I have done the calculations as below, but my major question is following the passive sign convention the power of the DC source is negative, so it is delivered, but in reality, i think it is absorbed.




The average inductor power is 0 as the inductor current is periodic.

 

Please explain, in detail, how you are concluding that the power of the DC voltage source is negative according to the passive sign convention. You aren't giving enough detail for us to see where the flaw in your reasoning is.

Start with explaining YOUR understanding of the passive sign convention, particularly as it applies to a source.

On another note, is there ANY way I can get you to stop ignoring units in your work and just throwing on whatever units you happen to want the answer to have onto the final result? Do you not WANT to get better grades and spend less time on your homework because you will find many of your errors almost immediately upon making them?

This is a recipe for disaster. It doesn't matter that your instructor and/or your textbook might be sloppy with their units. It's unfortunate, but a sad reality is that most instructors and most textbook authors have little to no real world experience. To them, a wrong answer that results from sloppy units is just some missed points on an assignment or exam. In the real world, mistakes that could have been caught but weren't because someone couldn't be bothered to track their units properly throughout their work can have disastrous consequences. An airliner with over sixty passengers ran out of fuel in midflight because several people couldn't be bothered to track their units, and so a simple mistake went uncaught repeatedly. A $125 million probe crashed into Mars instead of going into orbit because engineers and software developers couldn't be bothered to document and track units properly. I watched someone die, leaving behind kids and a pregnant wife, because they couldn't be bothered to track their units and just threw on the units that they expected the answer to have. I was within three feet of him when it happened and it could easily have been me.

 

Thank you for your time and reply.
As per the textbook this is the sign convention



It falls under case b but current direction is reverse so i am concluding it is absorbing.
Yes i will start following the units from the next works.

 

Thank you for your time and reply.
As per the textbook this is the sign convention
View attachment 362561

View attachment 362562
It falls under case b but current direction is reverse so i am concluding it is absorbing.
Yes i will start following the units from the next works.

Thanks. This is important because there are two common conventions of the passive sign convention.

The first one says that ALL two terminal devices are assigned polarities such that positive current flowing into the positive voltage node of the device. Under this convention, no distinction is made between loads and sources. This has the advantage that we don't have to guess whether any given device is a load or a source (sometimes it's not obvious). We apply the convention and if the result is positive, the device is absorbing power and if it is negative it is sourcing power. The downside is that we end up working with a lot more negative numbers, which is something that humans are good at screwing up.

The second one is the one you show above in which we assign each two terminal component to be either a "load" or a "source" and assign the currents so that a load acting like a load will has positive power and, likewise, a source will be acting like a source if it has positive power. The choice of "load" and "source" for each component is completely arbitrary -- we could do it by flipping a coin. But, if we can tell which components are normally loads and which are normally sources, then if we are correct we work with a lot fewer negative values.

Neither convention is right or wrong, we just need to be consistent in whichever one we use.

Using the convention you have, the current in the voltage source needs to the the current LEAVING the positive terminal.

The current 'I' in your diagram is ENTERING the positive termina.

So the current LEAVING the positive terminal is '-I' in order to conform to your passive sign convention.

Hence the power of the source is negative, meaning that it is absorbing power, when both I and V are positive.

 

Thank you for clarification. One doubt i have on passive sign convention is which i am trying to solve a problem, suppose inductor is getting charged during ON time and during OFF time power source is disconnected from the circuit, now inductor will supply current / power, should i now consider Inductor as source and accordingly determine if it is absorbing or delivering?

 

Thank you for clarification. One doubt i have on passive sign convention is which i am trying to solve a problem, suppose inductor is getting charged during ON time and during OFF time power source is disconnected from the circuit, now inductor will supply current / power, should i now consider Inductor as source and accordingly determine if it is absorbing or delivering?

Yes.
The voltage polarity across the inductor and the direction of the current through it determine whether it is absorbing (storing) or suppling energy.

 

Below is an LTspice simulation to show the inductor power when supplied from an AC source:
The inductor power (red trace) goes from positive to negative, depending upon the inductors relative voltage and current polarity.
(The LTspice convention is that positive is power being absorbed and negative is power being delivered.)

 

Thank you for clarification. One doubt i have on passive sign convention is which i am trying to solve a problem, suppose inductor is getting charged during ON time and during OFF time power source is disconnected from the circuit, now inductor will supply current / power, should i now consider Inductor as source and accordingly determine if it is absorbing or delivering?

If you are talking about changing your current assignments depending on whether it is charging or discharging, you can certainly to this if you want. You just have to be very careful that you are consistent in all of your work, which basically means treating the analysis as two different problems with unrelated variable assignments. Sometimes this can make the analysis simpler, but you have to be very careful not to mix things up.

It is generally better to just define a single current polarity for the inductor, usually treating as a load since it is a passive device, albeit a reactive one, even though you know that sometimes it is absorbing energy and other times it is supplying it.

 

View attachment 362557


View attachment 362558

Hi,

Where did you get that average power expression for the resistor from?
With a current of 2 amps and resistor of 4 Ohms, we've already got 2^2*4=16 watts average power.

When dealing with sinusoidal waveforms you can't always integrate over the full period, sometimes you have to go over one-half period and adjust the period also.
Even more to the point, DC power is i^2*R not i*R. If time dependent, it is i(t)^2*R over the appropriate period.

See if that matters with your problem.

 

Yes my mistakes i will recalculate.

 

Hi,

Where did you get that average power expression for the resistor from?
With a current of 2 amps and resistor of 4 Ohms, we've already got 2^2*4=16 watts average power.

When dealing with sinusoidal waveforms you can't always integrate over the full period, sometimes you have to go over one-half period and adjust the period also.
Even more to the point, DC power is i^2*R not i*R. If time dependent, it is i(t)^2*R over the appropriate period.

See if that matters with your problem.

Few doubts i have as per the text book

The current is \[ i(t) = 2 + 6\sin(2\pi60t) Amps \]
Hence can i consider I1 = 2 Amps DC component, I2 = \[ 6\sin(2\pi60t) Amps\] , the RMS value of I1_rms = 2 Amps same as DC value, I2_rms = \[ \frac{6}{\sqrt2} = 4.24Amps \]
Avg Power (Watts) = \[ (I1_{rms}^2 +I2_{rms}^2)*R Watts = (2^2 + 4.24^2)Amp^2*4Ohms = 88 Watts \]
Can you please confirm if it is correct or not? I avoided integration this time.

 

i will try to simulate and update result.

 

The result seems to match closely.

 

hi V123,
This is my plot.
E

 

Remember I told you that not tracking your units properly will come back to bite you sooner or later -- and that I said it would likely be sooner? Well, it bit you right here.

ASIDE: Back when I was a junior and taking Intermediate Electricity and Magnetism, the professor (who eventually became the president of the university) walked into class the first day and started off by saying, "There are two things that separate you and me from Nobel Prize winning physicists. The person walking across the stage in Stockholm religiously does two things. They always, always, always track their units, and they always, always, always ask if the answer makes sense." That is quite possibly the most impactful advice that any professor has ever given me and it has paid huge dividends throughout my career. I made a point of telling him so some ten years later and I've had two of my own students come back after several years to say something similar to me. So, why don't I have a Nobel Prize, you might ask. Simple, as diligent as I am, I'm obviously still not religious enough about it.

So let's consider the "does it make sense" part.

What is the power of a 2 A DC current flowing through a 4 Ω resistor? That's an easy one, it's 16 W.

That's before we add in a 6 V amplitude sinewave. The RMS value of a 6 A amplitude sinewave is (6/√2) A and so the power being dissipated by it alone would be 72 W.

So does it make ANY kind of sense that, when combined, the resulting power being dissipated in the resistor is going to go down by more than two orders of magnitude compared to the DC current alone????

While we might not know exactly how the combined components will affect the resultant power, it's reasonable to expect that it will be no lower than the smaller of the two (16 W) and probably no larger than the simple sum of them (88 W). If we don't get an answer that is somewhere within that range, we need to spend some serious effort figuring out whether we made a mistake, or whether we are overlooking something when we came up with our bounds.

Okay, so let's shift focus to units. Normally, I avoid plugging in specific values until the end of a problem and prefer to work symbolically. Symbols carry units just like the actual values do. So R1 has units of resistance and Vo has units of voltage and I·R has units of voltage. It is NOT proper to write the units that you want out to the side like you did. Those are not only useless, they are wrong since the symbols carry their own units.

Your integrand has a current multiplied by a resistance multiplied by time. What units does that work out to?

(A)(Ω)(s)

From Ohm's Law, we have V=IR, so the units are (V)=(A)(Ω), hence resistance has units of (Ω)=(V/A)

So that gives us:

(A)(Ω)(s) = (A)(V/A)(s) = V·s

Is volt-seconds a unit of power?

Nope! So ALL of your work that depends on this equation is guaranteed to be wrong right from the beginning. But since you just ignore the units and just tack on whatever units you want the answer to have and then assume that you didn't make a mistake, you end up with a wrong answer that goes uncaught. You lose points on the problem and your grade suffers. Out in the real world, you calculate a power that is wrong and size your resistor based on that wrong value that is a couple orders of magnitude too small, then the resistor burns up and the flight control system it is in catches fire causing the plane to crash and kill three hundred people. All because you think it is not worth your time track your units properly. I'm being harsh not to be mean, but to try to drive the point home after having failed to do so several times.

Even if you had gotten the integrand correct, your answer still would have been wrong because to find the average, you need to integrate over the period and then divide by the period -- just like you do when you add up a bunch of numbers and you divide by the number of values you summed up. This is a common mistake that is easy to make -- and it messes up the units! But ONLY if the units are there to be messed up. If they are, you will catch the mistake at some point. In fact, I made this exact mistake in the work that follows and only caught it because it became obvious that I was headed for an answer that was going to have units of A²Ωs and not A²Ω. So I took a pause and thought about it for a few seconds, slapped my head, and went back and patched up the equations, which took literally twelve seconds (yes, I timed it because I was curious) despite having to edit all of the LaTex code (which is always royal pain).

So what might this work have looked like, with proper units tracking?

Well, let's do it -- and let's start from first circuit principles.

I'm going to go through it in excruciating detail so that you can walk through it step by step at whatever level of detail you need. Also, if you don't follow something, you can point it out and we can go through how to get from where you did understand to where you didn't by looking at what changed between those lines (which could just result in me making a mea culpa for a typo in my LaTeX code or some other mistake), so I'm keeping the changes small. Normally, you would do this in a small fraction of the number of lines that I'm going to show here.

We want the average power dissipated in the resistor, which since it is a periodic current, is the instantaneous power averaged over one complete cycle.

\(
P_{avg} \; = \; \frac{1}{T_0} \int_0^{T_0} \; p(t) \cdot dt
\)

The power dissipated in a resistor is the product of the voltage across the resistor and the current through it:

\(
p(t) \; = \; v_R(t) \cdot i_R(t)
\)

We are given the current and we can get the voltage from Ohm's Law

\(
v_R(t) \; = \; i_R(t) \cdot R
\)

Bringing this all together, we have:

\(
P_{avg} \; = \; \frac{1}{T_0} \int_0^{T_0} \; p(t) \cdot dt \\
P_{avg} \; = \; \frac{1}{T_0} \int_0^{T_0} \; v_R(t) \cdot i_R(t) \cdot dt \\
P_{avg} \; = \; \frac{1}{T_0} \int_0^{T_0} \; i_R(t) \cdot R \cdot i_R(t) \cdot dt \\
P_{avg} \; = \; \frac{1}{T_0} \int_0^{T_0} \; {i_R(t)}^2 \cdot R \cdot dt
\)

As you become comfortable with the fundamental concepts, you will be able to just start with that last equation.

The next thing is to verify and, if necessary, fix the units on the given information. It's sad that we often have to do this, but that's life.

You are given that the current is

\(
i(t) \; = \; 2 \; + \; 6 sin\left(2 \pi 60t\right) \; A
\)

This has a few problems with it. First, units act like multiplying factors and obey the usual rules of precedence, so this is really:

\(
i(t) \; = \; \left( 2 \right) \; + \; \left( 6 sin\left(2 \pi 60t\right) \; A \right)
\)

This means we are adding a pure number to a current, which can't be done. We can either put parens around the numerical expression so that the units distribute across the terms, or we can put them on each term explicitly. Personal preference.

\(
i(t) \; = \; \left( 2 \; + \; 6 sin\left(2 \pi 60t\right) \; \right) \; A \\
i(t) \; = \; 2\;A + \; 6 sin\left(2 \pi 60t\right) \; A
\)

Next, the arguments to any transcendental function, such as sin(), must be dimensionless. As written, the argument has units of time since 2, π, and 60 are pure numbers and t has units of time. The correct way is:

Frequency f has units of Hz, in which

\(
1\;Hz \; = \; 1\; \frac{cyc}{s}
\)

The 2π seems like a dimensionless number, but it really isn't. It is there because there are 2π radians in one cycle. So its proper units in this context is

\(
2 \pi \; \frac{rad}{cyc}
\)

I'll say a bit more about this at the end, but for now I want to be very explicit to show you how everything works out when you track them correctly.

The end result is that our current should really be expressed as

\(
i(t) \; = \; \left( 2 \; + \; 6 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \; \right) \; A \\
\)

We also need To, which is the amount of time that is equivalent to one cycle.

\(
T_0 \; = \; 1 \; cyc \cdot \frac{1}{60\;Hz} \\
T_0 \; = \; 1 \; cyc \cdot \frac{1}{60\;\frac{cyc}{s}} \\
T_0 \; = \; \frac{1}{60}\; s
\)

We'll leave it as To for now until we need it near the end.

Going back to our prior equation for Pavg, we have:

\(
P_{avg} \; = \; \int_0^{T_0} \; \left( i_R(t) \right)^2 \cdot R \cdot dt \\
P_{avg} \; = \; \int_0^{T_0} \; \left( \left( 2 \; + \; 6 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \; \right) \; A \right)^2 \cdot 4\;\Omega \cdot dt
\)

The next thing we need to do is square the expression for the current (and remember, units act like any other factor -- treat it just like it were a variable as you manipulate things):

\(
{i_R}^2 \; = \; \left( \left( 2 \; + \; 6 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \; \right) \; A \right)^2 \\
{i_R}^2 \; = \; \left( 2 \;A \; + \; 6 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \; A \right)^2 \\
{i_R}^2 \; = \; \left( 2 \;A \right)^2 \;
+ \left(\; 2 \cdot 2\; A \cdot 6 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A \right) \;
+ \; \left(6 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A\right)^2 \\
{i_R}^2 \; = \; 4 \;A^2 \;
+ \;24 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2 \;
+ \; 36 sin^2\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2 \\
\)

Now we can plug this back into the integrand and break up our integral into three integrals:

\(
P_{avg} \; = \; \frac{1}{T_0} \int_0^{T_0} \; \left( i_R(t) \right)^2 \cdot R \cdot dt \\
P_{avg} \; = \; \frac{1}{T_0} \int_0^{T_0} \; \left(
4 \;A^2 \;
+ \;24 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2 \;
+ \; 36 sin^2\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2
\right) \cdot 4\;\Omega \cdot dt \\

P_{avg} \; = \; \frac{1}{T_0} \left( \int_0^{T_0} \; 4 \;A^2 \cdot 4\;\Omega \cdot dt \;
+ \; \int_0^T \; 24 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2 \cdot 4\;\Omega \cdot dt \;
+ \; \int_0^T \; 36 sin^2\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2 \cdot 4\;\Omega \cdot dt \right) \\

P_{avg} \; = \; \frac{1}{T_0} \left( \int_0^{T_0} \; 16 \;A^2\Omega \cdot dt \;
+ \; \int_0^{T_0} \; 96 sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2\Omega \cdot dt \;
+ \; \int_0^{T_0} \; 144 sin^2\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \;A^2\Omega \cdot dt \right) \\
\)

Now let's pull out constants from the integrands (something we could have done, at least partially, earlier). At the same time, let's simplify the units since, using Ohm's Law, it easy to establish that

\(
1 \; A^2\Omega \; = \ 1 \; A^2\frac{V}{A} \; = \; 1\;W
\)

This is also something that you will quickly internalize, namely that volts·amps, volts²/ohms, and amps²·ohms are all equal to watts, because you will know that P = V·I = V²/R = I²·R

\(
P_{avg} \; = \; \frac{1}{T_0} \left( 16 \int_0^{T_0} dt \;
+ \; 96\int_0^{T_0} \; sin\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \cdot dt \;
+ \; 144\int_0^{T_0} \; sin^2\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \cdot dt \right) \;W
\)

Now we can evaluate the integrals. The first is trivial since the second one is identically zero since it is the integral of the sine function over one complete period. This gets us to

\(
P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \;
+ \; 144\int_0^{T_0} \; sin^2\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \cdot dt \right) \;W
\)

For the third, we can use the following trick, exploiting the fact that, by symmetry, the integral of sin²θ over one period is equal to the integral of cos²θ over one period.

\(
\int_0^\pi sin^2\left( \theta \right) d\theta \; = \; \int_0^\pi cos^2\left( \theta \right) d\theta \;
\)

Thus, we can play the following game:

\(
\int_0^{2\pi} sin^2\left( \theta \right) d\theta \; = \; \frac{1}{2} \left( \int_0^{2\pi} sin^2\left( \theta \right) d\theta \; + \; \int_0^{2\pi} sin^2\left( \theta \right) d\theta \; \right) \\
\int_0^{2\pi} sin^2\left( \theta \right) d\theta \; = \; \frac{1}{2} \left( \int_0^{2\pi} sin^2\left( \theta \right) d\theta \; + \; \int_0^{2\pi} cos^2\left( \theta \right) d\theta \; \right) \\
\int_0^{2\pi} sin^2\left( \theta \right) d\theta \; = \; \frac{1}{2} \int_0^{2\pi} \left( sin^2\left( \theta \right) \; + \; cos^2\left( \theta \right) \right) d\theta \\
\int_0^{2\pi} sin^2\left( \theta \right) d\theta \; = \; \frac{1}{2} \int_0^{2\pi} 1 \; d\theta \\
\int_0^{2\pi} sin^2\left( \theta \right) d\theta \; = \; \frac{1}{2} \left. \theta \right|_0^{2\pi} \\
\int_0^{2\pi} sin^2\left( \theta \right) d\theta \; = \; \pi
\)

I derive it each time because I don't have the world's best memory -- that's why I majored in physics and not chemistry. I've also seen way too many people "recall" it as just 1/2 or π/2. Since it takes just a minute to do it on a piece of scratch paper (in a LOT less detail than I showed above), it's not worth the risk (to me).

This now gets us to

\(
P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \; + \; 144 \pi \right) \;W
\)

At this point, I looked at this equation when I previewed the rendering and guess what I noticed? The units don't work!

First and most obvious is that the two terms being added together have inconsistent units with the first having units of time and the second being dimensionless. So right here there is something wrong. Since this sum is divided by time and then multiplied by 1 W and I need the final units to be of power, I need the sum to have units of time. While the first term may or may not be correct, the second term is guaranteed to have an error in it. So that guides my efforts to find and fix the error. I start by going back and finding when the units went from working out to not working out. They worked out fine before I replaced the third integral with what it evaluated to and didn't work out after that, so my error must be in evaluating that integral.

Instead of just fixing it and editing the work, which is what I would do on something for grading or for a customer, I'm leaving this here intentionally to show how I fixed it -- and note that this was the THIRD mistake I made that tracking units caught! In my defense, I'm spending most of my time dealing with how to write the LaTeX code combined with it being hard for me to visualize the units from the code, so I get a lot further than I normally would on paper before catching them because it has to wait until I render, usually several new lines at once. Normally, I'm in the habit of asking if the units still work out AS I'm writing each line, so I usually catch mistakes immediately (if they are mistakes that mess up the units). In fact, remember that I said that initially I made the same mistake of not dividing the integral of instantaneous power by the total period to get the average? Well, on paper I would have caught that immediately because the integral is p(t)·dt, which has units of power·time, which is not the units of power on the left-hand side. Humans are not good at multitasking, regardless of what people might think or claim, so dealing with the LaTeX typesetting was enough distracting to me that I failed to do the line-by-line checks that I normally do, especially at the very start of any work. Normally (like when I was writing conference papers or working on my dissertation) I will split these apart and do the work on paper, where I'm very good and diligent about tracking units at every step without even realizing that I'm doing it. Then, once I'm happy with that, I can deal with the LaTeX code to reproduce the equations that I already have on paper. By combining them when writing a post like this, I can lose that units focus -- which is why I'm not likely to ever be in the running for any of those Nobel Prizes.

So how did I resolve this? Hopefully, you already see the mistake.

First, I looked at my derivation of my trick so make sure that nothing was amiss there. Everything looked fine. The mistake is that I assumed (and it's a common mistake and one I've made more than once) that just the fact that I'm integrating over one period means that I didn't need to worry about the argument to sin²(). In fact, I almost commented on this before writing that last equation. But this isn't the case because I need the differential, dθ, to be the differential for the argument of the sin²() function, and it's not. In essence, I have

\(
sin^2(\theta(t))dt
\)

So we need to do a change of variables.

\(
\theta(t) \; = \; 2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t \\
d\theta(t) \; = \; 2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot dt \\
dt \; = \; \frac{d\theta(t)}{2 \pi \frac{rad}{cyc} \cdot 60 \; Hz}
\)

This gives us

\(
sin^2(\theta(t))dt \; = \; \frac{sin^2(\theta(t))}{2 \pi \frac{rad}{cyc} \cdot 60 \; Hz} d\theta(t) \\
sin^2(\theta(t))dt \; = \; \frac{sin^2(\theta(t))}{120 \pi} \; s \; d\theta(t)
\)

Note that that 's' before the dθ is the unit seconds, not some variable that somehow appeared. It is what survived from combining the units in the denominator. Also note that 'radians' is dimensionless (it's basically units of circumference over radius), so we can be flexible in putting it in and taking it out as needed to remind us when a quantity represents an angle in radians.

Whenever we do a change of variables in an integral, we need to change the limits so that they reflect the change. We could do it explicitly -- and it would be a good exercise for you to do so to confirm that you get the same result -- but we can handwave this one because we are going from t at the end of one cycle (which is To) to θ at the end of one cycle (which is 2π). Note that, again, this is a place that it is very easy for people to make a mistake and forget to change the limits. Had we done so, the units would not have worked out because units of time are not the same as units of radians, so our units tracking would have come to our rescue.

Now we can go back and do things correctly, starting from:

\(
P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \;
+ \; 144\int_0^{T_0} \; sin^2\left(2 \pi \frac{rad}{cyc} \cdot 60 \; Hz \cdot t\right) \cdot dt \right) \;W \\

P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \;
+ \; 144\int_0^{T_0} \; sin^2\left( \theta(t) \right) \cdot dt \right) \;W \\

P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \;
+ \; 144\int_0^{2\pi} \; \frac{sin^2(\theta(t))}{120 \pi} \; s \; d\theta(t) \right) \;W \\

P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \;
+ \; \frac{144 \;s}{120 \pi} \int_0^{2\pi} \; sin^2(\theta(t)) \; d\theta(t) \right) \;W \\

P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \;
+ \; \frac{144 \;s}{120 \pi} \pi \right) \;W \\

P_{avg} \; = \; \frac{1}{T_0} \left( 16 \left. t \right|_0^{T_0} \;
+ \; \frac{6 \;s}{5} \right) \;W
\)

The last thing is to evaluate things at the limits:

\(
T_0 \; = \; \frac{1\;s}{60} \\
P_{avg} \; = \; \frac{60}{1\;s} \left( 16 \left( \frac{1\;s}{60} \; - \; 0 \right) \; + \; \frac{6 \;s}{5} \right) \;W \\
P_{avg} \; = \; \left( 16 \; + \; 72 \right) \;W \\
P_{avg} \; = \; 88\;W
\)

So, your answer was off by a factor of 662 -- nearly three orders of magnitude. That's in the realm of being-fired territory if you don't catch it because your employer absolutely cannot trust the work you produce and you very likely just lost a customer if they find out about it.

ASIDE: Think of it another way. Imagine you take someone dear to you, let's say a ten-year-old daughter, to a doctor with a complaint. The doctor diagnoses her problem correctly, but when calculating the dosage of the medication just multiplies her weight by the dose/kg of body weight value given in their physicians reference. Because they didn't bother to write down that her weight was in pounds instead of kilograms, they prescribe a dosage that is 2.2 times too much (they failed to track their units). Nor did they bother to consider whether it was reasonable for this ten year old girl to weigh 90 kg as opposed to 90 lb (they failed to ask if things make sense). As a result, your daughter dies. To what degree to you think this doctor should be held liable? Should they lose their medical license? Should they face prison? When people are posed this question, a very typical response is that their actions amounted to gross incompetence and/or negligence and that that cannot and should not be tolerated -- after all, if a doctor makes a mistake, they could kill someone. Okay. Fine. Now consider that when an incompetent or negligent doctor makes a mistake, they are usually limited to killing one person at a time. When an incompetent or negligent engineer makes a mistake, they can easily kill people in job lots. Hence my general position that the engineering profession should be held to a higher standard of professionalism and accountability than the medical profession and that failing to properly track units amounts to willful and gross negligence -- and while there is no widespread agreement on this in the legal community, courts and juries have taken that same position from time to time in specific cases.

Getting back to the problem, we are actually seeing a result that, ideally, you should have seen before being assigned this problem -- namely that when a combined DC and AC signal is applied to a resistor, the resulting power is simply the sum of the power that each would deliver separately. But, if you hadn't seen this, or didn't remember it, you can see how you don't really need it because you can always fall back on the fundamentals. In fact, I didn't remember this quite correctly -- I was expecting the powers to add like the square-root of the sum of the squares (it's been a quarter-century since I did this last), but it's the RMS voltage/current values that add this way (i.e., the resultant voltage and current is the Pythagorean sum of the individual voltages and currents).

Finally, I said I would revisit the units on 2πft. What I showed here is the strictly correct way, but ω=2πf is so pervasive that this is the one spot where I don't get tweaked if someone doesn't track the units on that 2π factor in their work -- unless they make a mistake that doing so would have caught (and I have seen it happen, but it's pretty rare). But you should still mentally verify your units in this case, making sure that the arguments to any transcendental functions (trig, exponential, logarithm, etc) truly is dimensionless. I have seen numerous errors that came down to a failure to do this, sometimes resulting in significant costs -- for example, an ASIC that didn't work and, as a result, we had to pay for the new mask set and fab run to the tune of about $20k (this was back when these costs were cheap, today you could be looking at well over a million dollars to redo a single mask).

EDIT NOTE: Fixed typos and expounded a bit on the asides.

 

I am not sure what to say but such a detailed mail covering each and every point in so much detail, thank you very much for the time and effort. It is lessons for me how to avoid mistakes and put in more effort to actually understand the problem not just solving, for that off course I have to be more mature and confident i am not sure if am near to that, but i can try my best. Thank you to all of you once again for supporting and guiding who wants to learn and gain knowledge.

 

View attachment 362557

I have done the calculations as below, but my major question is following the passive sign convention the power of the DC source is negative, so it is delivered, but in reality, i think it is absorbed.

View attachment 362558

View attachment 362559
The average inductor power is 0 as the inductor current is periodic.
View attachment 362560

Note how the starting equation in all three parts of your work aren't dimensionally consistent -- every one of them has units of power·time because of not dividing by the total time. Tracking units would have caught each of them -- and just catching one would have been a red flag that would probably have led you to fix the others.

Also, you assertion that the average inductor power is 0 because the inductor current is periodic is weak and may or may not reflect a correct understanding.

Would you have made a similar assertion about zero average power if the component had been a capacitor instead of an inductor?

Would you also have made this assertion if you had been given a periodic inductor voltage that had both an AC and a DC component?

Think about both of these situations and see if you can give a reasoned and supported answer to each.

 

Few doubts i have as per the text book
View attachment 362724
The current is \[ i(t) = 2 + 6\sin(2\pi60t) Amps \]
Hence can i consider I1 = 2 Amps DC component, I2 = \[ 6\sin(2\pi60t) Amps\] , the RMS value of I1_rms = 2 Amps same as DC value, I2_rms = \[ \frac{6}{\sqrt2} = 4.24Amps \]
Avg Power (Watts) = \[ (I1_{rms}^2 +I2_{rms}^2)*R Watts = (2^2 + 4.24^2)Amp^2*4Ohms = 88 Watts \]
Can you please confirm if it is correct or not? I avoided integration this time.

Hi,

That was a good idea to turn to RMS values.

Using integration, I took the current squared times the resistance to get the instantaneous power:
P(t)=[2+6*sin(2*pi*60*t)]^2 * 4

then plotted it to make sure it was all above zero from 0 to 1/60 seconds (you can also use the 2nd derivative to check that), then did the integration:
Px=integrate(P(t),t,0,1/60)=22/15

then divide by the period 1/60:
Pavg=Px*60
Pavg=88 watts

This problem was easy because they told you what the current was beforehand due to the series current source, and because it is a current source the current must be the same in every element of the series circuit. Usually you have to calculate that first if you need it.

 

View attachment 362737
View attachment 362738
The result seems to match closely.

Hi,

I did not try this yet, but why does your plot look like it ends before 180ms yet the calculation period looks like it is 180ms.
That could be why you got 81w instead of closer to 88w. If you fix that I would bet you get much closer to 88w.

 

hi Al,
The earlier LTS plot with Power label added.
Resistor Watts = 87.96W
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