# Initial Conditions of Switched Circuits

Discussion in 'Homework Help' started by jegues, Oct 14, 2010.

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

Sep 13, 2010
735
43
See figure for question and my attempt.

I redrew the circuit at a time just before time 0.

Doing a KVL in loop 1, 2 and 3 I concluded that V1(0-) = 24V. I'm not 100% confident about this result, is it correct? If not, how do I go about finding V1(0-)?

I redrew the circuit again at a time just after time 0.

I have no clue how I'm suppose to find I2(0+) and V1(0+) here. Can someone get me started?

Thanks again!

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• ###### 7.8-7A1P2.JPG
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Last edited: Oct 15, 2010
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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V1(0-) won't be 24V.

The circuit has stabilized before t=0.

At this time ...

The inductor will look like a short and the capacitor will charge to a value determined by the voltage divider formed by the 24V supply and the series combination of the 20Ω and 80Ω resistors. The value of V1 and the capacitor voltage will be the same, since there will be no steady state voltage drop across the inductor.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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For the t=0+ case one approach is to treat the inductor as a current source and the capacitor as a voltage source - just for that instant. Then say apply superposition (or whatever method suits you) to find the unknown quantities.

The notional current source value will be the steady state inductor current at t=0-.

This approach only works because you are only required to solve for values at the instant the switch closes.

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

Sep 13, 2010
735
43
Okay so my BIG mistake was that I interpreted/read the question wrong.

The circuit is already at steady state when the switch closes. So at t(0-) my capacitor and inductor should be their steady state equivalents respectively. (Open circuit, short circuit)

Then once the switch closes, t(0+), I can model my capacitor and inductor accordingly.

I'm going to try this problem again from scratch and I'll post my results!

Thanks again tnk, your help is always appreciated!

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

Sep 13, 2010
735
43
UPDATE: I've reattempted the problem and came up with a final solution.

See the figures attached for my 2nd attempt at this problem.

Okay so basically what I did was I redrew the correct circuits for t(0-) and t(0+).

In the t(0-) circuit I solved for V1(0-) using a voltage divider as tnk mentioned, and clearly one can see that i2(0-) is 0.

I also obtained Vc and Il.

Then I went to my t(0+) circuit and applied mesh analysis. From these results I was able to solve for V1(0+) and i2(0+).

Does anyone see any errors in my work/results?

I'm really hoping I finally got this one down! Let me know what you think!

Thanks again!

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6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I to check you result for V1(0+) write this nodal equation

$240mA = \frac{V1}{80} - \frac{(24V - V1) }{80}$

And when I solve this I get this result for V1(0+)
V1(0+) = 21.6V

Your mesh equation is not correct becaues of a current source in the circuit

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

Sep 13, 2010
735
43
Okay I agree that my mesh equation must be wrong, but I'm not sure where exactly I went wrong. Can you see it? I don't see how the current source is causing any problems.

EDIT: I found my mistake. I had an algebra error in my equation *,

It should be,

$160i_{2} = 4.8 \rightarrow i_{2} = 0.03A$

Then one will find that,

$V_{1}(0+) = 21.6V$

Last edited: Oct 15, 2010
8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,019
1,123
good job, but for this circuit the nodal analysis is much faster.

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

Sep 13, 2010
735
43
Yup I realize that now, but if I didn't notice it at the time. I need more practice I guess...