Circuit problem involving sources and resistors

[Santilopez10]

Homework Statement

The currents $I_{{a}}$ and $I_{{b}}$ of the circuit have values 4A and -2A in that respective order.
A) Find $I_{{g}}$
B) Find the power dissipated by each resistance
C) Find $V_{{g}}$ (voltage drop across the current source)
D) Show that the power delivered by the current source is equal to that dissipated by every other circuit element.
See attachment "primero" for the circuit image.

Homework Equations

Kirchoff’s DC circuits laws only, no resistive simplification neither nodal analysis.

The Attempt at a Solution

By assuming that the negative current means that the current is flowing in the other direction, and some time trying to try me best to figure out how current circulated across all the circuit, I got attachment "segundo".

Now to my attempt, first, I wrote 4 KCEQ and 2 KVEQ: $$N_{{1}}:4A+2A=I_{{c}}, I_{{c}}=6A$$ $$N_{{2}}:I_{{d}}+I_{{c}}=I_{{g}}$$ $$N_{{3}}:I_{{d}}+2A=I_{{e}}$$ $$N_{{4}}:I_{{g}}=I_{{e}}+4A$$ $$V_{{4214}}:-V_{{g}}+30V+100V+60V=0, V_{{g}}=190V$$ $$V_{{4234}}:-190V+I_{{d}}*30\,\Omega+I_{{e}}*4\,\Omega+I_{{e}}*16\,\Omega=0$$

As you can see, $V_{{g}}$ and $I_{{c}}$ were easy to obtain. Then using $N_{{3}}$ I sustituted it in $V_{{4234}}$ to obtain that $I_{{d}}=3A$, then plugged in $N_{{2}}$ to get the value of $I_{{g}}$, 9A.
Until now, A and C are done, then by doing $P=I^2*R$ in each resistance I got 1310W for answer B.
For D, I calculated the current source power generated, which was $-I*V$ that got me 1710W, which is the answer from B plus the power consumed by the voltage source, then all the exercises are correct, BUT: My real question, and problem, is if I correctly indicated all the current circulations in the second attachment, because I want to know if it really matters after all to get the correct answer!

Thanks.

 

[phinds]

Now to my attempt, first, I wrote 4 KCEQ and 2 KVEQ: $$N_{{1}}:4A+2A=I_{{c}}, I_{{c}}=6A$$ $$N_{{2}}:I_{{d}}+I_{{c}}=I_{{g}}$$ $$N_{{3}}:I_{{d}}+2A=I_{{e}}$$ $$N_{{4}}:I_{{g}}=I_{{e}}+4A$$ $$V_{{4214}}:-V_{{g}}+30V+100V+60V=0, V_{{g}}=190V$$ $$V_{{4234}}:-190V+I_{{d}}*30\,\Omega+I_{{e}}*4\,\Omega+I_{{e}}*16\,\Omega=0$$

?

 

[Santilopez10]

?

What is the problem?

 

[phinds]

What is the problem?

You must not be seeing what I'm seeing:

EDIT: this is my bad. My browser has stopped running JavaScript for some reason.

 

[CWatters]

Your equations look ok to me but are hard to follow.

Firstly, it's very easy to make a sign error doing KVL and KVC so I prefer to stick rigidly to the following:

1) Define current into or out of a node as +ve and stick with that when writing KCL.
2) Write both the KCL and KVC equations to sum to zero.

Your first two eqns assume out is +ve and the third that in is +ve. You get the right answer but its a more error prone approach.

Again when writing KVL equations it's good practice to state where you are starting and which way around the loop you are going. Initially I assumed that V₄₂₁₄ meant you were starting at node 4 and going anti clockwise to node 2 etc. In which case the first term should be +Vg. But looking again I think you went clockwise and jump about which again makes it error prone. Despite that I think this equation is correct!

Likewise for V4234. It implies clockwise this time but you have a +(I_d*30) term which means you went anti-clockwise through R8. Again I think this equation is correct.

So I think all of the equations are correct if hard to follow.

 

[CWatters]

Until now, A and C are done, then by doing P=I2∗RP=I2∗RP=I^2*R in each resistance I got 1310W for answer B.

B asks for the "power dissipated by each resistance" not the total.

BUT: My real question, and problem, is if I correctly indicated all the current circulations in the second attachment, because I want to know if it really matters after all to get the correct answer!

Yes and no...

It is important that the arrows and the sign of the answer are consistent with each other (eg I_d points to the right and it's value is +3A).

It is not important which direction it points. (eg An equally valid answer would be I_d points left and the value is -3A.)

The trick to solving these problems is not to try and guess the direction of a current unless its obvious from the info given. For example the direction of I_b is specified in the problem, however it's not obvious at the start which way I_d flows. The correct procedure is just to choose a direction for I_d and mark it on the diagram, then ensure your KCL and KVL equations are consistent with that choice. Later on the polarity of I_d will become clear and it will be consistent with that choice of direction.

As an exercise you could work this problem again. This time mark the arrow next to I_d pointing to the left. Redo all the equations and this time you will find that Id turns out to be -3A instead of +3A.

 

[CWatters]

As an exercise you could work this problem again. This time mark the arrow next to Id pointing to the left. Redo all the equations and this time you will find that Id turns out to be -3A instead of +3A.

PS...

This particular problem doesn't ask you for the value of I_d but if it did then the arrow you mark on the circuit diagram at the start would form part of your answer. The examiner should accept either +3A or -3A depending on which way you drew the arrow.

For this reason if you ever get a negative value for a current (implying it flows the other way to your original prediction) then do not go back and change the direction of the arrow on your drawing. That would make your drawing inconsistent with your working.

 

[Santilopez10]

PS...

This particular problem doesn't ask you for the value of I_d but if it did then the arrow you mark on the circuit diagram at the start would form part of your answer. The examiner should accept either +3A or -3A depending on which way you drew the arrow.

For this reason if you ever get a negative value for a current (implying it flows the other way to your original prediction) then do not go back and change the direction of the arrow on your drawing. That would make your drawing inconsistent with your working.

got it, thanks a lot!