Engineering Superposition principle applied to a linear circuit

[Gabry89]

Homework Statement

I have to find I₃, I've already easily found the first component I₃^| but i don't understand how i can get I₃^||.
I thought about a Wheatstone bridge but doesn't work looking at the data.

The result is I₃=-7 A

 

[vela]

You drew the two sources the same way, but you're treating them differently. Is one a voltage source and the other a current source?

Your analysis of the second case is completely wrong. R1 and R2 aren't in series, nor are R3 and R4.

 

[Gabry89]

Is one a voltage source and the other a current source?

actually they're not draw in the same way and yes they're a voltage and a current source.
Then i don't understand how to get the second component cause I've tried again the exercise and I've found the correct result thinkin' about R1 and R2 in parallel and the same for R3 and R4.

I've forgotten to say that after having found I₃ i must find also the power generated by E and the same for J

 

[Redbelly98]

Moderator's note: thread moved to "Engineering, Comp Sci, & Technology".
Please use the Engineering, Comp Sci, & Technology forum for electronics problems.

 

[jegues]

actually they're not draw in the same way and yes they're a voltage and a current source.
Then i don't understand how to get the second component cause I've tried again the exercise and I've found the correct result thinkin' about R1 and R2 in parallel and the same for R3 and R4.

I've forgotten to say that after having found I₃ i must find also the power generated by E and the same for J

You found the correct result by thinking about R1//R2 and R3//R4, that's because they are indeed in parallel!

What are you confused about?

 

[vela]

actually they're not draw in the same way and yes they're a voltage and a current source.

The symbols you're using aren't typical for either current or voltage sources. See, for instance, http://en.wikipedia.org/wiki/Voltage_source.

Then i don't understand how to get the second component cause I've tried again the exercise and I've found the correct result thinkin' about R1 and R2 in parallel and the same for R3 and R4.

What you're saying here doesn't match what you posted originally, nor does it contradict what I said earlier, which was, again, based on what you posted originally. It does sound like you did end up figuring the problem out correctly, though.

I've forgotten to say that after having found I₃ i must find also the power generated by E and the same for J

 

[Gabry89]

Your analysis of the second case is completely wrong. R1 and R2 aren't in series, nor are R3 and R4.

i've studied that when I'm trying to get R_eq i don't have to consider the presence of the sources as they would have been switched off. According to this consideration here's why they're in series.

Moderator's note: thread moved to "Engineering, Comp Sci, & Technology".
Please use the Engineering, Comp Sci, & Technology forum for electronics problems.

Thank you very much, here's a better place.

You found the correct result by thinking about R1//R2 and R3//R4, that's because they are indeed in parallel!

What are you confused about?

is that correct now?

The symbols you're using aren't typical for either current or voltage sources. See, for instance, http://en.wikipedia.org/wiki/Voltage_source.

well, i know that these symbols are the most common in this case but other equivalent symbols are those i drew. E.g. here's a pic I've found in wikipedia http://commons.wikimedia.org/wiki/F...c_symbols_of_current_source.png?uselang=en-gb

 

[jegues]

i've studied that when I'm trying to get R_eq i don't have to consider the presence of the sources as they would have been switched off. According to this consideration here's why they're in series.



Thank you very much, here's a better place.



is that correct now?


well, i know that these symbols are the most common in this kind case but other equivalent symbols are those i drew. E.g. here's a pic I've found in wikipedia http://commons.wikimedia.org/wiki/F...c_symbols_of_current_source.png?uselang=en-gb

-5A is what I get for the second case as well.

 

[Gabry89]

-5A is what I get for the second case as well.

Then it should be right.

Now i have to find the power of the current source and the same for the voltage source.

I've found that:

I₁ = I₁^| + I₁^|| = 2 A
I₂ = I₂^| + I₂^|| = 14 A
I₃ = I₃^| + I₃^|| = 9 A
I₄ = I₄^| + I₄^|| = 7 A

P_e+P_J= I₁²R₁ + I₂²R₂ + I₃²R₃ + I₄²R₄ + E²/R₅

But how i get them separately?

 

[jegues]

Then it should be right.

Now i have to find the power of the current source and the same for the voltage source.

I've found that:

I₁ = I₁^| + I₁^|| = 2 A
I₂ = I₂^| + I₂^|| = 14 A
I₃ = I₃^| + I₃^|| = 9 A
I₄ = I₄^| + I₄^|| = 7 A

P_e+P_J= I₁²R₁ + I₂²R₂ + I₃²R₃ + I₄²R₄ + E²/R₅

But how i get them separately?

If you want to find the power the current source is supplying, simply take the voltage across the current sources times the current the source is supplying the circuit with.

Same with the voltage source, figure out what current the source is supplying the circuit with and multiply it by the voltage.

 

[vela]

i've studied that when I'm trying to get R_eq i don't have to consider the presence of the sources as they would have been switched off. According to this consideration here's why they're in series.

Regardless of whether the source is present or not, R1 and R2 are not in series; same with R3 and R4.

 

[Gabry89]

Regardless of whether the source is present or not, R1 and R2 are not in series; same with R3 and R4.

then how would you get the R_eq? and why if i consider them in series for calculating R_eq the result is correct?

Talking about P_Jand P_E i tried doing like this:

P_J = J*Vab = 819.2 W

P_E= E*I_E = E² / R_eq = 320

which are not the correct results. The correct results are: P_J= 1392 W, P_E = 288 W

 

[vela]

then how would you get the R_eq? and why if i consider them in series for calculating R_eq the result is correct?

In your calculations, you multiply by R_eq to find V_AB and then later divide by it to find I₃, so it ends up canceling out.

You used

V_{AB} = R_{\mathrm{eq}} J

followed by

V_{R_3R_4} = V_{AB} \left(\frac{R_3 || R_4}{R_{\mathrm{eq}}}\right) = R_{\mathrm{eq}} J \left(\frac{R_3 || R_4}{R_{\mathrm{eq}}}\right) = J(R_3 || R_4)

You could have used any value of R_eq and gotten the correct answer. Note that your final answer turns out to be

I_3 = \frac{R_4}{R_3+R_4} J

Do you recognize that formula?

For two elements to be in series, all the current that goes through one element must go through the other. That's certainly not true for the pair R₁ and R₂ nor for the pair R₃, which carries 7 A, and R₄, which, by KCL, must carry 9 A.

In series, elements are connected end-to-end with nothing else connected the nodes. With R₁ and R₂, they're connected both on the left and right ends. Since both nodes have other circuit elements connected to them, R₁ and R₂ are not in series.

On another note, you have R₅ in parallel with a short circuit when you remove the voltage source. What's the equivalent resistance of that combination?

 

[Gabry89]

I_3 = \frac{R_4}{R_3+R_4} J

Do you recognize that formula?

Sure, it's the current divider, i thought about that but i was confused by the short circuit in the middle, well thanks now it's clear.

On another note, you have R₅ in parallel with a short circuit when you remove the voltage source. What's the equivalent resistance of that combination?

Well, when a resistor is in parallel with a short circuit the voltage across the resistor is 0 then the equivalent resistance of that combination is 0.

 

[vela]

Well, when a resistor is in parallel with a short circuit the voltage across the resistor is 0 then the equivalent resistance of that combination is 0.

Right. I wasn't sure if you realized that because in your first attempt, you kept R₅ and got rid of the short, and in your subsequent attempt, you still included R₅ even though you could remove it.

Also, since you asked earlier about how to find R_eq when E=0, I thought I'd mention that in your second attempt, you reduced the circuit to the point where you should be able to see what the equivalent resistance is between A and B.

 

[Gabry89]

too bad i don't find the correct answer, i tried it again with my colleagues but I've found

P_J = 1360 W rather than P_J = 1392 W which is the correct answer.

 

[vela]

Then it should be right.

Now i have to find the power of the current source and the same for the voltage source.

I've found that:

I₁ = I₁^| + I₁^|| = 2 A
I₂ = I₂^| + I₂^|| = 14 A
I₃ = I₃^| + I₃^|| = 9 A
I₄ = I₄^| + I₄^|| = 7 A

P_e+P_J= I₁²R₁ + I₂²R₂ + I₃²R₃ + I₄²R₄ + E²/R₅

But how i get them separately?

Didn't you say I₃=-7 A was the correct answer earlier? Why are you using 9 A now?

too bad i don't find the correct answer, i tried it again with my colleagues but I've found

P_J = 1360 W rather than P_J = 1392 W which is the correct answer.

It's not really informative if you just say you got the wrong answer. How did you calculate the power? In particular, how did you find the voltage across the current source?

Note that you can't use superposition when finding the power since power isn't linear. You need the voltage and current through the sources in the original circuit.

 

[Gabry89]

Didn't you say I₃=-7 A was the correct answer earlier? Why are you using 9 A now?

Sorry i wrote it wrong in the post, i wrote I₃ instead of I₄ and viceversa

The correct currents are:

I1 = I1^| + I1^|| = 2 A
I2 = I2^| + I2^|| = 14 A
I3 = I3^| + I3^|| = 7 A
I4 = I4^| + I4^|| = 9 A

It's not really informative if you just say you got the wrong answer. How did you calculate the power? In particular, how did you find the voltage across the current source?

yes, sorry, here's what i mean

R_eq = (R1 || R2) + (R3 || R4) = 5,3125 \Omega

P_J = J² * R_eq = 1360 W

 

[vela]

To calculate the power correctly, you need to find the currents and voltages in the original circuit. You can't calculate the powers individually with the other source set to 0, which is what you are effectively doing by using R_eq.

Why? Consider the current source, for example. The voltage across it will change depending on whether you have E=0 or E=32 V, but in both cases, it will source 16 A of current. Since P=IV, the power provided will be different between the two cases.

 

[Gabry89]

done! I've found out how to manage that problem:

(KCL) I_E^| = I_R2^| + I_R4^| + I_R5^| =10 A

(KCL) I_E^|| = I_R1^| - I_R3^|| = 1 A

P_J = J²*R_eq + (I₃^|*R₃ - I₁^|*R₁) * J = 1392 W

P_E = E*I_E^| - I_E^||*E