Discussion in 'Homework Help' started by kristapsdreija, Apr 15, 2014.

1. ### kristapsdreija Thread Starter New Member

Apr 15, 2014
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Hi! Can you please help me with my electronics homework? I've been trying to solve it for hours, but with no luck. Can You please help me?

The switch is between the power source and R1.
http://puu.sh/89ojd.png
Given:
---------
V=330V
R1=2k
R2=1.3k
C3=5uF
switch: closed-->open
-----------
i2-?

Thank you very much, friends!

2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
i2 is undefined.

What was the question and what approach have you tried?

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3. ### WBahn Moderator

Mar 31, 2012
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If you've been trying for hours, then you should be able to show us your best attempt so that we have something to work from. We won't just work your homework for you.

I do agree with the statement

i2-?

You don't give any indication what i2 is, so it is very much an unknown.

4. ### kristapsdreija Thread Starter New Member

Apr 15, 2014
18
0
http://puu.sh/89rF2.fpg
Here is the circuit i drew.
And here is the whole page.
http://puu.sh/89rPg.fpg
I tried simulating the circuit in LTSpice, but could not get into conclusions. I understand the fundamentals of the topic, but can not seem to understand how to calculate the things needed in this assignment.
p.s. the task is to get a current function of time for i2 (current through R2) and to plot it.
Thank you very much for trying and sorry if my forum-posting skills are bad, as this is my first post. Thanks.

5. ### WBahn Moderator

Mar 31, 2012
20,076
5,666
I don't read your language, so I'm doing a bit of guessing here, but I think your problem is that you are computing your time constant by placing R1 and parallel with R2. If so, what is your basis for this? Remember, the switch is now open.

Apr 15, 2014
18
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7. ### MrChips Moderator

Oct 2, 2009
14,298
4,198

"Please Help" does not tell me what you need help with.

8. ### kristapsdreija Thread Starter New Member

Apr 15, 2014
18
0
Help me find i2 !

9. ### Alec_t AAC Fanatic!

Sep 17, 2013
7,037
1,457
You state R1=2k, R2=1.3k but your post #1 schematic shows the reverse. Which is correct?
Do you need to know i(R2) while the cap charges (switch closed) or discharges (switch open after being closed)?

10. ### MrChips Moderator

Oct 2, 2009
14,298
4,198
Where did you lose it?

11. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Lots of folks here are very happy to help, although please help is a very poor title since no one knows what your thread is about and many may not bother to read it as a result.

A much better title would be

Help finding current function in RC circuit.

A few lines of text helps us understand.

And you need to provide a proper indication of what i2 is by marking it on a diagram as an arrow and labelling it.

I say an arrow because that defines a guessed direction. If the guess is wrong the result will be numerically correct but negative.

Have you read the header instructions for posting in Homework Help?

And can you honestly say that you have followed them?

You say you have spent hours on this, so can you tell us what the start value of i2 is, before the switch is changed, and why?

12. ### kristapsdreija Thread Starter New Member

Apr 15, 2014
18
0
Sorry, the schematic resistor names are in reverse!
The first resistor coming from the voltage source is 2k, and the second, parallel = 1.3k.
The voltage applied is 330 V, and i need to know the current function through second resistor (1.3k) when the switch OPENS after being closed (cap discharges)

13. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Drawing a diagram is good, but it is also very important to get things correct.

The diagram does not have to be pretty, just accurate.

I am still not sure what your circuit is so I have drawn a diagram to include the switch and the current etc.

In fig1 the switch is closed.
These provide the intial conditions.

You state that you have to find the current through the capacitor, and you state that it is the same as the current through the 1300Ω resistor.
What is your justification for this?

I asked you a question about the initial current, but you have not answered??

So let us start at the beginning, in fig1

The switch is closed.

What is the voltage across the capacitor ie between points A and B?
What is the current through the capacitor that is what is i2?

Now let us open the switch as in fig2

What starts to happen?

You should not play with computer programs until you understand the electrics of the circuit action. Spice is overkill for this.

• ###### itwo.jpg
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14. ### MrChips Moderator

Oct 2, 2009
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You have made an assumption on where is i2.

I have no idea where is i2, hence I cannot find it.

Edit: Reread op's post. i2 is the current flowing through the second resistor. I have to go back and look for the "second" resistor.

15. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
I am simply trying to lead the OP into good habits so that his (her?) next question will not still be unclear after 16 posts.

16. ### WBahn Moderator

Mar 31, 2012
20,076
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I was trying to help you. I looked over your work, even though it is in a foreign language (posted on an English-language forum) and identified what I believe might be a key mistake that you have made. You totally ignored it. From that I can only conclude that by "help" you mean that you want someone to do your work for you. So goodbye.

17. ### kristapsdreija Thread Starter New Member

Apr 15, 2014
18
0
The start value of i2 is: i2 = i1 = U/(R2+R1) = 330/3300 = 0.1A. Because the capacitor is fully charged and is not passing the DC current through.

The voltage across the capacitor between A and B at the time of switching initially is the same as the voltage across R2: U(C3) = U(R2) = i2(initial) * R2 = 0.1 * 1300 = 130 V . So, when t = 0, U(C) = 130 V .

When the switch is being opened, the capacitor starts to discharge through R2, thus increasing (?) the current through resistor R2.
Am I right thus far?

Please do not get offended, my friend, I am sorry I did not reply to your post, as I am fairly new to this forum and I just did not see your reply.

My reply: if the switch now open, it means that the current stored in the cap will flow through only resistor R2, so the time constant tau will be: tau = R2 * C (?) (not by taking R1 || R2 (?)).

Thank you studio and everybody else for helping me, I appreciate that very much!

P.S. I tried to make a more explanatory topic name, but I guess it was too long and so the forum did not accept it for a few times, so I got frustrated and made a short one...

Last edited: Apr 15, 2014
18. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Not really, but that is the point of homework help.

So let us take things a bit at a time and start with some corrections.

First let us set zero time (t=0) to the opening of the switch.

Now the usual assumption is that the capacitor is fully charged at t=0.
The key phrase in any question about this will be something like "the switch has been closed for a long time" or " Initially all the circuit values have assumed their steady state values"

There are two situations in which the current through a capacitor is zero.

1) When there is zero voltage across it. (we know that is not the case here)

2) When it is fully charged. That is the case here. (Actually when the voltage across it is not changing)

So there is zero current in the capacitor at t=0. that is I2 = 0 at t=0.

Also at t=0

Note I am concentrating on the capacitor because you did state your task is to find the discharge current of the capacitor.

Because I2 (as I have shown it is zero it does not add to or subtract from the current in R2.

So your calculation of the current in R2 (and of course R1) is correct. But only because I2=0 at t=0. At any other time you cannot say this.
this was why I asked for your justification that the discharge current flows purely through R2.
So also your calculation of the voltage between A and B is correct.

OK so now the big moment, we throw the switch to open.

Now the switch is open is there a path for any further current to leave a battery terminal, pass round a complete circuit and back to the battery?

19. ### kristapsdreija Thread Starter New Member

Apr 15, 2014
18
0
No, the circuit is broken, the power supply stops supplying current to the circuit and the only remaining current in the circuit is that which is stored in the capacitor.
P.S. does it makes a difference weather the switch is on the positive side of the power supply or the negative? (in my task it is between the power source and R1) . Oh, and I have to find the current function through the resistor R2 !

Thank you very much, I sincerely did not expect to get so much help! You guys are awesome!

20. ### WBahn Moderator

Mar 31, 2012
20,076
5,666
Not true at all. In a capacitor with a sinusoidal signal, the current through the capacitor is at its peak when the voltage across it is zero. For a capacitor,

i = C(dv/dt)

Setting i=0 only forces dv/dt to be zero and places no constraint at all on the magnitude of v, just that whatever it is, including zero, that it not be changing. Similarly, setting i≠0 only imposes the constraint that the voltage does have to be changing while imposing no constraint on what it's magnitude is, including it being zero (but changing).

The ONE situation in which the current through a capacitor is zero is when the voltage across it is not changing.

It's confusing when you are using a schematic that you've made up that doesn't use the same definitions of the variables that the OP's schematic uses. He is looking for HIS I2, not YOUR I2. His I2, per his schematic (supplied in Post #4) is the current flowing downward through R2, which is the 1.3kΩ resistor in parallel with the cap.