Circuit help - resistor and capacitors... time too!?!?

hgmjr

Joined Jan 28, 2005
9,029
b.) At ten seconds, I assume that the capacitor is fully charged.
Going from what you guys are all saying, the ten ohm resistor should have the same amount of potential voltage difference as the capacitor ( :( again, I don't follow why this is true ) and from above, the current flowing across would be 6; multiplying that by 10 ohms, we get 60 Volts for the voltage built up across the 6 micro farad capacitor.
Your conclusion and therefore your answer is correct for Part B.

hgmjr
 

shteii01

Joined Feb 19, 2010
4,647
b.) At ten seconds, I assume that the capacitor is fully charged.
Going from what you guys are all saying, the ten ohm resistor should have the same amount of potential voltage difference as the capacitor ( :( again, I don't follow why this is true ) and from above, the current flowing across would be 6; multiplying that by 10 ohms, we get 60 Volts for the voltage built up across the 6 micro farad capacitor.
It is true for the following reasons:
1) Capacitor is fully charged.
2) A fully charged capacitor act as an open circuit.
3) Another way to describe an open is circuit is to replace it with really really big resistor.
4) You now have a 10 Ohm resistor in parallel with another resistor.
5) One of the features of resistors in parallel is that they all have the same voltage across them.

That is why the voltage across 10 Ohm resistor will also appear across the capacitor in question.

Basically it is like this: Resistors in series have the same current. Resistors in parallel have the same voltage.
 

Thread Starter

Supervisor

Joined Jan 17, 2009
32
Supervisor,

Is it possible that the capacitors should be in Farads instead of MicroFarads?

hgmjr
No, they are micro farads - checked question again.

It is true for the following reasons:
1) Capacitor is fully charged.
2) A fully charged capacitor act as an open circuit.
3) Another way to describe an open is circuit is to replace it with really really big resistor.
4) You now have a 10 Ohm resistor in parallel with another resistor.
5) One of the features of resistors in parallel is that they all have the same voltage across them.

That is why the voltage across 10 Ohm resistor will also appear across the capacitor in question.

Basically it is like this: Resistors in series have the same current. Resistors in parallel have the same voltage.
Thanks!
C and D!?
 

syed_husain

Joined Aug 24, 2009
61
Sorry, you still have it wrong. When voltage tries to change it cannot. That means that the current actually surges. The current at the instant t=0+ increases briefly to 10 amps.

hgmjr
Vc(0-)=Vc(0+), this must be satisfied for the capacitor as voltage can't be changed instantaneously. from the question Vc(0-)=0 V, so is Vc(0+). Now plug this value in this equation:
Ic(0+)= C ((Vc(0+)- Vc(0-))/(t(0+)-(t(0-))
clearly, the numerator Vc(0+)- Vc(0-)=0 and the Ic(0+) becomes zero not 10 amps.

N.B. Vc(0-) = voltage accross capacitor before swtich one closed
Vc(0+)= voltage accross capacitor after swtich one closed
 

hgmjr

Joined Jan 28, 2005
9,029
I believe that Part D: involves the instantaneous opening of t2 and closure of t3. That means that the charge previously stored in the 6 microfarad capacitor will redistribute itself to the parallel capacitors. The voltage would therefore drop to 40V. Assuming ideal capacitors, there is no dc current path to discharge the two caps in parallel so after the specified time, the voltage would still be 40V.

hgmjr
 

hgmjr

Joined Jan 28, 2005
9,029
Vc(0-)=Vc(0+), this must be satisfied for the capacitor as voltage can't be changed instantaneously. from the question Vc(0-)=0 V, so is Vc(0+). Now plug this value in this equation:
Ic(0+)= C ((Vc(0+)- Vc(0-))/(t(0+)-(t(0-))
clearly, the numerator Vc(0+)- Vc(0-)=0 and the Ic(0+) becomes zero not 10 amps.

N.B. Vc(0-) = voltage accross capacitor before swtich one closed
Vc(0+)= voltage accross capacitor after swtich one closed
So if I understand your logic, you agree that the voltage cannot change instantaneously but you insist that the current is zero which means that the voltage across the capacitor would have to instantaneously jumps from 0 volts to 60 volts.

hgmjr
 

Thread Starter

Supervisor

Joined Jan 17, 2009
32
I believe that Part D: involves the instantaneous opening of t2 and closure of t3. That means that the charge previously stored in the 6 microfarad capacitor will redistribute itself to the parallel capacitors. The voltage would therefore drop to 40V. Assuming ideal capacitors, there is no dc current path to discharge the two caps in parallel so after the specified time, the voltage would still be 40V.

hgmjr
Hmm... So the 6 micro farad capacitor will drop to 40...
Then what about the 3? Will that have 20 (gaining the 20 that the 6 dropped)?

EDIT: The question says at time 30 seconds (10 seconds after switch two is opened and switch three is closed).
 
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hgmjr

Joined Jan 28, 2005
9,029
Hmm... So the 6 micro farad capacitor will drop to 40...
Then what about the 3? Will that have 20 (gaining the 20 that the 6 dropped)?
When you open switch 2 and close switch 3, the two caps are in parallel. That means the voltage across one must equal the voltage across the other capacitor. That voltage drops from 60V to 40V to preserve the total charge that was on the 6 microfarad capacitor right before switch 2 and switch 3 changed states.

hgmjr
 

Thread Starter

Supervisor

Joined Jan 17, 2009
32
When you open switch 2 and close switch 3, the two caps are in parallel. That means the voltage across one must equal the voltage across the other capacitor. That voltage drops from 60V to 40V to preserve the total charge that was on the 6 microfarad capacitor right before switch 2 and switch 3 changed states.

hgmjr
But... How?
How can both of them have the same amount of voltage across them?

If each are 40, then that means 80 total?
I thought it had to be conserved...
Hrmm...
 

syed_husain

Joined Aug 24, 2009
61
So if I understand your logic, you agree that the voltage cannot change instantaneously but you insist that the current is zero which means that the voltage across the capacitor would have to instantaneously jumps from 0 volts to 60 volts.

hgmjr
thanks a lot. i made a simple but crucial mistake forgetting about the 10 ohm resistor as that is connected to 6 u cap in parallel. as t=0, Vc=0 so is V10=0 as they r in parallel and hence they r briefly shorted and all 150 V appears accross 15 ohm resistor.

thanks again 4 pointing me the error.
 

hgmjr

Joined Jan 28, 2005
9,029
But... How?
How can both of them have the same amount of voltage across them?

If each are 40, then that means 80 total?
I thought it had to be conserved...
Hrmm...
Keep in mind that the two caps are in parallel when switch 2 is opened and switch 3 is closed. This means that the voltage across the two caps seeks an new equilibrium point. Your comments seem to assume that the caps are in series. In effect the two caps behave as a single cap that is 9 microFarads in value.

hgmjr
 
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Terp

Joined Jun 6, 2008
32
I = C (dV/dt) this is the relationship between capacitor current and voltage. i m now directly quoting from the book "Engineering Circuit Analysis" by Hayt, Kemmerly ( ed. 7th) from p. 217. the authors say with regarding the above equation " a constant voltage accross a capacitor results in zero current passing through it; a capacitorr is thus an "open circuit to dc" ".
That's the steady state solution to a capacitor.... But, it will always have a transient component nevertheless... In the context of this problem, the transient is the fact that for that one instant, the capacitor behaves as a short to ground, but as time progresses, it approaches its DC solution, i.e. behaving as an open circuit.
 

Terp

Joined Jun 6, 2008
32
at t=0 when the switch is closed, the capacitor tries to prevent the current flowing through it because voltage can't be changed instantaneously. what this means that "a sudden jump in voltage requires an infinite current and since it is physically impossible we will therfore prohibit the voltage accross a capacitor to cahnge in zero time" (Hayt, p.217). the moment switch is closed (t=0) no current will flow that means the capacitor is open circuit.
"Infinite Current" is relative.... it does not signify, million amps of current. It just means that "relative to other components in the circuit, that are parallel to the capacitor," the current is infinitely large... meaning the current through the resistor in parallel to the cap is negligible compared to the current through the cap itself.
 

Thread Starter

Supervisor

Joined Jan 17, 2009
32
Keep in mind that the two caps are in parallel when switch 2 is opened and switch 3 is closed. This means that the voltage across the two caps seeks an new equilibrium point. Your comments seem to assume that the caps are in series. In effect the two caps behave as a single cap that is 9 microFarads in value.

hgmjr
After a good rest I'm back --
I know that it acts as a 9 micro farad capacitor (because capacitors in parallel are like resistors in series), but why do both of them have to retain the same amount - namely 40 V? I need the voltage across each of them and from part (b) I saw that when the capacitor was fully charged, it would be the same as the 10 ohm resistor.
Would it not be the same in this case, thus making the capacitors have maybe 60 Volts each (if they are supposed to have the same amount of voltage)?
 

hgmjr

Joined Jan 28, 2005
9,029
Actually, when you simultaneously open switch 2 and close switch 3, you disconnect the two capacitors (which are now in parallel) from the 60V source. The 3 microFarad is assumed to be fully discharged at the time of the switches are changed. The only thing that can determine the voltage across the parallel combination is the initial charge that existed on the 6 microFarad capacitor at the time just prior to the switch state changes. You calculated that charge to be 0.000036 Joules. That same charge will be present before and after the switches change state. That means that the charge has to redistribute itself across the two caps. Just like two resistors in parallel, the voltage across the two caps in parallel will need to settle to a new voltage based on the charge that was there just prior to the switches changing state. This new voltage will be

\(V_{\small t_{\.0+}}=\frac{Q_{t0-}}{6uF+3uF}\)


hgmjr
 

Thread Starter

Supervisor

Joined Jan 17, 2009
32
Actually, when you simultaneously open switch 2 and close switch 3, you disconnect the two capacitors (which are now in parallel) from the 60V source. The 3 microFarad is assumed to be fully discharged at the time of the switches are changed. The only thing that can determine the voltage across the parallel combination is the initial charge that existed on the 6 microFarad capacitor at the time just prior to the switch state changes. You calculated that charge to be 0.000036 Joules. That same charge will be present before and after the switches change state. That means that the charge has to redistribute itself across the two caps. Just like two resistors in parallel, the voltage across the two caps in parallel will need to settle to a new voltage based on the charge that was there just prior to the switches changing state. This new voltage will be

\(V_{\small t_{\.0+}}=\frac{Q_{t0-}}{6uF+3uF}\)


hgmjr
You are saying this when the... err... switch is on for 10 seconds, right?
The question reads what will be the voltage on each after switch two has been opened and switch three, closed for ten seconds (at 20 seconds switch 2 is opened and switch three is closed but at thirty seconds [after ten seconds] will the charge have not distributed itself already?).
 

hgmjr

Joined Jan 28, 2005
9,029
You are saying this when the... err... switch is on for 10 seconds, right?
The question reads what will be the voltage on each after switch two has been opened and switch three, closed for ten seconds (at 20 seconds switch 2 is opened and switch three is closed but at thirty seconds [after ten seconds] will the charge have not distributed itself already?).
The redistribution of charge across the two capacitors happens very rapidly so that it is settled long before the time delay measured in seconds. The voltage across the two capacitors in paralled settles to 40V soon after the switches are thrown.

hgmjr
 
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