# RC Circuit Response (First Order)

Discussion in 'Homework Help' started by jegues, Oct 16, 2010.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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See figure attached for problem statement as well as my attempt.

I've never really done a question of this type before so I don't really know the procedure.

First I drew the circuit with the switch open, and the capictor as it's steady state equivalent(Open Circuit).

The whole circuit becomes an open circuit so no current is flowing. This mean's that V(infinity) = 24V.

I'm also asked for the time constant in this circuit. The only formula I know for solving the time constant in an RC circuit is,

$\tau = C \cdot R_{t}$

So I attempted to find the equivalent resistance and multiply it by the capacitance of the capacitor.

The answer they list for the time constant in part a is 3s. What am I doing wrong?

For the second part, I redrew the circuit with the switch closed and the capictor as it's steady state equivalent (open circuit).

I had lots of trouble finding v(infinity) for some reason, I just couldn't see what exactly was going on.

The only logical way I could reason to the correct answer was that we have a voltage source that splits into two branches. Now the total voltage in each parallel branch should be the same, so I have to split the 24 V between two 50ohm resistors, so they'll each get 12V.

Then I simply applied a KVL in the loop with v(infinity) and the first 50 ohm resistor.

Is there another more formal way of seeing that this v(infinity) is indeed 12V?

Also, I'm having trouble finding the time constant in this problem as well. Certainly if I'm making mistakes in trying to find it in the first half of the problem, I'm problem making the exact same mistake in this part of the problem as well.

Can someone clarify how v(infinity) is 12V in part B and how to correctly find the time constant in both part A and B?

Thanks again!

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2. ### Georacer Moderator

Nov 25, 2009
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1,266
There are two parts in each question.
a) Finding Vinf and
b) Finding the time constant

So let's get down to it:
a) For the calculation of Vinf, you open-circuit the capactitor and try to find the voltage across its terminals. For the first circuit, as you noticed, no current flows through the resisors, because the source is open-circuited. Therefore we have no voltage drop on the resistors and every point on them has the same voltage as V+. Thus, V+=24V and V-=0V and Vcap=V+-V-=24V. Your answer was correct.

For the second circuit, you were also correct, buy you didn't have to sweat so much to find it. Vs=24V is applied on a branch with 2 50Ω resistors. This constitutes a voltage divider with equal resistances. The voltage in the middle will be Vs/2=12V, as you said.

b) The way to calculate the time constant in a simple RC circuit is to nullify the source (here replace it with a short) and without removing the capacitor, try to merge all the resistanses into one single resistance.

You did fine until a tiny mistake kept you from your answer:
$R_{eq}=100||50=\frac{100 \cdot 50}{100+50}=\frac{ \mathbf{5000} }{150}$ - Note on the 5000

Try to confirm the time constant for the second circuit too.

P.S. Does anyone know how to bold characters inside the $brackets? \mathbf doesn't seem to work...$

Last edited: Oct 16, 2010
3. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,938
1,089
A trick for parallel resistors :
R||R = R/2
So if you have 100||50 and if you think of the 50Ω resistor as two 100Ω resistors in parallel. So the whole circuit is like tree 100Ω's in parallel
RT = 100||50 = 100/3 = 33.3Ω

So you have algebra error.

As for v(infinity) you overcomplicated such a simple circuit.
The two 50Ω resistor form voltage divider.
The upper 50Ω does not affect the voltage on the capacitor and time constant.

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4. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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One question here, why would be replace the voltage source with a short circuit? We've been taught to replace current sources with a short circuit to nullify them, and voltage sources with an open circuit.

Does it have to be a short circuit in this case?

EDIT: Whoops! I got it backwards! You're right!

Anyways I'm trying to do that, but I still can't get the proper time constant.

I have two 50 Ohm resistors that would be in series if it weren't for the capacitor inbetween them, and then I've got another resistor in parallel with those two resistors.

How do I simplify this? Do I simplify choose one of the resistors and simplify it with the other resistor in parallel and stop there?

Because if I do,

$50\Omega || 50\Omega = 25 \Omega$

Then,

$\tau = RC = 2.25s$

That's what the answer should be, but I don't know if this is the right way of concluding it.

How do I "reduce" my 3 resistors?

Last edited: Oct 16, 2010
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,938
1,089
To find RT you need to find resistance seen from the capacitor terminal.
With voltage source replace with a short circuit

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6. ### Georacer Moderator

Nov 25, 2009
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In which part do you have a problem now? With the switch open or closed?

I wouldn't say that. I think it is sufficient to reduce the circuit to a capacitor and a resistance, independently of where that resistance is. The result is the same and you save yourself the brainstorm.

That's for simple circuits, of course. If it involves dependant sources etc. I think you select the resistance as seen from the output terminal. Correct me if I 'm wrong.

7. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Yes, but how do I reduce this? I've got a 50ohm in parallel with another 50 ohm, that's going to give me 25 ohm.

Isn't this 25 ohm now in series with the last 50ohm?

The only way I can get the "right" answer is if my Rt = 25 ohm and I just don't see how they can do it with 3 50 ohm resistors.

With two you could, but not 3.

EDIT: Is it because it's the resistance from the CAPICTORS TERMINAL? And not just the total equivalent resistance?

8. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
Don't get desperate, you don't see the layout as you should, that's all. A little re-arranging always helps.

For the first case, switch open:

Edit:Rbc is in series with Rca. Their combination is parallel to Rab.

For the second case, switch closed:

Edit:Rab is parallel to Rbc. Rac has both terminals shorted, therefore it is removed from the circuit. No current flows through it.

Try once more.

Last edited: Oct 17, 2010
9. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I think you got some typos here,

In the first case,

Rbc is in series with Rca. Their combination is parallel to Rab.

For the second case,

Rab is parallel to Rbc. Rac has both terminals shorted.

I think the part I'm having a hard time seeing is what Rac looks like with its terminals shorted. Could you give me an example of something like this?

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,938
1,089
Yes, you mus look from capacitor terminal

Find current in this circuit, if voltage source is equal 1V, then
Rt = V/ I = 1V/ I

11. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
ARGH! That's the second error today! Thanks for the correction in spelling. That happens when you write the text first and add the image later.

You got the idea about how the resistanses should be merged, so finding the time constant shouldn't be a problem.

Now, about why we can ommit Rac. You can clearly see that both its terminals are connected through the lower part of the outer mesh by nothing but thin wire. That means they are shorted, right? That means they have the same voltage, right? Therefore $I_{ac}=\frac{0V}{R_{ac}}=0$. No current flows through the resistance. It serves no purpose being there.

You can see it in another way. Rac is in parallel with a short-circuit, the lower half of the outer mesh. A short circuit can be described as a resistance of 0Ω. Thus the equivalent resistance is Rac||0=0Ω. That means you replace the two of them with a wire (or you can just erase the Rac and leave the wire).