How Does a Capacitor Charge to Negative Voltage in an RC Circuit?

[ace8888]

Homework Statement

At initially, the switch is open meaning there is no current flowing. So how does the capacitor is initially charge to -5V? and what does a negative voltage capacitor means?

I thought it is only when the switch is closed (t=0) that when the capacitor is being charged.

Homework Equations

Vc = Vs*(1-e^-t/RC)
Vc = Vs*(e^-t/RC)[/B]

The Attempt at a Solution

2.5 = 10*(1-e^-t/RC), solving for t get me .000002s but this is not correct. The correct answer is .000004159s[/B]

 

[kuruman]

At initially, the switch is open meaning there is no current flowing. So how does the capacitor is initially charge to -5V? and what does a negative voltage capacitor means?

With the switch open, imagine connecting a second, 5 V battery to the capacitor such that the negative terminal is on the top capacitor plate. Once the capacitor is charged, the second battery is disconnected leaving the capacitor in the initial state of a negatively charged top plate. Thus the starting value of the voltage is -5 V. When the switch is closed, the voltage across the capacitor becomes less negative, goes through zero and then charges up to the full value of 10 V.

Positive voltage is when the top plate has positive charges and negative voltage when it has negative charges. How do I know? Because the 10 V battery has its positive terminal connected to the top plate.

 

[ace8888]

With the switch open, imagine connecting a second, 5 V battery to the capacitor such that the negative terminal is on the top capacitor plate. Once the capacitor is charged, the second battery is disconnected leaving the capacitor in the initial state of a negatively charged top plate. Thus the starting value of the voltage is -5 V. When the switch is closed, the voltage across the capacitor becomes less negative, goes through zero and then charges up to the full value of 10 V.

Positive voltage is when the top plate has positive charges and negative voltage when it has negative charges. How do I know? Because the 10 V battery has its positive terminal connected to the top plate.

The voltage will begin to charge the capacitor to the match the source when the switch closed for a long period of time. They wanted to know how long it would take for the capacitor voltage to equal 2.5V, I am taking it that they are asking how long it would take the voltage source to charge the capacitor to 2.5V. The answer i get is not correct though.

 

[haruspex]

The answer i get is not correct though.

The equation you used is inappropriate. If you plug in t=0 and t infinite you can see that it is for the case of a capacitor starting at 0V and finishing at 10V.
How do you think you could modify the equation to fit the given data?

 

[ace8888]

The equation you used is inappropriate. If you plug in t=0 and t infinite you can see that it is for the case of a capacitor starting at 0V and finishing at 10V.
How do you think you could modify the equation to fit the given data?

It said that the switch has been open for a "long time". Shouldn't the capacitor at t=0 (meaning right after the switch closes), still sitting at 0V at the start and if T>>0 (like infinity) the capacitor will charge up to the source voltage which is 10V so isn't the equation i used is correct? capacitors do not jump up in voltage at an instant the switch is closed. The thing that i can't get around is the -5V initially charged of the capacitor and how it play into the this problem.

 

[haruspex]

It said that the switch has been open for a "long time".

I interpreted that as meaning it is in steady state, no current.

 

[ace8888]

I interpreted that as meaning it is in steady state, no current.

No current b/c the switch is open. After the switch closes, there is current flowing but there is no voltage across the capacitor yet. I still don't understand where you said the equation that I've used is wrong.

 

[Asymptotic]

No current b/c the switch is open. After the switch closes, there is current flowing but there is no voltage across the capacitor yet.

No. -5V is across the capacitor when the switch is closed.
How much time does it take for voltage to increase from -5V to 0V?

 

[SammyS]

No current b/c the switch is open. After the switch closes, there is current flowing but there is no voltage across the capacitor yet. I still don't understand where you said the equation that I've used is wrong.

Here is your attached file with the problem statement. Clearly, there is a potential difference of 5 V across the capacitor prior to he switch being closed.

 

[ace8888]

No. -5V is across the capacitor when the switch is closed.
How much time does it take for voltage to increase from -5V to 0V?

Here is your attached file with the problem statement. Clearly, there is a potential difference of 5 V across the capacitor prior to he switch being closed.
View attachment 205709

Ok, so when t=0 the capacitor is initially charged to -5V but the source is 10V which mean it will try to charge the capacitor to match it value. How does the discharge and charging work at the same time? If it can;t be done, how does the capacitor discharge while the voltage source is still connected to it and has a higher potential value than itself. Where is the potential energy of the capacitor goes? In other word, where is the capacitor going to discharge the -5V when it is connected to a voltage source that has a higher potential.

 

[haruspex]

No current b/c the switch is open. After the switch closes, there is current flowing but there is no voltage across the capacitor yet. I still don't understand where you said the equation that I've used is wrong.

It is the combination of switch open and long time that guarantees no current. When devices like capacitors and inductors are present in a circuit there can be residual oscillatory currents in an open circuit.
Your equation must satisfy the given conditions: at t=0 the voltage difference is -5 and at infinity tends to +10. In the general form A+Be^-t/RC, what values of A and B achieve that?

Ok, so when t=0 the capacitor is initially charged to -5V but the source is 10V which mean it will try to charge the capacitor to match it value. How does the discharge and charging work at the same time? If it can;t be done, how does the capacitor discharge while the voltage source is still connected to it and has a higher potential value than itself. Where is the potential energy of the capacitor goes? In other word, where is the capacitor going to discharge the -5V when it is connected to a voltage source that has a higher potential.

There is no conflict. Instead of of charging from 0 to +10 it charges from -5 to +10. If you are overdrawn $5 at the bank and pay $15 in it takes you from -5 to +10.

 

[gneill]

Ok, so when t=0 the capacitor is initially charged to -5V but the source is 10V which mean it will try to charge the capacitor to match it value. How does the discharge and charging work at the same time?

They are not different events, but part of a single process. Current drives charge onto the capacitor and changes its potential difference. The process is continuous and there's no "glitch" or change in behavior as the charge passes through zero. What establishes the curve is the total change in potential difference. In this case the capacitor's potential difference is going to go from -5 V to + 10 V, for a total change of +15 V. Consider the following curve:

V_start does not have to be zero; it can be any value, positive or negative.

If it can;t be done, how does the capacitor discharge while the voltage source is still connected to it and has a higher potential value than itself. Where is the potential energy of the capacitor goes? In other word, where is the capacitor going to discharge the -5V when it is connected to a voltage source that has a higher potential.

It's simply going to follow an exponential curve from -5 V to +10 V.

In a circuit, current is driven by potential difference. It might help you to think of the circuit first in terms of the current that will be driven through the resistance.

Choosing the negative end of V_s as the zero reference, at time zero the left end of the resistor R will be at 10 V while its right end will be held at -5 V by the initial capacitor charge. That will produce some initial value for the current (what is it?). Eventually steady state will be reached and current will stop flowing (the capacitor's voltage will match that of the source voltage). So the current will start at some initial value at time t = 0 and end at zero after some time. That current will follow the standard exponential decay curve, and you should be in a position to write its equation drawing on the "stock" exponential decay expression (akin to your second Relevant Equation, but for current rather than voltage).

Now, knowing the current w.r.t. time you could write an expression for the charge on the capacitor over time and hence its voltage over time. Or, use that current equation to determine the potential drop across R w.r.t. time, and then use Kirchhoff to find V_c in the circuit. But a simpler idea is to just realize that the capacitor will be experiencing a total change in potential of 10 - (-5) = +15 volts. That is, it will follow an exponential "charging" curve starting at -5 V and ending at +10 V, of a total change of +15 V. You should be able to write an expression for a capacitor charging from 0 to 15 V. Then it's just a matter of "offsetting" that by the initial value of -5 V

 

[ace8888]

Choosing the negative end of V_s as the zero reference, at time zero the left end of the resistor R will be at 10 V while its right end will be held at -5 V by the initial capacitor charge. That will produce some initial value for the current (what is it?). Eventually steady state will be reached and current will stop flowing (the capacitor's voltage will match that of the source voltage). So the current will start at some initial value at time t = 0 and end at zero after some time. That current will follow the standard exponential decay curve, and you should be in a position to write its equation drawing on the "stock" exponential decay expression (akin to your second Relevant Equation, but for current rather than voltage).

Now, knowing the current w.r.t. time you could write an expression for the charge on the capacitor over time and hence its voltage over time. Or, use that current equation to determine the potential drop across R w.r.t. time, and then use Kirchhoff to find V_c in the circuit. But a simpler idea is to just realize that the capacitor will be experiencing a total change in potential of 10 - (-5) = +15 volts. That is, it will follow an exponential "charging" curve starting at -5 V and ending at +10 V, of a total change of +15 V. You should be able to write an expression for a capacitor charging from 0 to 15 V. Then it's just a matter of "offsetting" that by the initial value of -5 V

This is my understand, initially the current will be the maximum value when t=0 and when the capacitor getting up to the value of the source, there will not be any current (time to infinity). I've graphed the current vs. time

As for your question about the value of current when t=0. This is where I get stuck,

KVL:

q=c*v; I=dq/dt

E - IR - V(c) =0

E- (dq/dt)R - q/c =0

Solving the DE equation:

Vc(t) = Vf (1-e^-1/RC); where vf is when the capacitor has been charged for long time and it is now act as an open circuit. No current flow, which mean the resistor voltage is zero and the sum of the voltage drop across the circuit is now just E = Vf (final voltage across the capacitor equal to the source).

I(t) = Io*e^(-t/RC); the Io (I initial) is when t=0 the switch just closed. The charge across the capacitor is still zero as it can't change instantaneously. therefore, the initial current Io is just the Io=E/R. However this is not what we have in this case. We have a capacitor that was charged initial at -5V so they will be some current flowing to across the resistor from the capacitor as you have stated. I'm not sure how to modify the derived equation to account for this initial current of the capacitor.With the charging equation above: Vc(t) = Vf (1-e^-1/RC)

t=0, e^0 =1

Vc(t) = 1-1 = 0 (this need to be -5V to account for the particular problem that we have).

t= infinity, e^infinity = 0

Vc(t) = Vf (which is 10V)

I can do trial & error with the coefficients to get the result but is there an more systematic approach like an derived equation that take account for the initial value of the capacitor (positive or negative)? did i miss something that i could have done to account for this in my DE equation?

thank you so much.

 

[haruspex]

derived equation that take account for the initial value

If you look at the detailed steps in solving the differential equation, one of them is an integratIon. Whenever you integrate you should consider the unknown constant of integration. This is generally determined by the boundary conditions. The initial voltage on the capacitor is your boundary condition at t=0.

 

[gneill]

This is my understand, initially the current will be the maximum value when t=0 and when the capacitor getting up to the value of the source, there will not be any current (time to infinity). I've graphed the current vs. time

It will be a "standard" exponential decay:

As for your question about the value of current when t=0. This is where I get stuck,

What's the potential difference across the resistor? It's got 10 V at one end and -5 V at the other...

I can do trial & error with the coefficients to get the result but is there an more systematic approach like an derived equation that take account for the initial value of the capacitor (positive or negative)? did i miss something that i could have done to account for this in my DE equation?

It's simple, and I gave you big hints in my previous post. You know the total ΔV that will occur, and that's the only part that applies to the exponential curve. You also have an initial "offset", that is, the initial potential across the capacitor. Those two facts alone (along with the circuit's time constant of course) are enough to pin down the equation.

 

[kuruman]

To expound on post # 14 by @haruspex, the solution of the diff. eq. is $V(t)=V_A+V_B(1-e^{-t/RC})$. What should the values of $V_A$ and $V_B$ be to match the initial and final conditions?

 

[ace8888]

To expound on post # 14 by @haruspex, the solution of the diff. eq. is $V(t)=V_A+V_B(1-e^{-t/RC})$. What should the values of $V_A$ and $V_B$ be to match the initial and final conditions?

$V_A$ = -5
$V_B$ = 15

The general DE eq after you account for the initial not equal to zero is

v(t) = $V_s$ + ($V_o$ -$V_S$)$(1-e^{-t/RC})$

 

[haruspex]

$V_A$ = -5
$V_B$ = 15

Right. So what do you get for the answer now?

 

[ace8888]

Right. So what do you get for the answer now?

4.1micro seconds.

 

[haruspex]

4.1micro seconds.

Which is almost the correct answer, yes?

 

[ace8888]

Which is almost the correct answer, yes?

yes, it is. Thank you for your help.