Show that power supplied = power dissipated

[courteous]

Show that "power supplied" = "power dissipated"

Homework Statement

For the circuit of Fig. 3-42, show that the power supplied by the sources is equal to the power dissipated in the resistors.

Ans. P_T=940.3\text{ }W

[PLAIN]http://img269.imageshack.us/img269/2010/dsc01005kn.jpg

Homework Equations

P=VI

The Attempt at a Solution

The 3\text{ }\Omega resistor gets current I_{3\Omega}=\frac{500}{500+3}\times 10\text{ }A=9.94\text{ }A

Correct so far? From this I get V_R=I_{3\Omega}\times R=29.82V and also the power dissipated P_{3\Omega}=\frac{V_R^2}{R}=296.4\test{ }W

So far so good?

Now I can also get the value for the dependent source:
3V_R=3\times 29.82\text{ }V=89.46V ... to which 1.75\text{ }V is added for total of V_T=91.21\text{ }V.

The power dissipated on the large resistor is P_{500\Omega}=\frac{V_T^2}{R}=16.64W.

Since the "power dissipated" by the two resistors don't add up to P_T (and sources have their polarities aligned), it is obvious I've done some mistake up to this point. What am I doing wrong?

 

[berkeman]

Homework Statement

For the circuit of Fig. 3-42, show that the power supplied by the sources is equal to the power dissipated in the resistors.

Ans. P_T=940.3\text{ }W

[PLAIN]http://img269.imageshack.us/img269/2010/dsc01005kn.jpg

Homework Equations

P=VI

The Attempt at a Solution

The 3\text{ }\Omega resistor gets current I_{3\Omega}=\frac{500}{500+3}\times 10\text{ }A=9.94\text{ }A

Correct so far? From this I get V_R=I_{3\Omega}\times R=29.82V and also the power dissipated P_{3\Omega}=\frac{V_R^2}{R}=296.4\test{ }W

So far so good?

Now I can also get the value for the dependent source:
3V_R=3\times 29.82\text{ }V=89.46V ... to which 1.75\text{ }V is added for total of V_T=91.21\text{ }V.

The power dissipated on the large resistor is P_{500\Omega}=\frac{V_T^2}{R}=16.64W.

Since the "power dissipated" by the two resistors don't add up to P_T (and sources have their polarities aligned), it is obvious I've done some mistake up to this point. What am I doing wrong?

I don't understand how you came up with your first equation for the current through the 3 Ohm resistor.

I'd write the KCL equations instead, to get equations for the currents and voltages involved. How many KCL equations do you think you should write? I'd probably put the ground node in the upper righthand corner...

 

[courteous]

I don't understand how you came up with your first equation for the current through the 3 Ohm resistor.

The factor of 500/503 is due to the current divider ... guess I didn't do this step correctly? Or is the current source ideal (so that it represents zero resistance; same for 2 voltage sources) and so no current flows through 500\text{ }\Omega resistor (and the whole of 10 A then flow through the 3\text{ }\Omega resistor)?

I'd write the KCL equations instead, to get equations for the currents and voltages involved. How many KCL equations do you think you should write?

As there are only 2 "principal nodes", I'd say 2 KCL equations? EDIT: Just 1 equation? So, is there a formula that says: \text{\# principal nodes - 1}=\text{\# KCL equations} ?

I'd probably put the ground node in the upper righthand corner...

This is just an intuitive thing? Would the placement "show" in the KCL equations?




EDIT: Another try. First, I wrote the KCL for the bottom "principal node": I=I_{10A}+I_{500\Omega}=10A+\frac{(1.75+3V_R)V}{500\text{ }\Omega}

The same I also comes "out of" the upper "principal node" and into the 3\text{ }\Omega: I=\frac{V_R}{3\text{ }\Omega}
Combining both equations gets us V_R=30.56\text{ }V


Now, all is set for finding out P_{supplied} and P_{dissipated}:

\begin{align*}<br /> P_{supplied} &amp; = &amp; P_{3V_R}+P_{1.75\text{ }V}+P_{10A} \\<br /> &amp; = &amp; -(3V_R\times I)-(1.75V\times I) - (V_{10A}\times I_{10A})<br /> \end{align*}


Now, what is the voltage on the current source? Is V_{10A}\stackrel{?}{=}3V_R+1.75\text{ }V?


BTW, with only P_{3V_R} and P_{1.75\text{ }V} I get very close to the given solution: P_{3V_R} + P_{1.75\text{ }V} = -934.3\text{ }W

 

[KingNothing]

Your assumption that you can use a "current divider" is incorrect. Those two resistors are not in parallel. They are ONLY in parallel when they share exactly the same two nodes.

 

[courteous]

You're correct. In my second try (last post) I didn't use this incorrect assumption and have relied only on KCL ... but, I still don't get the "correct" solution? Can you help me out, please?

 

[berkeman]

What you wrote for a KCL equation doesn't look right to me. I'm also used to writing KCL equations of the form:

Sum of Currents Leaving a Node = 0

So I would write 1 KCL equation at the node at the top of the 500 Ohm resistor (I think that's all you need). Could you try putting it in that form to see how it comes out?

 

[courteous]

The KCL for the upper principal node:
I_{10A}-I_{3\Omega}-I_{500\Omega}=0


I_{3\Omega}=\frac{V_R}{3\text{ }\Omega}}

I_{500\Omega}=\frac{V_{500\Omega}}{500\text{ }\Omega}}

Now, here is (my) difficulty: what is V_{500\Omega} ?

I'd say there is no voltage drop over the 500\text{ }\Omega resistor, as there is an ideal current source with 0\text{ }\Omega resistance in parallel to it. Is my reasoning sound?

(1) If it is, then I_{10A}=I_{3\Omega} \Rightarrow V_R=30\text{ }V. From this, we get the power supplied by the two voltage sources by the P=V\times I equation.

By the point (1), the power dissipated by the 500\text{ }\Omega is 0\text{ }W. But what is the power supplied by the ideal current source? Its resistance is 0 (R_{10A}=0) and so is the voltage drop over it (V_{10A}=0), correct? But it seems illogical to say that it doesn't give power! ​

(2) If it is not, that is, there is a current through the 500\text{ }\Omega ... well, then what is it? Or, by the same token, V_R=?


If I still didn't come even close to the solution, would you please put me out of misery, and show me the solution?

 

[berkeman]

V500 is just the difference between that node's voltage and the sum of the independent and dependent voltage sources...

 

[courteous]

But there isn't any current through the 500\text{ }\Omega resistor, is there?

 

[berkeman]

But there isn't any current through the 500\text{ }\Omega resistor, is there?

Sure there is. Maybe that's where your confusion is coming from.

The ideal current source is an open circuit, so the voltage across it is defined by other circuit elements in parallel with it. There is a voltage difference across the 500 Ohm resistor, so there will be a corresponding current through it.

 

[courteous]

If I should have any hope of ever solving this, I should go step-by-step. (Please point out at which step the first mistake occurs.)

1) OK, so the current through the 500\text{ }\Omega resistor is:

I_{500\Omega}=\frac{V_{500\Omega}}{500\text{ }\Omega}=\frac{(3V_R+1.75)-V_R}{500\text{ }\Omega}

I_{3\Omega}=\frac{V_R}{3\text{ }\Omega}

I_{10A}=10A



2) \begin{align*}&amp;I_{10A}-I_{3\Omega}-I_{500\Omega}=0 \\<br /> &amp;\Rightarrow V_R=29.634\text{ }V \\<br /> &amp;\Rightarrow I_{500\Omega}=0.122 A \text{ and } I_{3\Omega}=9.878 A<br /> \end{align*}



3) P_{supplied}=P_{3V_R}+P_{1.75\Omega}+P_{10A}

\begin{align*}P_{3V_R}&amp;=3V_R\times I_{3\Omega} \\<br /> P_{1.75\Omega}&amp;=1.75V\times I_{3\Omega} \\<br /> P_{10A}&amp;=\text{ }?<br /> \end{align*}

What is the current source's power? P_{10A}\stackrel{?}{=}V_{500\Omega}\times 10A

 

[courteous]

Please help me out. Are the steps {1,2,3}) in previous post OK?

Here is the picture, so that we are on the same "page".
[PLAIN]http://img171.imageshack.us/img171/2010/dsc01005kn.jpg

 

[berkeman]

The steps look okay. Do you get the power to balance? The power supplied by the current source should be its current multiplied by the voltage across it, I would think.

 

[courteous]

OK, so the voltage across the current source is the same as that over the 500\text{ }\Omega resistor, right?

Then, P_{10A}=V_{500\Omega}\times 10A=\left(3V_R+1.75-V_R\right)\times 10A=57.518\text{ }V\times 10\text{ }A=575.18\text{ }W, correct?

As I've already (hopefully correctly) written:

\begin{align*}P_{supplied}&amp;=P_{3V_R}+P_{1.75\Omega}+P_{10A} \\<br /> &amp;= (3V_R\times I_{3\Omega}) + (1.75V\times I_{3\Omega}) + (V_{500\Omega}\times 10A) \\<br /> &amp;= 1470.64\text{ }W \neq 940.3\text{ }W<br /> \end{align*}

Where did I make a mistake?

 

[berkeman]

OK, so the voltage across the current source is the same as that over the 500\text{ }\Omega resistor, right?

Then, P_{10A}=V_{500\Omega}\times 10A=\left(3V_R+1.75-V_R\right)\times 10A=57.518\text{ }V\times 10\text{ }A=575.18\text{ }W, correct?

As I've already (hopefully correctly) written:

\begin{align*}P_{supplied}&amp;=P_{3V_R}+P_{1.75\Omega}+P_{10A} \\<br /> &amp;= (3V_R\times I_{3\Omega}) + (1.75V\times I_{3\Omega}) + (V_{500\Omega}\times 10A) \\<br /> &amp;= 1470.64\text{ }W \neq 940.3\text{ }W<br /> \end{align*}

Where did I make a mistake?

I think I now see a sign error in post #11 for the current through the 500 Ohm resistor. You wrote the KCL with the current flowing away from the node, but solved for the current flowing into the node. I'm not sure if that is enough to cause the power mismatch you see.

With the voltage at the bottom of the 500 Ohm resistor being larger than at the top, the current will flow up through the resistor and combine with the 10A current to give more than 10A through the 3 Ohm resistor, not less.

 

[ideasrule]

I have a few suggestions. First, before writing out any of Kirchoff's laws, label all currents and give all of them a direction. If you get the direction wrong, hell won't break loose; you'll just get a negative answer for that current, which is perfectly fine.

That should take care of the sign issues with Kirchoff's laws. The other issue is:

What is the current source's power? P_{10A}\stackrel{?}{=}V_{500\Omega}\times 10A

The current source's power is actually negative. The voltage sources contribute positive power because they take electrons at low potential and raise them to a higher potential, which injects energy into the circuit. On the other hand, the current source takes electrons at high potential and drops them to a lower potential, which uses power for the same reason that a voltage drop across a resistor means the resistor uses power.

You should get consistent values after addressing these two issues.

 

[berkeman]

Thanks ideasrule!

 

[courteous]

First, before writing out any of Kirchoff's laws, label all currents and give all of them a direction. If you get the direction wrong, hell won't break loose; you'll just get a negative answer for that current, which is perfectly fine.

Yes, the I_{500\Omega} is entering the resistor from the top ... so the V_{500\Omega} (from the post #11) has a minus before it:
I_{500\Omega}=\frac{V_{500\Omega}}{500\text{ }\Omega}=\frac{-\left[(3V_R+1.75)-V_R\right]}{500\text{ }\Omega}

From this follows that V_R=30.375\text{ }V and I_{3\Omega}=10.125\text{ }A.

The current source's power is actually negative. The voltage sources contribute positive power because they take electrons at low potential and raise them to a higher potential, which injects energy into the circuit. On the other hand, the current source takes electrons at high potential and drops them to a lower potential, which uses power for the same reason that a voltage drop across a resistor means the resistor uses power.

I've re-calculated the P_{3V_R}+P_{1.75\Omega} which now gives 940.3\text{ }W=P_T, the textbook-given solution.

But ... that means that (by the equation P_{supplied}&amp;=P_{3V_R}+P_{1.75V}+P_{10A}) the current source's power P_{10A}=0. Or is it that the current source doesn't "belong" to the P_{supplied} equation (but rather to the P_{dissipated})?

In short, I got closer to the solution, but still having conceptual difficulties regarding the current source's role in power. Can you please explain further?BTW, I found this http://circuits.solved-problems.com/309/problem-1-9/" : if (as you said) the current source takes energy, is such a circuit only a theoretical model?

 

[vamshi_ece]

convert that current source(10A) and shunt resistance(500\Omega) pair to voltage source(5000V) with a series resistance(500\Omega) using source transformations.then problem will be simple.
by applying KVL in circuit we find
current flowing in circuit =9.769A(from + to - of V_R)
then,
power supplied=5000*9.769 + 1.75 * 9.769 =48862.3
power dissipated=503 *9.769² +[3* (3*9.769)]*9.769=48861.7

 

[courteous]

convert that current source(10A) and shunt resistance(500\Omega) pair to voltage source(5000V) with a series resistance(500\Omega) using source transformations.then problem will be simple.
by applying KVL in circuit we find
current flowing in circuit = 9.769A (from + to - of V_R)
then [...]

Then the circuit would look like this, right:
[PLAIN]http://img852.imageshack.us/img852/7263/dsc01026b.jpg
And KVL would be like this: 5000V-(500+3)\Omega\times I+3V_R+1.75V=0
Now, how did you find the I_{3\Omega}? There are 2 unknowns in the KVL equation, you can't just take V_R from before "source transformation", can you?

Can you also help us with the problem without the "source transformation"?