Finding voltages in a linear circuit

[JessicaHelena]

Homework Statement

A linear circuit is shown in the figure.
The elements in this circuit have the following values: R = 100 Ohms, R2 = 200 Ohms, R3 = 350 Ohms, V = 5V, and I = 0.004 A.
1. Determine the potentials of drops v_a and v_b across resistors R1 and R2 respectively.
2. Now let us determine if the answers you came up with satisfy the laws of physics.
(a) What is the power (in Watts) dissipated in resistor R1?
(b) What is the power (in Watts) dissipated in resistor R2?
(c) What is the power (in Watts) dissipated in resistor R3?
(d) What is the power (in Watts) coming out of the voltage source V?
(e) What is the power (in Watts) coming out of the current source I?

Relevant Equations

KVL
KCL
V/I = R

From the circuit I have:

$-v_b + v_a + V = 0$
$v_b - V = v_a$
$i_1 = (v_b - V)/R_1$
$I + i_2 = i_1$
$(v_b - V)/R_1 = I + v_b/R_2$
From this last equation I get $v_b = 10.8$ and hence $v_a = 5.8$.

However, apparently that is wrong. (And hence my answers to #2 were all wrong as well.) Why is that so? What might I be doing wrong

 

[gneill]

Could you mark up your circuit diagram with the currents you're working with, showing their assumed directions? I suspect that the issue you're having may have to do conflicting or inconsistent assumptions about current directions.

 

[JessicaHelena]

@gneill

I also have this equation written down: $i_1 + i = i_3$.

 

[gneill]

Okay! Take a look at how you're calculating $i_2$. Note the assumed polarity of $V_B$ and the direction of $i_2$.

 

[JessicaHelena]

Oh, should it actually be $I + i1 + i2 = 0$ (bc they all come into the node with potential $v_b$)?

 

[JessicaHelena]

But I didn't really use this equation to get the voltages...

 

[gneill]

I believe that you used the equation:

$(v_b - V)/R_1 = I + v_b/R_2$

But $v_b/R_2$, given the defined polarity of $v_b$, describes a current flowing from the node through $R_2$ to ground, opposite of what you defined $i_2$'s direction to be. So, change your equation to:

$(v_b - V)/R_1 = I - v_b/R_2$

and see what you get.

 

[gneill]

Quick followup. If I were to define the currents for the circuit, I'd first take note of the predefined polarities of the voltages across the resistors to make my current direction choices. That way my choices will be consistent with the given definitions that we have to live with:

The direction of $i_3$ is a no-brainer, since it's being driven by the 5 V source.

 

[JessicaHelena]

I did think i3 would be driven by the 5V source, but I was confused bc I thought that i3 would be different from the original current from the 5V source (which I labelled in my diagram as $i$) due to the "addition" of the $i_1$ source at the first junction. Is that way of thinking wrong?

 

[gneill]

I did think i3 would be driven by the 5V source, but I was confused bc I thought that i3 would be different from the original current from the 5V source (which I labelled in my diagram as $i$) due to the "addition" of the $i_1$ source at the first junction. Is that way of thinking wrong?

Nope. $i_3$ is a separate current from your $i$. Labeling and analyzing are two different steps. Labeling sets your assumptions in place so that the analysis can proceed without logical contradictions creeping in.

Your $i_1$ and $i$ should indeed sum to $i_3$. But it is (hopefully) obvious that $i_3$ can be calculated quite simply by dividing the 5V source voltage by $R_3$.

 

[JessicaHelena]

@gneill
Also, with your correction from #7, I was able to get that $v_b = 3.6$ and $v_a = -1.4$.

Then the power dissipated in R1 is Va^2/R1 = 0.0196;
the power dissipated in R2 is Vb^2 / R2 = 0.0648;
the power dissipated in R3 is V^2/R3 = 0.071429;
the power coming out of V is V*i3 = -0.071429;
the power coming out of I is I * v_b = -0.0144

However, when I sum all the powers, they don't sum to 0. What may I be doing wrong?

 

[gneill]

The power coming out of V should include both of the currents through $R_1$ and $R_3$. Note the directions of the currents. Both are flowing out of V, so they should both be making a positive contribution to the power supplied. You may be using a convention where "power supplied" is actually a negative "power consumed", so I won't argue that point. But you still must consider the current directions figuring the total.

 

[JessicaHelena]

$i_1 = (v_b-V)/R_1$

Hence recalculating the power out of V, I have V*(i3+i1) = 5*(5/R3 + -1.4/R1) where R1 = 100 Ohms and R3 = 350 Ohms

Then I have that the power out of V is -0.00142857143.
I'm not sure if it's my calculations that's the problem, but I think it's even more wrong than before.

 

[gneill]

Notice that $V_a$ as defined is actually contrary to the reality of the circuit conditions. Your $i_1$ should be flowing from the 5 V source towards $R_2$. So $i_1$ should be a positive value for power supplied.

Your $v_b$ is now 3.6 V, right? So the current through $R_1$ is flowing to the left since the 5 V source is greater than $v_b$. The power supplied to that subcircuit must be a positive value.

 

[JessicaHelena]

Oh, my mistake... from what you told me, I should have added (not subtracted) 1.4/R1.

That gives me 0.141429 = power out of V

Still, I don't understand why it shouldn't be -1.4 since that's the v2 I found.

However,

 

[JessicaHelena]

Hold on, let me read your post from #14 first!

 

[JessicaHelena]

@gneill

I've read your previous post & now I understand why they should add up—thank you.
In the meantime, I've checked my answers, but $-0.141429$ Watts for power out of V (via $V*(i3+i1)$) and $I \cdot v_b = -0.0144$ Watts for power out of I are apparently wrong.
Do you think it is the signs, my calculations, or my equations, that're wrong?

 

[gneill]

Your value magnitudes look okay to me (I get the same numeric digits). It is perhaps a sign convention for power produced versus absorbed that's tripping you up.

 

[JessicaHelena]

@gneill
oh indeed, yes you are right—it was the sign convention. Then is it usually the case that power produced (which I believe is power coming out) is positive, and the power absorbed (power going in) is negative?

 

[gneill]

There is a standard definition for the power associated with a component, but authors/instructors sometimes choose their own interpretation. In this case the problem asked for the power coming out of the sources. Since both sources are indeed producing power that will be absorbed by the other components, I would interpret this to mean that the values should be positive.

 

[Jody]

I hate bookkeeping too. It looks like you pretty much have your answer, but you might like this approach I posted on another thread (here).

Draw all of the current going the same direction and set it to zero. The negative sign will "correct" your answer in the final solution. Here's an example:

I later presented in the thread solving a simple voltage divider. The point of it was to show that the direction of the arrows didn't matter.

It's not to say that predefined directions are wrong (this is really the same thing), but to suggest make everything the same predefined direction so there's no need to draw diagrams or remember direction... just be consistent.

 

[Babadag]

I=I1+I2

I2*R2-I1*R1=V

I3=V/R3

Ptot=i1^2*r1+i2^2*r2+I3^2*R3

IV+I1-I3=0

PV=V*IV

PI=Ptot-PV

 

[pkholling]

please see attached pdf file with a solution

 

[gneill]

@pkholling : For future reference, providing complete solutions to homework problems is not permitted unless the Original Poster has come to a complete and correct solution through their own efforts first. Only guidance and advice is permitted until then.

Fortunately in this instance the OP had managed to solve the problem.