No need to keep apologizing. You are doing fine and making progress.

Are you comfortable with the units problems that were in the radical (the square root) in your denominator?

THEVENIN RESISTANCE

Remember that the Thevenin resistance is the equivalent resistance as seen looking into the terminals of our Thevenin equivalent network with all supplies turned off. The circuit you have shown in the upper right corner of your figure is correct. But what does this mean?

Imagine taking the circuit and putting it into a black box with just those two wires sticking out. Those are the terminals that you want the equivalent resistance between. Now, take your finger and put it at the end of the top wire. Can you trace a path (any path) with your finger that makes it to the end of the bottom wire that goes through one of the resistors without going through the other? Is so, then they are not in series. They are in series ONLY if EVERY path from the end of the top wire to the end of the bottom wire that goes through one resistor also goes through the other.

To see if they are in parallel, pick one of the resistors and put your left index finger on one end of it and your right index finger on the other. Now see if you can move your two fingers so that they are at the two ends of the other resistor withou lifting your fingers or leaving the wire they are tracing. If you can, they are in parallel. If not, then they aren't.

This might help you see things better. Look at that figure again (the top right one with just the two resistors and the big arrow looking into it from the right). Have you changed anything if you were to slide that 1kΩ resistor on the top to the left and then down along the left-most leg of the circuit?

TRIG UNITS

The trig functions (both forward and inverse) all take a dimensionless number as an argument and return a dimensionless argument as a result. In the forward case, the argument (using geometry concepts) is a ratio of the length along the circumference of a circlular arc to the radius of that arg while the output is the ratio of two of the three sides of a right triangle having the same angle as the arc. Units such as "degrees" are an artificial unit and you can think of a trig function that uses these units as, under the hood, having a converter to convert between radians and degrees as needed.

So when you have a = arctan(b), 'b' has to be dimensionless and so will 'a' be. The same true for a = log(b) and a = exp(b).

So what about when you are operating in "degrees mode" in your calculator? All that is happening is the following process:

You punch in b_deg (and angle in degrees) and hit 'tan'. Your calulator computes:

a = tan(b_deg*(pi/180deg))

The degrees cancel out and you have a dimensionless number as the argument.

Now you punch in 'a' (this must always be dimensionless regardless of mode) and want to get 'b_deg' back out. Your calculator performs the following:

b_deg = arctan(a)*(180deg/pi)

 

With apologies to WBahn for not liberally including units, but I can says that all of my units are in ohms or volts, unless otherwise noted. I'm posting my solution for the SS inductor voltage, since this has gone around a few times.

Using phasor notation, I get something different from the book answer for the inductor voltage. My Thevenin resistance is 750 Ω. I used 8000 rad/sec to get an inductive reactance of 800 Ω. Then I used the Thevinin voltage of 30 V with the voltage divider formed by the equivalent resistance and the inductive reactance.

\(V_L=\frac{(30\angle0^o) \times (800\angle90^o)} {(750\angle0^o)+(800\angle90^o)}=21.9\angle43.2^o=15.97+j14.97 V\)

It looks like the book answer is only the real part for the inductor voltage. The same is true for the inductor current answer.

 

No, the book answers are not the real part; the book answers are the instantaneous voltages at time t=0s (which is what the problem is asking for). When you come up with a phasor solution to a voltage or a current, the phase angle is the phase at the reference time (t=0s). That this turns out to be the real part of the phasor is then a consequence of the driving votlage being a cosine and hence, if you assign zero phase to the voltage source, you have to use the real part of the phasor to get the cosine. Had the voltage been a sine and you had assigned it zero phase, then you would have had to use the imaginary part of the phasor. This happens automatically as long as all driving sources are written either as cos(wt+something) OR sin(wt+something) and the answer is expressed using which ever one was chosen. Note also that had the book asked for the values at any other time, the answers would, in general, not be just the real part.

Since you chose cosine (you had to to get a voltage of 30V @ 0deg), the conversion back to the time domain for your phasor is

\(
V_L = 21.9\angle 43.2^o V
v_L(t) = 21.9 \cos \left( 8000\frac{r}{s}t + 43.2^o \right)V
\)

Note that it is going around a few times not because of anything assocated with AC or phasors, but rather with getting a firm conceptual handle on what it means to find the resistance as seen between a particular pair of points in a circuit.

 

You're right. I forgot the book was asking for the value at t=0. If I multiply the peak voltage (21.9 V) times cos(43.2°), then I get the book answer. It would have been clear why if I had converted the phasor back to the time domain equation. However, I do think working in phasors has some merit -- as long as the conversion back to time domain is done correctly.

 

Sir, yes i understood that the configuration would yield a parallel circuit, but then i have one question that always bothers me. You say we have to consider the path of the current from the top to the bottom of the circuit only , so meaning then the diagram is not to be considered a circle ,with two resistors in series? since the disconnected wire cannot carry any current at all? the current which would flow through one resistor would also flow through the other, and i remember solving many questions assuming the same. So pls clear this out. i have posted an image to clearly state my question, pls take time to verify it

I get the point sir, that trigonometric functions return dimensionless quantitles, and also that the calculator will calculate for only radians(ratio between the arc and the radius)

And the units problem in the denominator is (Sqrt) ( Ohms^2 + rad^2) , should i just take it as Ohms? since radians is dimensionless?

thank you

 

With two isolated resistors as you have shown, they have to be in series because that is the only way to have a circuit. This would be true even if one point in the circuit were connect to ground (or anything else, for that matter). But it's not as simple as that because, starting at the lower-left corner, if the current is flowing up through the left resistor, then the voltage on the top node is lower than the voltage on the bottom node. But now the current must flow down through the right resistor and more voltage is dropped requiring that the bottom node be at a lower voltage than the top node. The end result is that if there is any current flowing in this manner, the bottom node must be at a lower voltage than the bottom node, which is not possible because it violates KVL. In actually, it is possible as long as the electric fields involved are nonconservative since KVL only applies to conservative electric fields, which we always assume we are dealing with in problems like this.

But we are not talking about isolated resistors. In calculating the Thevenin resistance you are analyzing the circuit as if there were a voltage source attached to the two ports and looking at how current gets from one port back to the other. Any current that were circulating around the two resistors as if they were in series never makes it to either port.

 

thank you sir, that cleared a lot of things regarding this, i will then from now on analyse using the points you just said. I feel relieved now, thanks a lot sir

 

And the units problem in the denominator is (Sqrt) ( Ohms^2 + rad^2) , should i just take it as Ohms? since radians is dimensionless?

 

And the units problem in the denominator is (Sqrt) ( Ohms^2 + rad^2) , should i just take it as Ohms? since radians is dimensionless?

No, because you can only add something in ohms to something else that is in ohms. Something that is dimensionless is not in ohms.

Think of it this way. I can add two numbers together and that is it. I can't add units. But think of the units as just another factor. If both terms have the same factor, you can factor it out. So, for instance,

2A + 3A = (2 + 3)A = 5A

But

2A + 3V ... can't get it to factor out.

How about:

2A + 3000mA = (2 + 3000m)A = (2 + 3000m*(1/1000m))A = (2 + 3)A = 5A

Here I multiplied the 3000m by the fraction 1/1000m and since 1000m is simply 1, the whole fraction is 1 and I haven't changed the value of anything. But now I have an 'm' in both the numerator and the denominator so, like any such factor, they cancel out.

Since the terms being added do not match in terms of units, go back to the expressions that resulted in each of those terms and see if there is an error there that resulted in the wrong units on one or the other. If not, keep working your way back.

 

thank you sir, i do understand the basics of what you have said but the formula does require us to add the values , without considering the units, even checked an example with the same formula. So , what is the procedure then? and sir, i am aware of what you just said. my question is a honest one, since i just checked with a similar example. they have just added the two quantities.

 

Go back to the figure you posted:

http://forum.allaboutcircuits.com/image_cache/httpimg710.imageshack.usimg7107936tempzp.jpg

Now look at the symbolic expression at the bottom left:

\(
i(t)=\frac{V_m}{\sqrt{R^2 + (\omega L)^2}} \cos \left( \omega t + \tan^{-1} \left( \frac{\omega L}{R} \right) \right)
\)

What are the units on \(\omega L\)?

\(
(\omega)(L) = \left( \frac{rad}{s} \right) \left( \frac{Vs}{A} \right)
(\omega)(L) = \left( \frac{(rad)(V)(s)}{(s)(A)} \right)
(\omega)(L) = \left( \frac{V}{A} \right)
(\omega)(L) = \Omega
\)

Remember, radians are dimensionless so you can pretty much bring them in or take them out as makes sense.

The point here is the frequency times inductance is its reactive impedance, which as units of ohms. Similarly, if you work on the units on the reactive impedance of a capacitor, 1/(wC), you find it also has units of ohms.

Do you see how this properly makes the argument of the arctan function dimensionless, as well?

Also, I just noticed that in the left hand equation where you plug in the values, you included the cos() term associated with the source. That was incorrect. The Vm in that equation is the magnitude of the forcing voltage and that equation assumes that it is a cosine function with zero phase.

As I think I mentioned earlier in this thread (I have to point it out a lot, so I don't know if I did here or not), the writers of textbooks are notoriously sloppy with units. Frankly, I suspect that is because so few of them have ever done any real-world engineering in which units mistakes cost time, money, and sometimes lives. So expect to see it and, when you do, curse at them a bit and then do what they were too lazy to do, which is add in the necessary units and then track them properly from there.

 

Thank you for clearing that point sir, yes the book does specify that the reactive impedance and the capacitive impedance have a unit of Ohms. So that answers the units problem.
As the product of Omega*t becomes zero, the degree still remains, so cos of that degree is still accountable ? i do not quite understand that point sir. Is the degree the reference?

 

I actually need to take back what I said. When I was looking at your expression on the right hand side I didn't see the parentheses correctly (an artifact of the vision problems I have that are only getting worse). Other than the units, your expression is fine.

Now, the equation on the left still assumes a cosine voltage with no phase offset. The best way to see what is happening and how it would show up in the textbook equation is to assume that there is a phase angle on the source of Φ and then do the analysis and see where that Φ end up in the final result. Then do the same for a sine wave with an offset. Then see if you can reconsile them with the fact that a sine wave can be written as a cosine and vice versa.

 

Sir thank you for your response i have a doubt, regarding conversion of a current source to a voltage source. Am i correct when i convert a 12*cos(2000t) A current source in parallel with a 200 Ohm resistor and a 0.2 H inductor, to a voltage source of 12*200*cos(2000t) V in series with the same resistor and the inductor? and solve them in frequency domain? later converting them to time domain to find the power absorbed by the resistor at t=1ms? am i wrong sir? i did not find the correct answer through this method but i did find using nodal analysis using the current source, but i want to know where i went wrong in this method so that i can correct myself.

 

sir i have one more doubt, well cos((2000 rad/s * 0.001 rad) + x degrees) , how do i calculate this? since i have to add a radian and a degree component?

 

Sorry for troubling you sir, going through your previous comments i figured out that i have to convert the radian(here, 2 rad) to an equivalent degree and then add with the x degrees to find the cosine of that value. This simple issue reiterates the importance of checking units. thanks again.

 

Sir thank you for your response i have a doubt, regarding conversion of a current source to a voltage source. Am i correct when i convert a 12*cos(2000t) A current source in parallel with a 200 Ohm resistor and a 0.2 H inductor, to a voltage source of 12*200*cos(2000t) V in series with the same resistor and the inductor? and solve them in frequency domain? later converting them to time domain to find the power absorbed by the resistor at t=1ms? am i wrong sir? i did not find the correct answer through this method but i did find using nodal analysis using the current source, but i want to know where i went wrong in this method so that i can correct myself.

Without looking at the problem in detail, my guess is that you are making the common mistake of thinking that these conversions preserve information about what is going on "inside the box" -- they don't. Thevenin and Norton equivalent circuits are designed only to take the components inside a box that are connected to the rest of the circuit by two wires (ports) coming through the box and replace them with a very simple circuit inside the box that is the same as seen by the circuit outside the box. That's it. What is going on inside the box may be very, very different

As a real simple example, consider a Thevenin equivalent source that is a 12V battery in series with a 12Ω resistor. How much power is being dissipated in the 12Ω resistor if the load resistor if very large (like, say, an open circuit)? Now consider the Norton equivalent, which would be a 1A current source in parallel with a 12Ω resistor. How much power is dissipated in the 12Ω resistor now when their is no load resistor? There is no reason to think that either of these represent the actual power being dissipated by the source in the portion of the real circuit that these equivalent sources are modelling.

 

sir i have one more doubt, well cos((2000 rad/s * 0.001 rad) + x degrees) , how do i calculate this? since i have to add a radian and a degree component?

How would you add 1 foot + 3 inches? You would convert one to the other using the fact that there are twelve inches in one foot. Do the same here using the fact that a full circle is both 360 degrees and 2pi radians.

But you have a bigger problem (which is, perhaps, just a typo) in that ((2000 rad/s * 0.001 rad) has units of square-radians per second. Is that supposed to be 0.001s, perhaps?

 

A current source of 12cos(2000t) A., a 200-ohm resistor, and a 0.2-H inductor are in parallel. Assume steady-state conditions exist. At t = 1ms., find the power being absorbed by the
(a) resistor
(b) inductor
(c) sinusoidal source

^ this being the solution i solved, my question is can i convert the parallel current source to a series voltage source equivalent? am i wrong here?
And i got your point about the thevinin and norton equivalents, but that is not my question sir and also about the units problem, i see it clearly now thanks to you. But will be kind enough to answer my query here?

 

You cannot use converted sources if you want to find out anything about the voltage, current, or power of any of the components that have been replaced by an equivalent source. Think about it. Those components are gone! They are no longer part of the circuit; they have been replaced with something else! You can only use the equivalent circuits to find out information about components that have NOT been replaced. Now, if you want to do conversions to find voltage and currents in the other components and then go back to the original circuit and apply that information to help solve for the rest, fine.

Again, look at the example I gave you. If you can't get the same power results in the components of a Thevenin and Norton equivalent source that's sitting there looking stupid with no load, then clearly you can't use them to get the power in the actual circuit because the results you get will be different depending on whether you used the Thevenin or the Norton source.

I've looked over your work and it is time for me to put my foot down. I will not help you any further until you start tracking your units throughout your work.

Also, when you switch to a new problem, please start a new thread for that problem. It will make it a lot more likely that others will look at it and participate and it makes it a lot easier for me to refer back to the original problem because it is in the first post, not in the Nth post of the Mth page.