Homework question: Find the total energy developed in the circuit.

The first thing I need help understanding is: is this question supposed to be asking for power instead of "energy." How can you calculate the energy developed without some type of time constraint?

Ok, so assuming we are talking about power developed, here is what I know:

First, I know that i = -8 A so the dependent current source is always equal to -24 A.

Second, I must calculate the power absorbed by each element by using the formula I*V. (Of course we will use the formula -I*V if the current is moving in the direction of the voltage rise). So the power absorbed by the 20V voltage source and the 100V voltage source is -160 Watts and -1600 Watts. And since these numbers are negative that tells us that power is being delivered.

So I figured out that these two elements are developing 1760 W of power, but I can't find the power created or absorbed by the current sources because I can't create 3 independent equations to find the three remaining voltages across the terminals of the current sources. (KVL only yield two such equations)

How do I find the power absorbed or delivered by these 3 remaining circuit elements?

P.S. the answer key says the answer is 220 W so I know my answer of 1760 W is already way off!

 

One thing you can say with certainty is that with the answer key stating the answer in terms of Watts then your initial question is answered outright.

hgmjr

 

One thing you can say with certainty is that with the answer key stating the answer in terms of Watts then you initial question is answered outright.

hgmjr

haha... good point.

 

Since power is additive and always a positive quantity, then I am at a loss as to how the answer can be 220 Watts since you already have 1760 Watts. This is well in excess of the value given in the answer key.

Either I have overlooked something or the answer in the answer key is for a different problem or the circuit as drawn has an error.

hgmjr

 

NichA,

I would say that no power is dissapated in the circuit because no resistance exists. Please post the solution when you receive it. Ratch

 

I don't suppose the thing that says 3i is in fact a 660V source?
Try that.

 

I would assume there are an infinite range of possible circuit conditions but the nett energy loss is always zero.

Assume any voltage from +∞ to -∞ across the 3i controlled source and you have always have power balance as the voltage magnitude increases with either polarity.

Take the case of +20 volts across the controlled source.

The 8A source would have zero voltage drop and the 16A source would have -80V drop.

Source Powers in that case

20V source = +160W [supplying power]
100V source = +1600W [supplying power]
8A source = 0 W
16A source = -1280W [absorbing power]
Controlled source = -480W [absorbing power]

So the total power input balances the total absorbed power.

Take the case of -50V across the controlled source.

The 8A source would have -70V drop
The 16A source would have -150V drop

Source powers in that case

20V source = +160W as always
100V source = +1600W as always
8A source = -560W
16A source = -2400W
controlled source = +1200W

Once again the supplied & absorbed powers balance.

Seems a rather pointless question.

 

Perhaps a final comment ....

Imagine a very high resistance value shunting the controlled source. Let's say it's 100MΩ.

No current would flow in this resistor since the controlled source just balances the total independent current source output. By the same reasoning the shunting resistor would add no power loss to the system.

Given there is no current then there would be zero voltage drop across the resistor. So the controlled source potential difference would be zero. The controlled source would neither absorb or generate any power.

In that case the two independent current sources would absorb the 1760W derived from the two independent voltage sources. No other losses would need explanation since none would exist.

One can probably equally well reason that the shunting resistor could be any value (including 0Ω ...?) and there would be no power loss due to the presence of the resistor, irrespective of its ohmic value.

The proposed answer alluding to the mysterious & unaccounted loss of 220W continues to puzzle me. One could, for instance, place a resistance of 0.382 ohms in series with the controlled source and thereby produce the required 220W.

I wonder if the OP has inadvertently omitted any circuit elements in transcribing the schematic...?

 

AFAIK an ideal current source has zero volts across it. Thus the left leg has 20V across it, and the middle leg has 100V across it.

But as these two legs are in parallel, there cannot be two different voltages.

Thus the question is in error in at least 2 places.

Addendum: Power in watts is equivalent to joules per second. Joules is energy. One *could* ask how much energy does it consume/emit in some given time. But without specifying how much time then the energy is indeterminate.

 

AFAIK an ideal current source has zero volts across it.

That's incorrect.

Imagine an ideal current source of 1A driving a 1Ω load. The source has 1V across it. An ideal current source adjusts its terminal voltage to keep the current in the load constant.

 

t_n_k, by assuming a fixed voltage across the current source you're simply calculating the situation where the controlled source is actually a voltage source. It's a different problem.

The power delivered and absorbed will always balance because that's a law of physics. Similarly I could put a resistor anywhere or add another independent source anywhere and have the same thing happen.

The question is very simple because i is entirely known right on the schematic. You don't even need an equation. i is the current in the branch containing the 8A source but opposite polarity, therefore it's -8A.

Then the 3i source becomes -24A.
Notice that 8A + 16A = 24A which is consistent. You don't even need to calculate 3i, you simply add 8 and 16 as per KCL.
Notice that in both of your calculations the current in the 'controlled source' is 24A. It's forced by the current sources.

But yes you're right, if you add a shunt element to the controlled source it will not contribute to power in either direction. This is because the controlled source current exactly balances the two independent sources by definition. That is, the situation by KCL is i + 2i - 3i + x = 0, where x is the shunt element current. x is always 0.

 

Hi Ghar,

I agree with your thoughts.

In fact one can argue that you could dispense with the controlled source altogether and replace it with a short circuit. Then there is no ambiguity in the problem.

The real issue with this problem is that it appears to have no unique solution, with respect to the voltage drop across the controlled source. If anyone can write the equation(s) which include this voltage as the unknown variable for which a unique solution may be found, it would be a useful addition to the discussion. My impression is that no such means exist.

The other issue with this problem is that we don't know the "history" of the circuit. We are simply given an incomplete picture of its current state.

I suspect one could mount an argument for saying that the circuit (albeit a closed system) will always tend to a minimum energy condition - though I'm not sure of the reasoning if any which would support that view.

Interestingly if one applied a constant DC source across the controlled source there would be no energy drawn from the imposing DC source - it would simply set the voltage across the controlled source to a fixed value. Another thought would be to place a charged capacitor (say 1 picofarad @ 100V) across the controlled source at "time zero" or just some arbitrary time. Since this capacitor could neither discharge or further charge (there being no excess current available) it would impose a constant 100V across the controlled source without any energy exchange. As ludicrous & as pointless an exercise as that might seem.

I'd appreciate the OP coming back with the worked solution from whomever posed the problem. Otherwise this one should probably be allowed to sink to the bottom.

 

That's incorrect.

Imagine an ideal current source of 1A driving a 1Ω load. The source has 1V across it. An ideal current source adjusts its terminal voltage to keep the current in the load constant.

That's quite correct. Oops, must have been one of THOSE days.

 

That's quite correct. Oops, must have been one of THOSE days.

I know what you mean - the older I get the more frequent they become!

 

If you solve a power balance equation you can get the voltage across the controlled source.
i.e.
20V*8A + 100V*16A +(x-20)*8A + (x-100)*16A + x(24) = 0
1760W - 1760W + 48x = 0
x = 0V

That actually makes sense, the controlled source is a short. That's consistent with the observation about additional shunt elements.

 

Thanks Ghar,

I have a different solution to the power balance problem ....

 

Hm you're right, I had a sign mistake.
The equation doesn't help...

I guess it really is ambiguous.
Which again makes sense (except maybe correctly this time?)... each branch contains a current source which can have any voltage across it whatsoever and not upset anything.

 

Yeah - it's a "mongrel" problem. Let sleeping dogs lie.

At least (given the views stat) the thread has generated some interest.

 

Which again makes sense (except maybe correctly this time?)... each branch contains a current source which can have any voltage across it whatsoever and not upset anything.

I don't think it's true that each current source can have any voltage across it and not upset anything.

As the OP mentioned, it's easy to see two equations; just write two KVL equations for the two obvious loops.

Denoting the voltage across the 8A source as V1, the voltage across the 16A source as V2 and the voltage across the 3I source as V3, we have for the two equations:

Eq1: 100 - V1 - 20 + V2 = 0
Eq2: -V2 + V3 -100 = 0

For a third equation, we can use power balance:

Eq3: 20*8 + 1000*16 + 8*V1 + 16*V2 - 24*V3 = 0

We can create a matrix description of this system with Eq 3 first:

\(\left[ \begin{array}{3}\8&16&-24\\1&-1&0\\0&-1&1\end{array}\right]*\left[ \begin{array}{4}V1\\V2\\V3\end{array}\right]=\left[ \begin{array}{4}-1760\\ 80\\ 100\end{array}\right]\)

As it happens, the coefficient matrix is singular which means that there is either no solution, or an infinite number of solutions.

One solution is:

\(\left[ \begin{array}{4}V1\\V2\\V3\end{array}\right]=\left[ \begin{array}{4}-20\\ -100\\ 0\end{array}\right]\)

This is the case where V3 is zero.

As it happens, the null space of the coefficient matrix is not empty; a vector that spans the null space is:

\(\left[ \begin{array}{4}1\\ 1\\ 1\end{array}\right]\)

Any multiple of the null vector can be added to the first solution, and that result is again a solution. Since any multiple of the null vector can be added, we have an infinite number of solutions. For example, add 100 times the null vector:

\(\left[ \begin{array}{4}V1\\V2\\V3\end{array}\right]=\left[ \begin{array}{4}-20\\ -100\\ 0\end{array}\right]+\left[ \begin{array}{4}100\\ 100\\ 100\end{array}\right]=\left[ \begin{array}{4}80\\ 0\\ 100\end{array}\right]\)

So, the voltages across the 3 current sources can't have just any values. They must be equal to the first solution vector above plus any multiple of the null vector.

The upshot of all this is that the solution is indeterminate.

 

Note: I've been trying to post a new thread for the past week however keep getting a server error (so I have posted my question here instead), Could the mod's kindly move this to a new thread?

Hi there, I have a question with circuit attached as Problem1.jpg.

View attachment 33839

The question requires to find the resistance of Zo for which maximum power transfer at this load resistor occurs.

I have attached my working as working.zip (containing 4 jpg files). Just wondering if my working is correct because I have one unused equation that obtains a different result.

The solution to this problem was not given, so I have no way of checking.

Thank you for your help in advance.