R2 by using equation P = V^2/R.
Does that mean that the voltage at R2 = voltage at C1 = voltage at C2 = voltage at R1 since I assume that the switch has been opened for a long time (no current flowing)? But I'm not sure if that's what the question asks. And, if I assume that there is no current flowing, how come there is P at R2 since P means that power dissipated because of currents goes to heat, right?

Please help
 
  • #6
terryds said:
Sorry, I missed a thing.
After a long time means that the capacitor has fully discharged.
Or charged. Capacitors will charge or discharge depending upon their initial state and circuit they're in. They reach whatever final charge is dictated by the potential difference presented to them by the circuit.
After the switch is open, the voltage between C1 equals C2, right? (assume that the time is after a long time after opening the switch)
Yes. They will end up having the same voltage.
But, still, I cannot determine the voltage at the capacitors. But, I think I can get the voltage at R2 by using equation P = V^2/R.
Does that mean that the voltage at R2 = voltage at C1 = voltage at C2 = voltage at R1 since I assume that the switch has been opened for a long time (no current flowing)? But I'm not sure if that's what the question asks. And, if I assume that there is no current flowing, how come there is P at R2 since P means that power dissipated because of currents goes to heat, right?
Don't confuse the two steady states that the circuit describes. Take them one at a time.

In the first case the switch has been closed for long time and there is a path that current can follow. It passes through R2. That's when it dissipates the power that's given in the problem statement. You should be able to determine several things given that information, including the voltage drops across R2 and R1 and the voltage of the battery. But first you need to be clear on what path the current takes in the circuit and what components are in parallel (so that they share the same potential difference).

In the second case the switch has been open for a long time. Then you have two parallel RC branches that are effectively isolated from each other, connected to a voltage source. Can current continue to flow in either branch after a long time? If not, what potential difference must the capacitors achieve?
 
  • #7
gneill said:
Or charged. Capacitors will charge or discharge depending upon their initial state and circuit they're in. They reach whatever final charge is dictated by the potential difference presented to them by the circuit.

Yes. They will end up having the same voltage.

Don't confuse the two steady states that the circuit describes. Take them one at a time.

In the first case the switch has been closed for long time and there is a path that current can follow. It passes through R2. That's when it dissipates the power that's given in the problem statement. You should be able to determine several things given that information, including the voltage drops across R2 and R1 and the voltage of the battery. But first you need to be clear on what path the current takes in the circuit and what components are in parallel (so that they share the same potential difference).

In the second case the switch has been open for a long time. Then you have two parallel RC branches that are effectively isolated from each other, connected to a voltage source. Can current continue to flow in either branch after a long time? If not, what potential difference must the capacitors achieve?

16804788_836191036523412_1908678309_o.jpg

First case,
V at R_2 = sqrt(P*R_2) = sqrt(2.4*7000) = 129,6 V

C1 is parallel to R1, means the voltage is equal at the time current was flowing.
C2 is parallel to R2, means the voltage is equal at the time current was flowing.

When the switch is closed for a long time (capacitor fully charged), there is no current flowing through the capacitors means no current in mesh 1, 2, and 3 (?), but how come there is current passing through R2?
I'm so confused, have no idea the path that current takes in.
 
  • #8
terryds said:
View attachment 113409
First case,
V at R_2 = sqrt(P*R_2) = sqrt(2.4*7000) = 129,6 V

C1 is parallel to R1, means the voltage is equal at the time current was flowing.
C2 is parallel to R2, means the voltage is equal at the time current was flowing.
Yes, the above is correct. So you now know the voltage on C2 for the first case (switch closed), right? Given that, what's the charge on C2 for this case?
When the switch is closed for a long time (capacitor fully charged), there is no current flowing through the capacitors means no current in mesh 1, 2, and 3 (?), but how come there is current passing through R2?
I'm so confused, have no idea the path that current takes in.
You probably don't want to apply mesh analysis until you've first determined where current can flow. At steady state, capacitors behave like open circuits (no current flowing through them). If you take your circuit diagram and erase the capacitors, that's how the circuit looks at steady state as far as the available current paths are concerned. You can apply whatever analysis method you wish then.
 
  • #9
gneill said:
Yes, the above is correct. So you now know the voltage on C2 for the first case (switch closed), right? Given that, what's the charge on C2 for this case?

You probably don't want to apply mesh analysis until you've first determined where current can flow. At steady state, capacitors behave like open circuits (no current flowing through them). If you take your circuit diagram and erase the capacitors, that's how the circuit looks at steady state as far as the available current paths are concerned. You can apply whatever analysis method you wish then.

At mesh 1, it will be counter-clockwise since if it's clockwise, the current can't get to the resistor (it will meet the dead-end a.k.a steady state capacitor first).
I think the circuit will just be like voltage - R1 - R2..

By voltage divider, the voltage at R2 is R1/(R1+R2) * V and the voltage at R1 is R2/(R1+R2) * V

So,

V at R_2 = R_1/(R_1+R_2) * V
sqrt( R_2 P) = R_1/(R_1+R_2) *V

By plugging P = 2,4 W, R_1 = 4000 ohms, R_2 =7000 ohms, we get V = 356.44 Volt.

Which means that V at R_2 is 129.6 V
V at R_1 is 226.8 V

After the switch is opened,
Charge at C_2 = Q_2 = C_2 * V_2 = 6* 10^-6 F * 129.6 V = 777.6 * 10^-6 C
Charge at C_1 = Q_1 = R_1 * V_1 = 3*10^-6 F * 226.8 V = 680.4 * 10^-6 C

Is it correct?

But, I still can't answer "b) When the switch is opened, after some miliseconds, how much does charge at C2 change?"
 
  • #10
A good attempt. Although the direction of your mesh current won't matter, you just have to identify the mesh first. In fact, with the capacitors eliminated from consideration there is only one loop. If you re-draw the circuit assuming that the capacitors are open circuits:
upload_2017-2-19_10-8-40.png
terryds said:
By voltage divider, the voltage at R2 is R1/(R1+R2) * V and the voltage at R1 is R2/(R1+R2) * V
Check your voltage divider equations. You have them backwards.

terryds said:
But, I still can't answer "b) When the switch is opened, after some miliseconds, how much does charge at C2 change?"

Re-draw the circuit without the switch. Consider the phrase "some milliseconds" to be equivalent to "after a long time". Given that you now know the source voltage, what voltage will end up on C2?
 
  • #11
gneill said:
A good attempt. Although the direction of your mesh current won't matter, you just have to identify the mesh first. In fact, with the capacitors eliminated from consideration there is only one loop. If you re-draw the circuit assuming that the capacitors are open circuits:
View attachment 113434
Check your voltage divider equations. You have them backwards.
Re-draw the circuit without the switch. Consider the phrase "some milliseconds" to be equivalent to "after a long time". Given that you now know the source voltage, what voltage will end up on C2?

Alright, I missed it.

V at R_2 is 129.6 V
V at R_1 is 74 V

Sorry, I still have no idea to determine the voltage at C2 when the switch is opened.
What should I do?
 
  • #12
terryds said:
Alright, I missed it.

V at R_2 is 129.6 V
V at R_1 is 74 V
Good.
Sorry, I still have no idea to determine the voltage at C2 when the switch is opened.
What should I do?
Can you determine the value of the source voltage from what you now know from part (a)?

Draw the circuit without the switch branch (it's open so it can be removed from consideration). What potential difference is across the C2 branch?
 
  • #13
gneill said:
Good.

Can you determine the value of the source voltage from what you now know from part (a)?

Draw the circuit without the switch branch (it's open so it can be removed from consideration). What potential difference is across the C2 branch?

The source voltage is 129.6*11/7 = 203.65 V

The circuit will be like voltage source - (C_1 and R_2) parallled with (R_1 and C_2)

The potential difference is the voltage source minus i_1 R_1, right?
By assuming there is no resistance in capacitors, it means that the current from the source is V/R_paralel = 203.65/(4000*7000/(4000+7000)) = 0.08 A

By using current divider, the current i_1 = i R_2/(R_1+R_2) = 0.0509 A

So, the voltage across C2 is 203.65 V - 0.0509 A (4000) = 0.05 V

Is it correct?
 
  • #14
terryds said:
The source voltage is 129.6*11/7 = 203.65 V
Good.
The circuit will be like voltage source - (C_1 and R_2) parallled with (R_1 and C_2)

The potential difference is the voltage source minus i_1 R_1, right?
By assuming there is no resistance in capacitors, it means that the current from the source is V/R_paralel = 203.65/(4000*7000/(4000+7000)) = 0.08 A

By using current divider, the current i_1 = i R_2/(R_1+R_2) = 0.0509 A

So, the voltage across C2 is 203.65 V - 0.0509 A (4000) = 0.05 V

Is it correct?
No. You're looking for the potential across the capacitor after a long time. What's the steady state current through a capacitor?
 
  • #15
gneill said:
Good.

No. You're looking for the potential across the capacitor after a long time. What's the steady state current through a capacitor?

The current becomes zero at the capacitor after a long time (capacitor becomes like an open circuit)
So, the voltage is supposed to be equal with the voltage source, right?

Or, is the voltage zero because V = I R where I is zero?

Please help
 
  • #16
terryds said:
The current becomes zero at the capacitor after a long time (capacitor becomes like an open circuit)
So, the voltage is supposed to be equal with the voltage source, right?
Right.
Or, is the voltage zero because V = I R where I is zero?
No, that just means that the potential drop across R is zero.
 
  • #17
gneill said:
Right.

No, that just means that the potential drop across R is zero.

So, the charge in C2 is (203.65 V)(6*10^-6 F) = 1.22 *10^-3 C , right?

Anyway, the voltage is C1 is also equal to the voltage source, right?
And, there is no current flowing through resistors since all the capacitors have reached the steady state condition?
Because the current is not flowing, it means that the potential of each resistor equal to each capacitor which is the same as the voltage source.
Is it correct?

I need help to understand this concept, sometimes it's confusing
 
  • #18
terryds said:
So, the charge in C2 is (203.65 V)(6*10^-6 F) = 1.22 *10^-3 C , right?

Anyway, the voltage is C1 is also equal to the voltage source, right?
And, there is no current flowing through resistors since all the capacitors have reached the steady state condition?
Right.
Because the current is not flowing, it means that the potential of each resistor equal to each capacitor which is the same as the voltage source.
Is it correct?
What do you mean by "the potential of each resistor"? If you mean the potential drop across each resistor, then no. With no current there is no potential drop across a resistor: ##V = I~R##. The capacitors attain the same potential difference as the source, that is, they end up with a voltage across them that is equal to the source voltage. The resistors have zero potential difference.
 
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  • #19
gneill said:
Right.

What do you mean by "the potential of each resistor"? If you mean the potential drop across each resistor, then no. With no current there is no potential drop across a resistor: ##V = I~R##. The capacitors attain the same potential difference as the source, that is, they end up with a voltage across them that is equal to the source voltage. The resistors have zero potential difference.

Thanks for good explanation. But, I want to ask some more, please
Is the voltage at capacitors equal to source because they are parallel, or is it just because of the steady-state condition?

I mean, if we say there is another capacitor in series with C_1 or C_2, does it also attain the same potential difference as the voltage source? Or will it be shared among the capacitors in series? If it's shared, how to determine the voltage at each capacitor since we can't use voltage divider equation?
 
  • #20
terryds said:
Thanks for good explanation. But, I want to ask some more, please
Is the voltage at capacitors equal to source because they are parallel, or is it just because of the steady-state condition?
Current flows until the potential across the capacitor equals the potential across the voltage source and the potential drop across the resistor is zero. When the potentials are the same there is no potential difference to drive current through the resistor.
I mean, if we say there is another capacitor in series with C_1 or C_2, does it also attain the same potential difference as the voltage source? Or will it be shared among the capacitors in series?
For a series connection the same amount charge flows onto both (since they are in series they must share identical current). If the capacitors have different values then they end up with different potentials across them (Q = CV).
If it's shared, how to determine the voltage at each capacitor since we can't use voltage divider equation?
There is a slightly different voltage divider equation for capacitors. Since every capacitor in series has the same charge, and since the voltage on a capacitor depends on its charge, smaller capacitors end up with larger voltages and larger capacitors end up with smaller voltages. Thus for two capacitors in series the voltages are inversely proportional to their capacitance. If you have two capacitors C1 and C2 in series and a total voltage V across them, then:

##V_{C1} = V \frac{C2}{C1 + C2}##

##V_{C2} = V \frac{C1}{C1 + C2}##
 
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  • #21
gneill said:
Current flows until the potential across the capacitor equals the potential across the voltage source and the potential drop across the resistor is zero. When the potentials are the same there is no potential difference to drive current through the resistor.

For a series connection the same amount charge flows onto both (since they are in series they must share identical current). If the capacitors have different values then they end up with different potentials across them (Q = CV).

There is a slightly different voltage divider equation for capacitors. Since every capacitor in series has the same charge, and since the voltage on a capacitor depends on its charge, smaller capacitors end up with larger voltages and larger capacitors end up with smaller voltages. Thus for two capacitors in series the voltages are inversely proportional to their capacitance. If you have two capacitors C1 and C2 in series and a total voltage V across them, then:

##V_{C1} = V \frac{C2}{C1 + C2}##

##V_{C2} = V \frac{C1}{C1 + C2}##

Thanks a lot. You're the best!
 

Related to Calculating Charge and Potential in a Series RC Circuit

1. How does an RC circuit charge?

An RC (resistor-capacitor) circuit charges by using a DC voltage source to charge the capacitor through the resistor. As the capacitor charges, the voltage across the capacitor increases until it reaches the same voltage as the source. Once this happens, the capacitor stops charging and reaches a steady state.

2. What is the equation for calculating the charge in an RC circuit?

The equation for calculating the charge in an RC circuit is Q = Q0(1-e-t/RC), where Q is the charge at time t, Q0 is the initial charge, R is the resistance, C is the capacitance, and e is the base of the natural logarithm.

3. How does the resistance affect the charging time of an RC circuit?

The resistance in an RC circuit affects the charging time by determining the rate at which the capacitor charges. A higher resistance will result in a slower charging time, while a lower resistance will result in a faster charging time.

4. What happens to the charge in an RC circuit when the capacitor is fully charged?

Once the capacitor in an RC circuit is fully charged, the charge will stop increasing and reach a steady state. This means that the voltage across the capacitor will be equal to the voltage of the source, and no more charge will flow through the circuit.

5. What is the time constant in an RC circuit?

The time constant in an RC circuit is the time it takes for the capacitor to charge to 63.2% of its maximum charge. It is calculated by multiplying the resistance (R) by the capacitance (C), or τ = RC.

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