This is a "homework" problem from the book Electric Circuit Analysis by Charles J. Monier. The reason I say "homework" problem is because I'm not taking a class... rather I'm studying on my own. This is the problem scanned from my textbook. It's from chapter 4 (KVL, KCL) so it's not super difficult. Solve for v: The problem with solving this circuit is that I can't reduce it to simpler forms -- nothing is in series or parallel. I also tried writing 5 equations from KVL for the 5 different loops. I then resorted to using a delta-wye transformation with the top 3 resistors (8, 4, 2 Ohms) -- this method took forever!. This got me the right answer, BUT I have this feeling that there was a much simpler, faster way to solve this problem! It's driving me crazy!
There are several techniques that can be employed to solve this problem. You may want to explore the information at the AAC Tutorial on Thevenin's Theorem. hgmjr
I don't want to resort to techniques that I haven't learned yet. I believe there is a simple way to solve this using voltage divider/current divider techniques..... Thevinens theorem hasn't been introduced yet.
there was a need for only three kvl equations.as three currents in three loops were not known then solve by simultaneous equation. is there a problem with this (maybe u shud post your attempt along with the way u chose direction of current. as Mr hgmjr said there are a lot of ways to solve the problem the reason why it took u long with star and delta transformation was u were not used to solving with these methods a little practice will make things better.
The book has covered this so far: 1. Ohm's law 2. Kirchoff's laws 3. Current and Voltage Division 4. Branch current method 5. Delta and Wye Circuits I used Delta-Wye knowledge to solve this problem.... but I have this nagging feeling that there is an easier way to solve it.
Yes I think you're right about the star and delta transform - it seem to take very long b/c it was my first practical use of it. I didn't think to write the KVL equations in terms of current. Instead I labeled each resitor v1, v2, v3 .. etc.... which generated 5 equations. I will now try the 3 equations with current instead. I'll post back my results.
Of the techniques listed, I would agree with recca02 and recommend Kirchoff's Laws. Out of curiosity, I would like to inquire if your textbook has a section on Millman's Theorem to be covered later in the course? hgmjr
Crap! I wrote 3 KVL equations in terms of current but it didn't yield the correct answer. Here is my attempt: As you can see, I got v = -41.9 V. The correct answer is -20 V. Any help would be great.... this is killing me!
I recommend that you look into Millman's Theorem at some point in your studies. I would wait until you have mastered the more common techniques that are taught. Only then will you be able to fully appreciate the power of the Theorem to take the pain out of circuit analysis. It is a very useful tool when combined with Thevenin's Method for analyzing circuits from the most mundane to the most complex. Don't be fooled by the simplistic examples provided to illustrate its application. hgmjr
I would draw your attention to the three loop equations that you have written and mention that you have failed to take into account the effect of all the currents flowing in those components that are common to more than one loop. You should have an example in your text book that would help you see what I am talking about. hgmjr
that is the only mistake u committed, the currents we assumed are not which actually flowing in components(they were taken as loop currents). it is the algebraic sum of the currents(sign matters with respect to direction) common to the component that flows thru it. like 80 = 2.(I1-I3) - 12.I1 (here both I3 and I1 flow in 2 ohm but with different directions which were assumed so -ve sign) did u get it? one more thing the 5 eqns u wrote using V b4 were i think KCL not KVL
I took into account your advice BUT I'm still getting the wrong answer. My three KVL equations are... (picture is not showing up, link to picture: http://jut.shanahan.googlepages.com/4.11b3KVLeqns.jpg/4.11b3KVLeqns-full.jpg I tried to solve it by hand and using the ti89, but yielded incorrect answers. I'm going insane!!
you made few mistakes in the eqns, second eqn shud have -4i2 not -4i1 third eqn shud be (14 i3 -2i1 +8 i2=0) remember to traverse in the direction of current and take voltage drops according to direction of current. i the third one you seemingly have taken wrong polarities of voltage drop, edit: i solved it and its -20 volts. be careful while writing down equations, u can always make a rule to choose a clockwise direction(so that u dont waste time deciding which direction the current is going to-sign will take care of that for u automatically) traverse a loop in direction of current and take voltages dropped by it (IR) as positive and voltages dropped by currents in opp direction as -ve. when u encounter a voltage source while traveling the loop take it as -ve if it supports the current direction else take +ve.then equate all these to zero. (all these rules i have made from my experience books may/may not help much but practice certainly does--and you are doing a good job already btw)
apply mesh equation for the 3 loops in the circuit. The equation for the 3 loops are: i1 (4+2+8) _ 8 i2 _ 2 i3 = 0 14 i1 _ 8 i2 _ 2 i3 = 0 (equation # 1) -8 i1 + 12 i2 = -80 (equation # 2) -2 i1 + 14 i3 = 80 (equation # 3) solve The 3 equations: we get i1 = -5 A i2 = -10 A i3 = 5 A To get the voltage (v) multiplay the current (i1) with the resistance (4 ohm) finally; V= i1*(4ohm)= (-5)*4= -20v V= -20 v
I don't understand how you arrived at (14 i3 -2i1 +8 i2=0) as the third equation. Are you using KVL around the top loop of the circuit? I substituted your equation and I got the right answer! But I still don't understand how you got that 3rd equation. According to my text book, "a voltage drop occurs when we travel from + to - through an electrical element ... a voltage drop is assigned a negative sign." I followed this rule for equations 1 and 2 and those were correct. Some day I will solve this damn problem. Thanks for all your help so far!
Thanks for the your help jameel, I appreciate it, but I haven't got to mesh equations yet in my book so it doesn't make sense to use them for this problem.
mesh equations are mesh analysis=loop analysis which implies KVL node for KCL i know its because u used same polarities of voltage drops for traversing different current in diff loops. as i suggested earlier think about it what makes sense while traveling in direction of current -this is why i never mark polarities. polarities are decided by the direction of current flow. for the third loop drop voltages in direction of the current in that loop (I3) and take opposite direction current voltage drop as -ve (as voltage rise) and equate to zero. do not concern yourself with polarities chosen earlier --life is much more easier without them. ask any doubt i'll try to help wherever i can. these never let me down. additionally u can also say if you encounter the negative terminal of battery first then take its voltage as -ve else +ve in the above method then equate as zero (note here i m taking voltage drop as +ve and voltage rise as -ve)
I finally solved it, HOORAY!! I was having trouble understanding your english so it was a little hard to understand your method of solving the problem. I finally understood what you were saying and applied your method and it worked! I then understood the mistake I was making... and corrected the method that I am used to and it worked too! I also solved the problem by labeling 5 separate current... and then solving 5 simultaneous equations, which worked too -- that method is nice because it gave me the currents in all of the branches. So to sum it up... I solved the problem three different ways: 1. Using a delta-wye transform on the top loop 2. Writing KVL equations in terms of 5 different currents. 3. Writing KVL equations in terms of 3 currents flowing in 3 loops. Here is my solution: I am very satisfied now! THANK YOU VERY MUCH RECCA for all of your consistent help... I couldn't have done it without you!