My professor has not posted the lecture notes for this question and my book doesn't really cover it so I figured I would try coming here. First off the switch has been open for a long time and is closed at t=0. So as I understand it the two resisters act as a voltage divider so the capacitor will hold 6 volts when fully charged with the switch open. Once the switch is closed the voltage divider will shift such that the first 2 Kohm resister will absorb 8 volts so there is 4 volts going to V0 from the source. Also at this time the total resistance is 3 Kohms so RC=.015 . There for my equation for V0(t)=4-6e^(-t/.015) Did I do this right?
capacitor discharge voltage help
- Thread starter Sean Reed
- Start date
When the switch is open the voltage across the capacitor will be 12 volts when fully charged
When the switch is closed the voltage across the capacitor will be 6 volts when fully charged
if you give a better explanation of the sequence the circuit works i would be able to hepl you better
I don't know how I can better explain it. The switch is open for a long time (ie the capacitor is fully charged) and then it is closed at which point stuff happens.
I guess I don't see how before the switch is closed the capacitor is at 12 volts if there is a 6 volt drop through the first resister. Also while I am not sure if there is a difference, I am looking for the voltage drop through the second resister, not the capacitor.
I guess I don't see how before the switch is closed the capacitor is at 12 volts if there is a 6 volt drop through the first resister. Also while I am not sure if there is a difference, I am looking for the voltage drop through the second resister, not the capacitor.
thingmaker3
- Joined May 16, 2005
- 5,083
If the switch has been open for a long time, there is not a six volt drop across the first resistor. There is a zero volt drop across all resistors. No current is flowing, so E=I*R=0.
When the switch is closed, the capacitor will begin discharging through the second two resistors.
Hint: At the instant the switch is closed, there are still 12V across the capacitor, but one end of the cap is at +6V with respect to battery negative. What, then, is the polarity of Vo at this instant?
When the switch is closed, the capacitor will begin discharging through the second two resistors.
Hint: At the instant the switch is closed, there are still 12V across the capacitor, but one end of the cap is at +6V with respect to battery negative. What, then, is the polarity of Vo at this instant?
as soon as the switch is closed I should have a negative V0 which I showed in my first post. At this point I am getting vary vary frustrated because my book does not cover this vary well, my professor has not posted the lecture notes on this topic yet, I have spent over 3 hours trying to find help from different web sites, and I am being given information that I have no idea where it is coming from.
Here is what I would like someone to explain to me
1) why the capacitor would be fully charged at 12 volts instead of 6 volts?
-My understanding is that there is a 6 volt drop in the first resister before the voltage even gets to the capacitor to charge it in the first place so why does the capacitor charge up to 12 volts?
2) Once the switch is closed will there be a 6 volt drop in the first resister, or an 8 volt drop?
-I ask this because I know if the capacitor was not there it would be an 8 volt drop so I do not see why adding the capacitor would make it a 6 volt drop.
3) What resistance do I use for the time content once the switch is closed?
-My first guess was to use the thevenin equivalent resistance of 3 Kohms but I am not sure if this is right.
Here is what I would like someone to explain to me
1) why the capacitor would be fully charged at 12 volts instead of 6 volts?
-My understanding is that there is a 6 volt drop in the first resister before the voltage even gets to the capacitor to charge it in the first place so why does the capacitor charge up to 12 volts?
2) Once the switch is closed will there be a 6 volt drop in the first resister, or an 8 volt drop?
-I ask this because I know if the capacitor was not there it would be an 8 volt drop so I do not see why adding the capacitor would make it a 6 volt drop.
3) What resistance do I use for the time content once the switch is closed?
-My first guess was to use the thevenin equivalent resistance of 3 Kohms but I am not sure if this is right.
thingmaker3
- Joined May 16, 2005
- 5,083
Six volts are dropped across the first resistor only when the cap first begins charging. The voltage across that first resistor drops to zero as the capacitor charges. The voltage across the capacitor increases. http://www.allaboutcircuits.com/vol_1/chpt_16/2.html
The fully charged capacitor isolates the two resistors.
Use the two resistors in series with the capacitor discharge path. The capacitor does not discharge through the power supply, so the first resistor may be ignored.
thingmaker3
- Joined May 16, 2005
- 5,083
I think you've got it!
thingmaker3
- Joined May 16, 2005
- 5,083
As would I.
The Electrician
- Joined Oct 9, 2007
- 2,970
It seems to me that 6-6e^(-t/.02) can't be correct. Since Vo is measured across a resistor, as soon as enough time has passed, the capacitor will be charged to a new voltage, and all currents will decrease to zero. This means that for t large, the voltage at Vo must be zero, but the above expression evaluates to 6 when t is large.
Also, at t=0, the above expression is zero, but the voltage Vo at t=o is not zero because that's when the current in the capacitor is a maximum.
Also, at t=0, the above expression is zero, but the voltage Vo at t=o is not zero because that's when the current in the capacitor is a maximum.
alwayslearning
- Joined Feb 27, 2008
- 22
as soon as the switch is closed I should have a negative V0 which I showed in my first post. At this point I am getting vary vary frustrated because my book does not cover this vary well, my professor has not posted the lecture notes on this topic yet, I have spent over 3 hours trying to find help from different web sites, and I am being given information that I have no idea where it is coming from.
Here is what I would like someone to explain to me
1) why the capacitor would be fully charged at 12 volts instead of 6 volts?
-My understanding is that there is a 6 volt drop in the first resister before the voltage even gets to the capacitor to charge it in the first place so why does the capacitor charge up to 12 volts?
2) Once the switch is closed will there be a 6 volt drop in the first resister, or an 8 volt drop?
-I ask this because I know if the capacitor was not there it would be an 8 volt drop so I do not see why adding the capacitor would make it a 6 volt drop.
3) What resistance do I use for the time content once the switch is closed?
-My first guess was to use the thevenin equivalent resistance of 3 Kohms but I am not sure if this is right.
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1) why the capacitor would be fully charged at 12 volts instead of 6 volts?
My understanding is that there is a 6 volt drop in the first resister before the voltage
even gets to the capacitor to charge it in the first place so why does the capacitor
charge up to 12 volts?
Solution for question 1)
Sean actually you are right; the voltage across each of the two resistors before the switch is closed is 6 volts. But this is true for only an instant in time during the start of the circuit transient, let me explain.
To answer your question we must examine the circuit before the switch is closed at two separate time intervals. First at the moment that the 12 volt battery source is applied (assume we have added a switch in series with this 12 volt source) then again at some time after the capacitor has completely charged. To begin, at the instant the 12 volts is applied, the capacitor appears as a short circuit to the system. For this reason there is for an instant in time, 6 volts across each of the two resistors. Now as the capacitor begins to charge, the 6 volts across each resistor begins to transfer to the cap. As you know, the cap will completely charge after a time of 5tau = 5RC = 5(2K+2K)5 uF = 0.1 sec or 100 ms. When the cap has charged to the supply voltage (steady state), it is here where your original question begins.
Another way to look at why there is 12 volts rather than 6 volts across the cap is to, for a moment take the cap out of the circuit. We then have only a 12 volt source in series with two 2 Kohm resistors and an open circuit where the cap was. Now a volt meter would read 12 volts between the open circuit terminals; since there is no current flowing through either resistor, their voltage is zero and according to KVL the 12 volts from the battery must appear across the open circuit terminals (resistors with no current flowing are still a conductor for voltage potential, same as a wire). Therefore when dealing with a DC series circuit, the voltage read across a fully charged series cap will equal the voltage read across the open circuit terminals where the cap was removed, and the power supply would see these two situations the same ( ideal component conditions of course, for example not including cap leakage). We can show this from the impedance (reactance) formula for a capacitor; z=1/jwC, where w=omega=2(pi)f, and f=frequency in hertz, in a DC circuit, f=0 which means the impedance is
z=1/(2(pi)0)=infinite, equivalent to an open circuit.
---------------------------------------------------------------------------------
2) Once the switch is closed will there be a 6 volt drop in the first resister, or an 8
volt drop? I ask this because I know if the capacitor was not there it would be an 8
volt drop so I do not see why adding the capacitor would make it a 6 volt drop.
Solution for question 2)
Yes, with the switch closed and after enough time has elapsed for the circuit to reach steady state (cap completely charged) there is 6 volts across the first resistor. Remember that once the cap charges there is no longer any current flowing through the cap path. This means that from the power supply point of view there is only two 2 Kohm resistors in series, with a constant series current flowing.
---------------------------------------------------------------------------------
3) What resistance do I use for the time content once the switch is closed?
My first guess was to use the thevenin equivalent resistance of 3 Kohms,
but I am not sure if this is right.
Again Sean, you are correct. Looking from the cap into the circuit after the switch
is closed is 3 Kohm .
--------------------------------------------------------------------------------
4) To conclude, what is Vout? Vout here according to the piture is taken across the 2
kohm load resistor. Using the knowledge we know about caps; we can determine the
following:
Before the switch is closed, at t=0-, Vout=0
As explained above, due to the fact that the cap is fully charged, the capacitor current is zero, which means the load resistor current
Is zero-making Vout=0
At circuit steady state, or at t=infinity, we know Vout=0 again for the same reason stated above.
Let us check your answer, Vout=6-6e^(t/0.02), this says that at, t=0, Vout= 0 volts, ok! And at t=infinity, Vout= 6 volts, therefore this formula is incorrect.
There are numerous ways to solve this question. By inspection we see that at the instant the switch is closed (t=0+) that there is 6 volts in series with 1 kohm, the cap and 2 kohm load resistor (cap and load resistor removed, circuit thevinized). Therefore there is a -6 volt drop divided between the 1 an 2 kohm.
Which means Vout=-4e(-t/.015)
Will confirm this with mesh/laplace analysis:
Solving for i(t), the series current after the switch is closed (t >= 0), with Vc(0)=12 volts. Vout can then be calculated:
Vout=( i(t))(2 kohm)
For the following note that I (current in laplace domain) and 1 look simular!
Loop 1:
-12/s + I1(s)2k + I2(s)2k +I1(s)2k = 0
Loop 2:
-12/s + I1(s)2k + I2(s)4k + [I2(s)/sC) +12/s] = 0 ;remember to add Vc(0)
I2(s) = -(24000/s)/((sC12*10^6+4000)/sC)
= -0.002/(s+66.67)
where (s+66.67), converts to in the time domain; e^-t66.67 or e^-t/.015
since 66.67=1/tau the circuit time constant.
Therefore:
i(t) = -0.002e^(-t/.015) Amps, (t >= 0)
Vout = (2 kohm)*i(t) = -4e^(-t/.015) volts, (t >= 0)
-----------------------------------------------------------------------------------
END
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Here is what I would like someone to explain to me
1) why the capacitor would be fully charged at 12 volts instead of 6 volts?
-My understanding is that there is a 6 volt drop in the first resister before the voltage even gets to the capacitor to charge it in the first place so why does the capacitor charge up to 12 volts?
2) Once the switch is closed will there be a 6 volt drop in the first resister, or an 8 volt drop?
-I ask this because I know if the capacitor was not there it would be an 8 volt drop so I do not see why adding the capacitor would make it a 6 volt drop.
3) What resistance do I use for the time content once the switch is closed?
-My first guess was to use the thevenin equivalent resistance of 3 Kohms but I am not sure if this is right.
---------------------------------------------------------------------------------
1) why the capacitor would be fully charged at 12 volts instead of 6 volts?
My understanding is that there is a 6 volt drop in the first resister before the voltage
even gets to the capacitor to charge it in the first place so why does the capacitor
charge up to 12 volts?
Solution for question 1)
Sean actually you are right; the voltage across each of the two resistors before the switch is closed is 6 volts. But this is true for only an instant in time during the start of the circuit transient, let me explain.
To answer your question we must examine the circuit before the switch is closed at two separate time intervals. First at the moment that the 12 volt battery source is applied (assume we have added a switch in series with this 12 volt source) then again at some time after the capacitor has completely charged. To begin, at the instant the 12 volts is applied, the capacitor appears as a short circuit to the system. For this reason there is for an instant in time, 6 volts across each of the two resistors. Now as the capacitor begins to charge, the 6 volts across each resistor begins to transfer to the cap. As you know, the cap will completely charge after a time of 5tau = 5RC = 5(2K+2K)5 uF = 0.1 sec or 100 ms. When the cap has charged to the supply voltage (steady state), it is here where your original question begins.
Another way to look at why there is 12 volts rather than 6 volts across the cap is to, for a moment take the cap out of the circuit. We then have only a 12 volt source in series with two 2 Kohm resistors and an open circuit where the cap was. Now a volt meter would read 12 volts between the open circuit terminals; since there is no current flowing through either resistor, their voltage is zero and according to KVL the 12 volts from the battery must appear across the open circuit terminals (resistors with no current flowing are still a conductor for voltage potential, same as a wire). Therefore when dealing with a DC series circuit, the voltage read across a fully charged series cap will equal the voltage read across the open circuit terminals where the cap was removed, and the power supply would see these two situations the same ( ideal component conditions of course, for example not including cap leakage). We can show this from the impedance (reactance) formula for a capacitor; z=1/jwC, where w=omega=2(pi)f, and f=frequency in hertz, in a DC circuit, f=0 which means the impedance is
z=1/(2(pi)0)=infinite, equivalent to an open circuit.
---------------------------------------------------------------------------------
2) Once the switch is closed will there be a 6 volt drop in the first resister, or an 8
volt drop? I ask this because I know if the capacitor was not there it would be an 8
volt drop so I do not see why adding the capacitor would make it a 6 volt drop.
Solution for question 2)
Yes, with the switch closed and after enough time has elapsed for the circuit to reach steady state (cap completely charged) there is 6 volts across the first resistor. Remember that once the cap charges there is no longer any current flowing through the cap path. This means that from the power supply point of view there is only two 2 Kohm resistors in series, with a constant series current flowing.
---------------------------------------------------------------------------------
3) What resistance do I use for the time content once the switch is closed?
My first guess was to use the thevenin equivalent resistance of 3 Kohms,
but I am not sure if this is right.
Again Sean, you are correct. Looking from the cap into the circuit after the switch
is closed is 3 Kohm .
--------------------------------------------------------------------------------
4) To conclude, what is Vout? Vout here according to the piture is taken across the 2
kohm load resistor. Using the knowledge we know about caps; we can determine the
following:
Before the switch is closed, at t=0-, Vout=0
As explained above, due to the fact that the cap is fully charged, the capacitor current is zero, which means the load resistor current
Is zero-making Vout=0
At circuit steady state, or at t=infinity, we know Vout=0 again for the same reason stated above.
Let us check your answer, Vout=6-6e^(t/0.02), this says that at, t=0, Vout= 0 volts, ok! And at t=infinity, Vout= 6 volts, therefore this formula is incorrect.
There are numerous ways to solve this question. By inspection we see that at the instant the switch is closed (t=0+) that there is 6 volts in series with 1 kohm, the cap and 2 kohm load resistor (cap and load resistor removed, circuit thevinized). Therefore there is a -6 volt drop divided between the 1 an 2 kohm.
Which means Vout=-4e(-t/.015)
Will confirm this with mesh/laplace analysis:
Solving for i(t), the series current after the switch is closed (t >= 0), with Vc(0)=12 volts. Vout can then be calculated:
Vout=( i(t))(2 kohm)
For the following note that I (current in laplace domain) and 1 look simular!
Loop 1:
-12/s + I1(s)2k + I2(s)2k +I1(s)2k = 0
Loop 2:
-12/s + I1(s)2k + I2(s)4k + [I2(s)/sC) +12/s] = 0 ;remember to add Vc(0)
I2(s) = -(24000/s)/((sC12*10^6+4000)/sC)
= -0.002/(s+66.67)
where (s+66.67), converts to in the time domain; e^-t66.67 or e^-t/.015
since 66.67=1/tau the circuit time constant.
Therefore:
i(t) = -0.002e^(-t/.015) Amps, (t >= 0)
Vout = (2 kohm)*i(t) = -4e^(-t/.015) volts, (t >= 0)
-----------------------------------------------------------------------------------
END
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my $0.02
lets call resistors R1, R2, R3 left to right
1) examine the charging current curve of a cap shows max I going to min I. therefore the current through any series components will do same VR1 and VR3 going from 6v to 0v during charge cycle (current decreasing), at that point cap is min current (I=0) and max voltage (12v).
2) close switch - cap is now a power source discharging in the same polarity of charge. as thingmaker said it will not discharge through power source, only R2 and R3. making VR3 reversed polarity as current is now going down throught it. Eventually the cap and R3 will no longer have current going through it as cap becomes charged to it's new voltage level, and VR3 = 0v.
3) discharge current is going through series R2 and R3 only, so I think Rt would be 2K Ohm + 2K Ohm.
lets call resistors R1, R2, R3 left to right
1) examine the charging current curve of a cap shows max I going to min I. therefore the current through any series components will do same VR1 and VR3 going from 6v to 0v during charge cycle (current decreasing), at that point cap is min current (I=0) and max voltage (12v).
2) close switch - cap is now a power source discharging in the same polarity of charge. as thingmaker said it will not discharge through power source, only R2 and R3. making VR3 reversed polarity as current is now going down throught it. Eventually the cap and R3 will no longer have current going through it as cap becomes charged to it's new voltage level, and VR3 = 0v.
3) discharge current is going through series R2 and R3 only, so I think Rt would be 2K Ohm + 2K Ohm.
The Electrician
- Joined Oct 9, 2007
- 2,970
Just before the switch is closed, the voltage at the left end of R1 is 12 volts, and the voltage at the right end of R1 is also 12 volts, because there is no current in R1.
As soon as the switch is closed, the voltage at the right end of R1 is no longer 12 volts, and a current flows in R1 as well as in R2 and R3. That current has to be taken into account.
Another way to look at the situation is to imagine the combination of R1, R2 and the 12 volt source being replaced (when the switch closes) by a 6 volt source in series with 1k ohms. This is the Thevenin equivalent.
With the switch open you have a 12 volt source in series with 2k, which charges the capacitor. When the switch is closed you suddenly have the 12 volts in series with 2k replaced with 6 volts in series with 1k (and the capacitor is still charged to 12 volts). Solve this circuit and you should get the desired answer.
In fact, you can get the solution without even writing anything down. Using the Thevenin equivalent, we have 6 volts in series with -12 volts, giving -6 volts in series with 3k ohms. The time constant will be 3K times 5 uF, or .015. SInce we have -6 volts in series with 3k ohms, we have a current at t=0+ of -2 mA. The initial voltage across Vo will be 2k * -.002 = -4 volts.
Thus the voltage across Vo will be -4e^(-t/.015) as alwayslearning said.
As soon as the switch is closed, the voltage at the right end of R1 is no longer 12 volts, and a current flows in R1 as well as in R2 and R3. That current has to be taken into account.
Another way to look at the situation is to imagine the combination of R1, R2 and the 12 volt source being replaced (when the switch closes) by a 6 volt source in series with 1k ohms. This is the Thevenin equivalent.
With the switch open you have a 12 volt source in series with 2k, which charges the capacitor. When the switch is closed you suddenly have the 12 volts in series with 2k replaced with 6 volts in series with 1k (and the capacitor is still charged to 12 volts). Solve this circuit and you should get the desired answer.
In fact, you can get the solution without even writing anything down. Using the Thevenin equivalent, we have 6 volts in series with -12 volts, giving -6 volts in series with 3k ohms. The time constant will be 3K times 5 uF, or .015. SInce we have -6 volts in series with 3k ohms, we have a current at t=0+ of -2 mA. The initial voltage across Vo will be 2k * -.002 = -4 volts.
Thus the voltage across Vo will be -4e^(-t/.015) as alwayslearning said.
The Electrician
- Joined Oct 9, 2007
- 2,970
I have a few criticisms of your work, Joe. First, the usual formula would deal with the voltage across the capacitor, not the charge. It's ok to use charge, but if you're going to do this, you should use charge, not voltage. You have the Qd as 12; that's the final voltage across the capacitor, not the charge. The charge is given by Q = C*V, so the final charge on page 2 should be 12*5*10^-6. I prefer to use voltages rather than charge.
You show an equation on page 2 for Vo, and it won't give the numbers you got for Vo if you really have (Qf - Q) as the expression in the numerator; you need to have voltages there, not charges. Even with this problem, you have gotten the correct voltages for Vo during the initial charging phase.
The graph you have on page 3 is misleading. You've shown a finite initial slope as the voltage rises toward 6 volts. This is not right; the voltage jumps up to 6 volts instantaneously (neglecting parasitic capacitances in the wiring). The first graph on page 6 has the same problem.
On page 5, you don't show your expression for Vo during the second phase of the problem, so I can't tell why you didn't get a peak voltage of -4 volts, rather than the -3 volts you did get.
Remembering that you got a peak voltage of -3 volts on page 5, why did you say on page 6, "...did not reach the plus and minus 6 volts."? It shouldn't have reached minus 6 volts according to your calculation on page 5.
Looking at the second graph on page 6, even though the first peak didn't reach 6 volts, I would expect that both peaks would be reduced proportionally. So if the second peak was as predicted by your analysis on page 5 (-3 volts), I would expect that second, negative going, peak to be half as high as the first one. In fact, it's substantially more than half as tall as the first peak, suggesting that the simulator is telling us that the negative going peak reaches more than the -3 volts your analysis on page 5 gave.
If you gave an expression for the calculations you show on page 5, perhaps we could see where you went wrong.
I've attached a simulation from MicroCap showing the initial charging of the capacitor to 12 volts and then the discharge when the switch in the OP's schematic closes. It verifies the analysis of alwayslearning and my shorter analysis.
The first curve is Vo, and the second is the voltage across the capacitor.
You show an equation on page 2 for Vo, and it won't give the numbers you got for Vo if you really have (Qf - Q) as the expression in the numerator; you need to have voltages there, not charges. Even with this problem, you have gotten the correct voltages for Vo during the initial charging phase.
The graph you have on page 3 is misleading. You've shown a finite initial slope as the voltage rises toward 6 volts. This is not right; the voltage jumps up to 6 volts instantaneously (neglecting parasitic capacitances in the wiring). The first graph on page 6 has the same problem.
On page 5, you don't show your expression for Vo during the second phase of the problem, so I can't tell why you didn't get a peak voltage of -4 volts, rather than the -3 volts you did get.
Remembering that you got a peak voltage of -3 volts on page 5, why did you say on page 6, "...did not reach the plus and minus 6 volts."? It shouldn't have reached minus 6 volts according to your calculation on page 5.
Looking at the second graph on page 6, even though the first peak didn't reach 6 volts, I would expect that both peaks would be reduced proportionally. So if the second peak was as predicted by your analysis on page 5 (-3 volts), I would expect that second, negative going, peak to be half as high as the first one. In fact, it's substantially more than half as tall as the first peak, suggesting that the simulator is telling us that the negative going peak reaches more than the -3 volts your analysis on page 5 gave.
If you gave an expression for the calculations you show on page 5, perhaps we could see where you went wrong.
I've attached a simulation from MicroCap showing the initial charging of the capacitor to 12 volts and then the discharge when the switch in the OP's schematic closes. It verifies the analysis of alwayslearning and my shorter analysis.
The first curve is Vo, and the second is the voltage across the capacitor.
