Guys can you please help me with this. I have to hand this in tomorrow.



I'm looking for the voltage at that capacitor in the middle. Usually I do an open source KVL on it. I'm not sure if it's 14 or 0?

Could someone just please tell me. I'm realllllly short on time guys

 

Sorry. That's not how it works here on AAC. We don't do your homework for you.
You show us your best attempt at arriving at a solution and then we can guide you.

 

lol
not even enough time to edit the pic to make it easy to read.

 

Guys can you please help me with this. I have to hand this in tomorrow.

I'm looking for the voltage at that capacitor in the middle. Usually I do an open source KVL on it. I'm not sure if it's 14 or 0?

Could someone just please tell me. I'm realllllly short on time guys

Hi,

You have to show some work so we know what way you already know to solve these, otherwise we might hand you some stuff you dont use or cant use.

Here's a little bit better drawing.
Note you first have to solve for the initial capacitor voltage due to the current source, and solve for the step response. You have to show some work so we know how to help, but if you have never done these kinds of circuits then there's no way you'll get it by tomorrow.

 

For a first order circuit you should already know that the transient response is going to be an exponential that goes from an initial value to a final value. So you need to determine three things, the initial value just after the change in the circuit, the final value a long time after the change in the circuit, and the time constant that governs how fast the circuit goes from the initial value to the final value.

Show your best attempt at all three of those things (you can do them independently) and then we will be in a position to help you identify what you have right and what you don't and help you move in the right direction.

 

My lecturer gave me a small extension till tomorrow. This is what I got. Please excuse egregious tracking of units

 

View attachment 126266 View attachment 126267
My lecturer gave me a small extension till tomorrow. This is what I got.

Let's put the correct units on just one of your equations.

You have
\(
-V \; + \; 4i_x \; - \; 6i_x \; = \; 0
\)

With the proper units tracking, that should be

\(
-V \; + \; \(4 \; \Omega \) i_x \; - \; 6i_x \; = \; 0
\)

The first term is a voltage. The second term is a resistance times a current, which is a voltage. The third term is a number times a current which is a current. You are trying to add two voltages and a current, which can't be done.

Had you tracked your units, you would have known immediately at this point that you had made some kind of mistake. Instead, because you obviously don't care about getting the right answer or a good grade or wasting time going down rabbit holes, you opted to slug on through a bunch of math that was (all but) guaranteed to produce a wrong answer and then just tack the units that you wanted onto the end result.

It turns out, in this case, you end up with the numerically correct answer by coincidence because the resistor that counts happens to be 1 Ω.

If you had let the units reveal this problem, you might have looked at your circuit a bit more carefully and seen that you are not summing up voltages around a loop plus you are asserting that the same current that is flowing in the 1Ω resistor is also flowing in the 4Ω resistor -- through an open circuit!

Please excuse egregious tracking of units

I can't force you to properly track your units. When you finally decide you want to stop wasting time, start getting correct answers more of the time, and improving your grades and employment prospects considerably, you might start tracking your units.

 

... another approach. It looks like you can use the current node approach by writing the individual branch currents into the two nodes at the top. You don't solve directly for Vc, but for the two node voltages, whose difference constitutes Vc. ... two equations, two unknowns.

 

... another approach. It looks like you can use the current node approach by writing the individual branch currents into the two nodes at the top. You don't solve directly for Vc, but for the two node voltages, whose difference constitutes Vc. ... two equations, two unknowns.

That's the better choice because, with a bit of experience, you can immediately see that the voltage on the right side of the capacitor will be 14 V and your voltage divider on the right will give you 2/3 of 14 V, so the voltage across the gap (left side relative to right side) has to be - (1/3) 14V.

 

I ended up with the voltage because 6ix is going across that 1 ohm resistor. 1 (6ix) is 6ix? Had it been 2ohms I would have calculated 12ix

 

I ended up with the voltage because 6ix is going across that 1 ohm resistor. 1 (6ix) is 6ix? Had it been 2ohms I would have calculated 12ix

It is not obvious at first, but you can convert the Ix component into terms of the node voltage ... Then you have two unknown node voltages and two equations.
... but you have to use the current node method, not KVL loops.

 

Hi,

Call in sick tomorrow for more time

One word of caution here...

We do have to solve for the initial cap voltage here, but it's more tricky than usual. Can you see why?
I'll give a hint...
Notice that the current source depends on the current through that one resistor, but at t=0- there is no current through that resistor. That's the only catch to this circuit i think. If that was a constant current source that would not be the case, but it's a dependent source.

 

I ended up with the voltage because 6ix is going across that 1 ohm resistor. 1 (6ix) is 6ix? Had it been 2ohms I would have calculated 12ix

Ah. I see. So I came down on you a bit harder than I should have. My apologies.

Though I still stand by my assertion that failing to track your units will result in you spending more time to get answers wrong than you would spend getting them right by properly tracking.

 

Not a problem my friend. Hey its probably good for me. In fact I know it is. Thanks for all the great advise. It's worth taking a lil brow beating for the teaching. I want peps to know though that I'm not lazy. I will do the work but I do like to ask silly questions. Thanks guys for all the direction through this thread. I shall investigate them all soon as I have a moment to breathe.

 

Hi again,

Are you trying to find the final voltage or answer the question as asked in that diagram?
I ask because it sounds like you are only trying to find the voltage after a long time has passed when the question is really about finding the entire time response.

 

Haha you're right MrAI. There's a page missing there, forgot to upload it. It is done and she's away now. Boom finiteo! Thank the Lord. Seriously there must have been 60 pages on that thing, some with sub questions.
I'm a bit older, some of the younger and definitely less wise students, decided it would be a good idea, to lodge a formal complaint about one of the lecturers. I'm thinking this was part of the outcome... Lul.
Intelligence and wisdom are definitely two different things...