# rlc circuit help

Discussion in 'Homework Help' started by Alex ryan, Apr 24, 2015.

1. ### Alex ryan Thread Starter New Member

Apr 24, 2015
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I kinda need help with some circuit analysis, or at least how to start. I'm pretty okay with everything in this course up until now. plz help, or at least lead me somewhere because I'm kinda lost with capacitors and inductors. Mostly because I had to miss the pass two lectures because i've been really sick.

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2. ### ericgibbs AAC Fanatic!

Jan 29, 2010
2,503
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hi alex,
If you told us what the question/s you are being asked to solve for your posted circuit, it would help us.
E

3. ### Alex ryan Thread Starter New Member

Apr 24, 2015
3
0
oh sorry. I need to solve for Vx, the voltage across the 12 ohm resistor

4. ### Papabravo Expert

Feb 24, 2006
10,140
1,790
The three types of passive components respond to voltage and current in different ways.
1. In a resistor the voltage across a resistor is proportional to the current through it. The resistance is the constant of proportionality.
2. In an inductor the voltage across it is proportional to the time derivative of the current through it. The inductance is the constant of proportionality.
3. In a capacitor the voltage across it is proportional to the integral of current with respect to time. The capacitance is the reciprocal of the constant of proportionality.
Does that help?
Can you translate my words into mathematical expressions?

5. ### Alex ryan Thread Starter New Member

Apr 24, 2015
3
0
alright yeah some of that makes sense. But what if i'm just given that it's connected for a "long time" what is a long time to integrate for capacitance

6. ### Papabravo Expert

Feb 24, 2006
10,140
1,790
In terms of inductors and capacitors a long time can be measured in time constants. Five or more is usually considered a long time.
1. For a capacitor and a resistance the time constant is R*C
2. For an inductor and a resistance the time constant is (L / R)
These are derived from the exponential response of a series RL or series RC circuit. After 1 time constant the response is $e^{-1}$ times the response at t = 0, and so on.

Last edited: Apr 24, 2015
7. ### WBahn Moderator

Mar 31, 2012
17,743
4,790
In the circuit you posted, what is the frequency of the supplies? DC? That would make sense in the context of the question you asked above.

By saying that something has been connected for a "long time" it means that whatever transients existed have died out and you are left with the steady state (the unchanging) solution.

For DC circuits that means that the voltages and currents are not changing anywhere in the circuit. Using the defining relationships for capacitors and inductors, since the voltage across an inductor is proportional to the rate at which the current is changing and since, in steady state, nothing is changing, that means that the voltage across an inductor has to be zero. It says nothing about the current, which can be whatever it needs to be, as long as it is not changing. Similarly, since the voltage across a capacitor is proportional to the integral of the current, the current has to be zero otherwise the voltage would be changing. In this case, the voltage can be anything (as long as it isn't changing), but the current must be zero.

If you think about these constraints, a short circuit (a wire) between two nodes imposes the constraint that the voltage difference must be zero while the current can be anything, while an open circuit between two nodes imposes the constraint that the current between those two nodes (along that path) must be zero while the voltage can be anything.

So leverage that when analyzing a circuit like you have. You'll see that you can drastically simplify the circuit and, assuming you have seen source transformations, can solve the circuit in about three lines of algebra, depending on how detailed they are.

So with all of this in mind, show your best attempt to solve the problem and we'll look it over.

8. ### ericgibbs AAC Fanatic!

Jan 29, 2010
2,503
380
Alex,
I would assume 'a long time' as already pointed out, is greater than 5 time constants, so I would say its a 'steady state'.

If you assume a steady state for the circuit you should now work out and post the simplified circuit diagram and solve for that.
E

9. ### MrAl Well-Known Member

Jun 17, 2014
2,430
490
Hi there,

Wow, quite a few inductors and capacitors, usually we dont see that many in one circuit in these forums. That creates probably a 6th order system, so be happy you dont have to solve for the time response of Vx

After a long time we either call it the steady state solution or the final value. Because this is excited by all DC sources, we'd most likely call the response after a long time the final value, or at least that is what we would be looking for, and that would be the steady state response.

If all of the poles of the function s*F(s) are in the left half plane, then the general solution for the final value is:
limit [s*F(s)] as s-->0
So you would first find F(s) and then compute that limit.

If you have not been taught to do it that way then if we assume that there is enough dissipation in the circuit to reduce all oscillations to zero after a long time, then we can use a shortcut. The shortcut is to consider all inductors a short circuit for DC and all capacitors an open circuit for DC. For this circuit that reduces it to a DC voltage source, a DC current source, and four resistors, because everything else is now either an open circuit or a short circuit. After that it's pretty simple to find the DC voltage Vx.

You can use this second method to check the previous method by comparing results.

BTW a long time is infinity, that's why we take the limit as the complex frequency goes to zero, but yes after 5 time constants of the longest time constant of the circuit it also gets close to the right result (99.3 percent of true final value) and after around 38 time constants it gets so close to the right value it takes special big number software to even compute the difference
But beware, the 5 time constant rule does not always work when there is more than one time constant unless the amplitudes are comparable.

Last edited: Apr 25, 2015
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