# RC circuit with dependent source

Discussion in 'Homework Help' started by Hitman6267, May 5, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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So here's what I'm thinking. Please point out any (even small) logic errors.

I need V(0). We can't have an disruption of voltage in a capacitor. So V(0) is the same as V just before we opened the switch. So my goal is to find that V.

The voltage of R1 and R2 is 12V each. And we can calculate the voltage of the dependent source by calculating ix.
So I'm looking for a way to calculate the voltage at R3 so I can do KVL and get the voltage I'm looking for. I haven't found any so far.

2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You don't need to calculate the voltage across R3 because you know that it will be zero. There will be no current in R3, since the capacitor is essentially an open circuit for DC.

3. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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ok but where does that leave me ?

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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12 volts is applied to R1, so the current in R1 (ix) is 12/8000; the current in R2 is irrelevant. Then the voltage out of the dependent source is r*ix = 3000*12/8000. So we have 12 volts on the left side of C, and 3000*12/8000 on the right side. The difference is the voltage on C at time T0+.

5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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Voltage should be Va-Vb
where
Va = voltage at R2 =12
Vb= V dependent source + V at R3

Do we consider that there is no current going through the circuit so R3 =0?

Edit: This post was a reply to some one, who I'm guessing deleted their post. But any way the last question still applies.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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And, be careful what you say. You don't really mean to say that "...R3=0"; what you should say is that "...the voltage across R3=0". R3 is not zero, R3 is 8kΩ. That kind of carelessness will lose you points on an exam.

7. ### Hitman6267 Thread Starter Member

Apr 6, 2010
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We've opened the switch.
=> no current passes through R1. So ix=0
=> doesn't that make V dependent =0 ?
So why does the calculation work ?

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Do you know how to use the nodal, mesh, or loop method to solve circuits?

What you're referring to is the technique where you apply a source to the terminals where you want to find Rth, and then solve the circuit with that source in place and any other independent voltage sources shorted. Rth is then the ratio of the voltage at the terminals to the current at the terminals.

9. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
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I know node analysis and mesh currents.

Edit:

A friend of mine told me how they calculated it but for some reason their method wont work for my values.

The idea:
Calculate V0 then use KCL to calculate ix and then use that to calculate Rth.

The problem appears when I'm calculating V0
V0= 12 - (r * ix) but r=6 and ix =2 so I get V0=0 which prevents me from calculating ix.

Last edited: May 5, 2010
10. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Because they want the voltage across the capacitor at time T0+. That means the instant of time just after the switch opens, but such a small time after that the voltage across the capacitor hasn't had time to change.

In other words, unless an infinite current (this would be what's called an impulse of current) can pass into the capacitor from a source that doesn't exist in this circuit, the voltage across the capacitor at time T0+ is the same as it was at time T0-.

So, if you calculate the voltage across the capacitor before the switch opens, but after it has been closed for a long time (that's what the problem means when it says the circuit is in the steady state), that will be the same voltage across the capacitor just after the switch opens.

You don't need to calculate anything about the circuit after the switch opens to get the answer to this question.

But, if you want to calculate the voltage across the capacitor some finite time after the switch opens, then you will need to do a different calculation. I think you have that problem also.

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You really must be more precise in your language; get your units right.

r is not 6; r is 6000.
ix is not 2; ix is .002

It would not be a good thing to get in the habit of assuming that when you say a resistor is 6, you are assuming units of kΩ. Use the right units all the time and you won't get messed up on an exam.

You weren't being prevented from calculating ix; you even said that ix=2 (which was wrong; ix =.002). It's Ith you need to calculate.

It's entirely possible to have a Vo of zero. What would you have done if Vo hadn't been zero? Go ahead an do that with Vo=zero and see if that makes sense.

You might be ahead of the game to take the time to set up some network equations.

12. ### The Electrician AAC Fanatic!

Oct 9, 2007
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This is an unusual problem. It looks like Vth is indeed zero, as is Ith, but Rth is not zero.

13. ### hgmjr Moderator

Jan 28, 2005
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I merged the two duplicate posts to avoid confusion.

hgmjr

14. ### Ghar Active Member

Mar 8, 2010
655
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Initial capacitor voltage isn't 0.

Ix is 12 / 8k = 1.5 mA
Then r = 3k, so rix = 4.5 V

Capacitor voltage is 12 - 4.5 = 7.5 V

And.... the numbers in these posts don't fit with the schematic. Could this have been the merging? I realize all the schematics look the same but with different numbers.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I'm afraid that merging has added to confusion rather than avoiding it!

Can they be un-merged?

You're quite right. The values of the components in the two threads is different. I didn't notice that at first, and was very puzzled why the OP's numbers didn't make sense.

16. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
I've made separate threads for each problems. I get HWs in sets of about 18 exercises so if I ask about them all in a thread that leads to a very confusing thread. So I took the habit of making a thread for each exercise (or values of a same circuit)
for example: I have a thread where I asked about the V and I. I learned the method and applied it to the different values I had for the same circuit.

The thread is a little confusing now but I got the point. Use correct units.
The only reason I got in the habit of of saying kohm is ohm is because in all previous exercises they would want I in mA. Considering kohm as ohm would prevent me from forgetting to convert and going crazy about why it's not working.

Thank you guys

I'm don't know what you mean by network equations

17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You said earlier that you know node analysis and mesh currents. The circuit in this thread has 3 obvious meshes, and you could set up 3 mesh equations and solve it easily.

This thread now has the wrong schematic in the first post.

You should probably re-post the correct schematic so that anyone who wants to help you will be working from the correct schematic.