# circuit with 2 voltage sources

Discussion in 'Homework Help' started by abyss, Jun 15, 2016.

1. ### abyss Thread Starter New Member

Jun 15, 2016
10
0
good day all
i am trying to learn about dc and following two different methods i found online to similar problems i still do not come up with the correct answer. i actually get two different answers. apparently the correct answer is a. how do i get there?

thanks

2. ### DGElder Member

Apr 3, 2016
348
87
One way is to use the principle of superposition. Or you could construct a Norton and then Thevenin equivalent to the right side source.

Last edited: Jun 15, 2016
3. ### abyss Thread Starter New Member

Jun 15, 2016
10
0
o.k thank you. i'll try those.

4. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
If you post your work for the two methods you've tried, we can look it over and point out where you are going wrong.

I would recommend using all three of the main techniques -- mesh current, node voltage, and superposition -- until all three give the same answer.

Also keep in mind that once you get an answer, regardless of how you got it, you should be able to confirm whether it is correct by simply assuming that it IS correct and then walking though the circuit checking if it is consistent with the problem.

For instance, you say that the correct answer is apparently (a), 2.3 A. So let's check it.

If the 10 V source has a current of 2.3 A in it (and I have to assume that this current is leaving the positive terminal -- but if that doesn't work then I would need to make the opposite assumption and check that because I can't just discard that possibility out of hand) then the 5 Ω resistor connected to it would drop 11.5 V across it. Let's declare the bottom-left node to be our 0 V reference, that means that the top-center node would be at -1.5 V. The current in the center 5 Ω resistor would then be 0.3 A (flowing upward) which would combine with the 2.3 A to give 2.6 A flowing to the right and then down into the 15 V supply and on into the bottom 5 Ω resistor. That resistor would drop 13 V making the voltage of the bottom-right node 13 V. The means that the voltage of the supply would need to be (13 V - -1.5 V) = 14.5 V, which is close to, but not exactly equal to, the 15 V that it is supposed to be. Thus the answer of 2.3 A is close to the correct value, but not exactly correct. In fact, the exact answer is 7/3 A and of the available choices, that is by far the closest.

5. ### abyss Thread Starter New Member

Jun 15, 2016
10
0
thank you WBahn
i'm still trying to get an answer using the mesh method i've seen in similar problems. i had used the superposition method earlier and came up with 2.6A at which point i think i got lost in the steps. this question is from a practice problem i found as i am looking to be able to understand dc. than move on to ac. it sounds like you saying that none of the answers (a,b,c,d) are correct. i will try to come up with 7/3A.

6. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
The answers are given to two sig figs. To that level, the correct answer is most definitely there. The exact answer is 7/3 A which, as a decimal, is 2.33333... A. Round that to two sig figs and you get 2.3 A. All I was saying is that if you use 2.3 A as a trial solution and plug it back into the problem you will not get an exact match. In general, answers will always have some roundoff error and so you can seldom expect an exact match, so don't reject a possible solution just because it is not an exact solution.

7. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Well, we can't help you very much unless you show us the steps you used. My crystal ball just isn't that good.

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,519
515
I got 2.3333 amperes for the current through the 10V battery. So A is correct.

9. ### DGElder Member

Apr 3, 2016
348
87
Show your work so we can show you where you went astray.

10. ### abyss Thread Starter New Member

Jun 15, 2016
10
0
using superposition this is what i came up with: 10v 15v
R1 R2 R3 R1//R2 R1//R2+R3 TOT R1 R2 R3 R1//R2 R2+R1//R3 TOT
E 2.16v 2.16v 6.67v 3.33v 10v E 5v 10v 5v 5v 15v
I .65a .65a 1.3a 1.3a 1.3a I 1a 2a 1a 2a 2a
R 5Ω 5Ω 5Ω 2.5Ω 7.5Ω R 5Ω 5Ω 5Ω 2.5Ω 7.5Ω
when using these values in the other tables for voltage and current the only time i came up with 2.3A was for the combined currents for the IR3. is this on the right track?

11. ### anhnha Active Member

Apr 19, 2012
776
48
I have just checked the current flowing through the 5Ω in series with 10V battery in two cases using superposition and your results is correct if R3 is the resistor in series with 10V battery.
You didn't designate what is R1, R2, R3 and your table above is very hard to read.
For me, I prefer nodal analysis for this circuit.

12. ### abyss Thread Starter New Member

Jun 15, 2016
10
0
yeah i'm not very computer savvy when it comes to making tables sorry. but even though the IR3 sum from the 10v and 15v tables add to give me 2.3A don't i have to take into account the IR1, and IR2 sums as well when looking for the 10v emf current?

13. ### DGElder Member

Apr 3, 2016
348
87
using superposition this is what i came up with: 10v 15v
R1 R2 R3 R1//R2 R1//R2+R3 TOT R1 R2 R3 R1//R2 R2+R1//R3 TOT
E 2.16v 2.16v 6.67v 3.33v 10v E 5v 10v 5v 5v 15v
I .65a .65a 1.3a 1.3a 1.3a I 1a 2a 1a 2a 2a
R 5Ω 5Ω 5Ω 2.5Ω 7.5Ω R 5Ω 5Ω 5Ω 2.5Ω 7.5Ω

Your kidding right? You expect someone to decipher that mess?

14. ### anhnha Active Member

Apr 19, 2012
776
48
Yes, actually when you calculate current through R3 in each case, R1, R2 (or IR1, IR2) are already taken into account.

15. ### dannyf Well-Known Member

Sep 13, 2015
2,196
417
I give you a different way of getting there.

Assuming that the voltage over the middle 5R resistor is V (from top to bottom).

It must be true then that the current through the bottom 5R resistor is (15v + V) / 5;

It must also be true that the current through the top 5R resistor is (10v - V) / 5. This is also the current through the 10v source.

The two current flows opposite of each other and creates a voltage drop of V over the middle 5R resistor so

[(10v - V) / 5 - (15v + V) / 5] * 5 = V.

Or -5 = 3V and V = -1.7v.

The current through the 10v source is (10v - (-1.7v)) / 5 = 2.3a.

Simple math.

16. ### abyss Thread Starter New Member

Jun 15, 2016
10
0
thank you anhnha i will try to wrap my head around that.
DGElder sorry for the mess. it was all spaced out and organized when i typed it. when i hit return it clumped together. there was no post preview option.

17. ### abyss Thread Starter New Member

Jun 15, 2016
10
0
thanks dannyf i'll try that too.

18. ### DGElder Member

Apr 3, 2016
348
87
"DGElder sorry for the mess. it was all spaced out and organized when i typed it. when i hit return it clumped together. there was no post preview option."

That happens to me as well, especially if I use Notepad to create the post and try to copy and paste it into here; but then I go in and edit my post to make it readable and, hopefully, intelligible.

Last edited: Jun 18, 2016
19. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Even if it were nicely spaced, I would have no idea how to interpret it.

To get things monospaced you need to use CODE tags.

You can preview your post by clicking "More Options..." and then there is a "Preview" button available.

Sadly, when typing in your post, there is no monospace option so you have to guess, preview, guess again. Also, when you copy/paste from a text editor, multiple spaces are reduced to a single space. Very annoying.

20. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
When you use superposition, you have to be particularly careful to properly track the directions of your currents in all components. This is most conveniently done by annotating your original diagram and then redrawing the circuit for each case and transcribing the current designators one it. Then solve the reduced circuit and tabulate the results (which I imagine is what you're trying to do with the table you posted).

A few notes that may or may not be obvious. I'm implicitly numbering the resistors R1, R2, and R3 to match the current designations. Similarly, the voltages across the three resistors are designated V1, V2, and V3 with their polarities established by the passive sign convention (i.e., the designation voltage is positive at the tail of the current arrow relative to the head). I intentionally assigned the direction of I1 and I2 so that they agree with the notion that a voltage source usually sources current, which will result in I1 and I2 being positive if that turns out to be the case. I assigned the directly of I3 arbitrarily, choosing downward only because that is how we normally think of currents in circuits. By inspection, however, I suspect the actual current will flow in the other direction and so I expect I3 to end up negative. If it doesn't, then I will want to consider why that is the case -- is my suspicion wrong or did I make a mistake?

When we turn off everything but the 10 V source, we have:

By inspection, we can see that the total resistance seen by the load is 5 Ω + (5 Ω || 5 Ω) = 7.5 Ω. Thus the total current from the source (which is equal to I1) is

I1 = 10 V / 7.5 Ω = 4/3 A

This current splits evenly in the other two resistors making

I2 = I3 = 2/3 A

The voltages across the resistors are obtained by Ohm's Law, so we have

Source I1 (A) I2 (A) I3 (A) V1 (V) V2 (V) V3 (V)
10 V 4/3 2/3 2/3 20/3 10/3 10/3
15 V
TOT

Now we can turn our attention to when only the 15 V source is on:

We again see that the total resistance seen by the source is 7.5 Ω and that the total current is 2 A. Furthermore, the current splits evening in R1 and R3 so has a magnitude of 1 A in each. But here is where it is critically important that we keep track of the polarities of the current designations because while I1 and I2 are positive, I3 is negative.

Source I1 (A) I2 (A) I3 (A) V1 (V) V2 (V) V3 (V)
10 V 4/3 2/3 2/3 20/3 10/3 10/3
15 V 1 2 -1 5 10 -5
TOT 7/3 8/3 -1/3 35/3 40/3 -20/3
TOT 2.33 2.67 -0.33 11.67 13.33 -6.67

We can now do some checks to see if we made any mistakes.

Applying KCL we know that

I1 = I2 + I3
2.33 A = 2.67 A + -0.33 A = 2.34 A

Which matches to within the roundoff error.

Now we can apply KVL around a couple of loops if we want. We'll settle for going around the outside loop:

10 V - V1 + 15 V - V2 = 0
10 V - 11.67 V + 15 V - 13.33 V = -0.01 V

Which, again, matches to within the roundoff error.