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schematic

simulate this circuit – Schematic created using CircuitLab

I'm not really sure how to solve this. I can't see any series or parallel connections.

I need to find the Total resistance, V1 and V2, and finally I3 with a direction.

Any help would be much appreciated.


(edit)

re-drawn after taking suggestions

Okay so now I have taken your suggestions and redrawn it and worked out the values. it is much easier now to work out, so thanks for all the great suggestions.

Feel free to correct me on any faults in my calculations.

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  • \$\begingroup\$ What is V1 and V2? Also I3? I would use Thevenin equivalent circuits to solve for V1/2 then use those to get I3 if it's related. \$\endgroup\$ – I. Wolfe May 21 '15 at 17:44
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    \$\begingroup\$ Redraw/rearrange the schematic. Keep redrawing until you see it. The schematic is simpler than it looks. \$\endgroup\$ – Nick Alexeev May 21 '15 at 17:44
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    \$\begingroup\$ @brhans I'm deleting your comment, because we don't do simply solve homeworks for the interwebs on EE.SE. \$\endgroup\$ – Nick Alexeev May 21 '15 at 17:46
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    \$\begingroup\$ Lol... I actually just glanced at it and didn't realize. Brilliantly done on the profs part. Not so brilliant on the students. \$\endgroup\$ – I. Wolfe May 21 '15 at 17:47
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    \$\begingroup\$ Mark every node in a different color, this might help you see it. \$\endgroup\$ – Sada93 May 21 '15 at 18:06
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This is a really easy circuit to solve. The reason it appears difficult is because the schematic is obfuscated by deliberately using unintuitive layout. First, this is a good lesson on why logical schematic layout matters. Second, you'll see this circuit is actually very simply once you draw it logically.

In general, draw schematics with power and ground voltages being horizontal lines sorted by ascending voltage bottom to top. Then show the rest of the circuit with logical flow (nothing to do with current flow) left to right.

In your case, draw the battery at left with the positive node going up from the top, then across the sheet to the right. Similarly, the negative node should go down from the battery, then across the sheet left to right. Now fill in the resistors vertically between the two horizontal power lines that run across the top and bottom of the sheet.

Once you do this, the circuit will be obvious and you'll wonder how you didn't see it before.

Hint: Look at how many nodes this circuit actually has. Yes, it really is that simple.

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  • \$\begingroup\$ Oh I just noticed now everything is in parallel \$\endgroup\$ – Omuse May 22 '15 at 4:39
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This is obviously a homework question. Please do us all a favor next time and actually tell us that right up front. We won't GIVE you the answer but we'll help you figure it out yourself.

I'll start off by making a suggestion. Follow the line from the battery (+) terminal and mark each end of every resistor that connects to that (+) terminal. Use the letter A to keep it simple.

Now do the same for the battery (-) and use the letter B.

Now re-draw the schematic with that information.

Modify your question if you are still having problems.

[Edit]

I'll add one more suggestion: re-draw the schematic with the battery on the left and all of the resistors on the right. Place all of the resistors vertically with all of the same letter at the top.

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  • \$\begingroup\$ That was my main intention to learn how to work this out. Answers mean nothing without the working out. \$\endgroup\$ – Omuse May 21 '15 at 18:26
  • \$\begingroup\$ Then you aren't trying. I'm NOT going to give you any more information until you post a re-drawn schematic based on my suggestion. \$\endgroup\$ – Dwayne Reid May 21 '15 at 18:27
  • \$\begingroup\$ Maybe I should have replied when I finished drawing the circuit not in the middle of it. whoops... \$\endgroup\$ – Omuse May 21 '15 at 18:40
  • \$\begingroup\$ Well it's done now. I also re-edited it with a new comment. \$\endgroup\$ – Omuse May 21 '15 at 18:41
  • \$\begingroup\$ Oh wait damn I redrawn it under someone else's advise... \$\endgroup\$ – Omuse May 21 '15 at 18:45
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As Olin said - Yes its really that simple.

enter image description here

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  • \$\begingroup\$ You really shouldn't just give out the answer to a homework question like this. \$\endgroup\$ – Olin Lathrop May 21 '15 at 22:29
  • \$\begingroup\$ @OlinLathrop: What the OP stated was: "I need to find the Total resistance, V1 and V2, and finally I3 with a direction." So where did Jim Dearden spill the beans? \$\endgroup\$ – EM Fields May 21 '15 at 23:07
  • \$\begingroup\$ Yeah I drew it before So now answers were given out. \$\endgroup\$ – Omuse May 22 '15 at 4:43
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If you pretty up your circuit it'll look something like the schematic below, so now all that's left is to figure out - knowing that V1 and V2 are merely the the battery voltage across the array - is I3, then the resistance of all the resistors in parallel and, finally, Is.

enter image description here

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  • \$\begingroup\$ You really should let the OP do the work of a homework question like this. By the way, you got I3 wrong. Also, we've been trying to teach a consistant way of drawing schematics with power busses horizontal sorted by descending voltage top to bottom. While your schematic is correct, I would really like it to be rotated 90 deg right to reinforce what we are trying to teach instead of undermining it. \$\endgroup\$ – Olin Lathrop May 21 '15 at 22:33
  • \$\begingroup\$ My schematic is perfectly legible and understandable, and I'm really not concerned with what you like or what you consider to be proper drawing procedure/aesthetics since I edited a real dog you once posted and you rolled it back to its original ugliness purely for egotistical reasons. As far as the homework thing goes, you really should look at posting times before you start coming to unwarranted conclusions. As for \$ I_{3} \$, conventional current is considered to exit a source and enter a load, so you must be thinking of something else. What might that be, pray tell? \$\endgroup\$ – EM Fields May 21 '15 at 22:57
  • \$\begingroup\$ @Olin Lathrop Don't worry about, it I found out already that everything is in parallel. I understand why you preferred it drawn your way. It is easier to guess the direction of flow of current. \$\endgroup\$ – Omuse May 22 '15 at 4:54
  • \$\begingroup\$ @EM Fields I also think I3 is across R3 like in the original schematic. If not anyone feel free to correct me. \$\endgroup\$ – Omuse May 22 '15 at 4:58
  • \$\begingroup\$ @Omuse: Physically, charge flows through a resistor while voltage appears across it, so I3, being a current, would go through R3 instead of appearing across it. \$\endgroup\$ – EM Fields May 22 '15 at 11:27

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