# NCEES 109



## chicago (Oct 19, 2007)

I understand that v(t) = *A*+*B*e^(-t/tau) is the general form of a homogeneous linear 1st-order diffy-q solution for an RC circuit.

Do I have the following information below correct?

- natural response (zero input response) is parameter *B*, V(0)

- forced response (zero state, steady state response) is parameter *A*, V(infinity)

Is the initial condition the B term (the voltage divider at t=0)?


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## jdd18vm (Oct 19, 2007)

chicago said:


> I understand that v(t) = *A*+*B*e^(-t/tau) is the general form of a homogeneous linear 1st-order diffy-q solution for an RC circuit.
> Do I have the following information below correct?
> 
> - natural response (zero input response) is parameter *B*, V(0)
> ...


I'll be waiting for an answer on this too. 109? I thought the first designation was the natural (initial?) response and the exponential portion the forced response. But the solution for 109 has the 4 volts as the B or second. Ugh

John


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## benbo (Oct 19, 2007)

Gosh, all that is right but you don't need to know the names of all that stuff to solve this. They might ask elsewhere I guess. I mean, A is zero because opening the switch gives the natural response, but I wouldn't think about it that way.

Just think about what happens.

The switch is closed for a long time. The capacitor is therefore fully charged to the voltage across it, which (as you said) is B, the intial condition given by the voltage divider === 6V*(2/3).

Then you open the switch. What is going to happen? The cap will discharge through the 2 ohm resistor. The time constant will by 2*3 = 6ms/

THe A term is more important when something is charging up (which is another way of saying forced response I guess).

Look at table 30.1 in the EERM.

I have a feeling you would get this right your way, and if that is easier for you, that is fine. But don't complicate things. This test is more abut practical things.


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## Flyer_PE (Oct 19, 2007)

I don't know if this helps or not. I've always just gutted my way through these problems based on the expected response. I'll use NCEES 109 as an example:

The value of tau is simply RC for an RC circuit.

The condition that exists prior to t=0 is that the capacitor is charged and no current will be flowing through it. v(t) prior to t=0 is then a simple voltage divider. v(0) = 4V

Once the switch opens. The response will be that the capacitor discharges through the resistor at a rate defined by e-t/tau.

At t=0, e-t/tau=e0=1

This results in A + B should be equal to 4 Volts.

As t heads for infinity, the value of esup]-t/tau approaches 0. Since there is no source in the circuit with the switch open, v(t) will also approach 0.

You end up with A + 0 = 0 : A=0

If A=0 and A+B=4 then B=4

For the simple RC and RL circuits, I try not to get bogged down in the diffy Q stuff. Most of them can be logically figured out just by figuring out the steady states for t=0 and t=infinity and making the equation match.

Jim


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## chicago (Oct 19, 2007)

Thanks guys for stepping through the transients before and after the switch is open. It helped me visualize what's actually occurring rather than getting too caught up in the equations and terminology.

I was drawing a blank in trying to figure out e^(t=infinity). But Jim cleared that up. Benbo, I must have not seen Table 30.1. Thanks for pointing it out.

Just for grins, suppose the switch was opened for a long time, and then closed at t=0. Help me go through how A and B terms would change if that happened. Hypothetically, of course.


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## benbo (Oct 19, 2007)

I may be wrong, so correct me anyone.

I'm a little rusty, but things are going to change if you do that. duh. The time constant will change because the whole resitive network will now be part of the circuit. The easiest way is to convert it into a Thev. equiv - a simple series rresistive charging circuit.. Anyway, the equivalent resitance would end up becoming 1 K parallel 2K or 2/3 kohm. THe equivalent voltage would still be 4 Volts (I think). This equivalnet circuit is actually the part I'm not sure about.

Now, what happens. Assuming the cap has no initial charge, v(t) would be charging from 0 up to 4 volts.

A would be 4 volts.

B would be -4 votls. This is because the exponential tem will go from 4 at t=0 to 0 at t=infinity. Therefore the cap will go from zero initial charge up to 4 volts.

tau = .67*3uF ms. Whatever that is.

THe actual equation is 4-4*e^(t/tau)

Look at the series RC charging circuit in that table.


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## Flyer_PE (Oct 19, 2007)

^^I think you nailed it.

Jim


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## benbo (Oct 20, 2007)

IFR_Pilot said:


> ^^I think you nailed it.
> Jim


I think I'm close - at least one small error -I believe that should be e^-t/tau as opposed to e^t/tau


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## Dark Knight (Oct 20, 2007)

You guys are good


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## mudpuppy (Oct 20, 2007)

I think I've said basically the same thing in another post a while back, but I'll repeat here:

There's three ways I know of to solve transient problems.

1. Write out the differential equations for the circuit and solve. This is a big PITA, if you ask me, but for some reason that's how they explain the solutions in EERM.

2. Transform the circuit into the frequency domain using Laplace and solve, then transform back to the time domain. Much easier than 1., but still takes some time, especially if you're rusty on LaPlace. And you have to be very careful with your initial conditions.

3. Use an ad hoc, by inspection method. IMO this is the preferred method because it is by far easier and quicker, but you still have to be very careful to think about what's happening in the circuit before and after t=0. Jim and benbo have done a great job of explaining this method so I won't go into any more detail.

The nice thing about these problems is if one method isn't working for you, you can always try one of the other methods. I would still definetly recommend method 3 first, though.


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## benbo (Oct 20, 2007)

The EB.com is so much better than the other places. A really good crop of examinees this time - asking lot's of technical questions. It is real good practice for me to try to refresh my memory, and in many cases see things solved by others that I have no idea how to solve. I keep learning off this board, I don't see why any workplace would ban it if someone was getting thier work done. Of course, lucky it's a little slow for me at work.


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## chicago (Oct 20, 2007)

Jim/Benbo,

If I were to incorporate Benbo's explanation for the switch closing scenario into Jim's setup of equations, I would have the following (correct me if I'm wrong):

switch open for long time prior to t=0:

V_th = 6V(2K/(2K+1K) = 4V (thevenin = voltage divider)

A+Be^(-t*tau)=v(0): A+0=4V ==&gt; A=4

switch close at t=0:

A+Be^(-t*tau)=v(infinity): 4+B*e^(0)=0V ==&gt; B=-4


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## Flyer_PE (Oct 20, 2007)

Looks like you have it to me. Benbo's final answer winds up being 4 - 4e-t/tau.

Jim


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## jdd18vm (Oct 25, 2007)

Way to GO! THIS is going in my binder.


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