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#1




basic loop analysis of circuits
Hello, I am trying to learn how to analyse circuits using loop analysis.
I have this circuit (attached file) and I want to calculate the three currents I1, I2 and I3. I think the procedure is as follows.  Divide into three mesh loops (I have labelled them 1, 2, 3)  Sum the voltages around each loop and equate to zero: eg. Loop 1: 7  I1 + 2(I2) = 0 Loop 2: 5  9(I3)  2(I2) = 0 Loop 3: 5  3(I3) = 0  solve the simultaneous equations to obtain values for the currents. The problem is, I am unsure if have calculated the three loop equations correctly. Can someone please tell me if I am going wrong here ? Thanks kindly for any help. 
#2




You have the right idea, but the loop equations are wrong as the loop and the labeled I's are not the same. Draw a clockwise circle around your 1 & 2. Loop 3 (also clockwise) is a given as the only voltage that can exists at its node is the 5 volts; and loop 3 is isolated as to not contribute to either loop 1 or 2. So I'll show the loop currents as lower case i's
loop 1: 7 = 1(i1) + 2(i1  i2) loop 2: 5 = 9(i2) + 2(i2  i1) loop 3: 5 = 3(i3) But note that these i's are not the same as the ones listed on your picture. Also note, the node at top of the 2 ohm resistes is: sumation of currents is 0 I1 + I2 = I3 (these I's are the one's labeled on your schematic). Hope that helps 
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fran1942 (04272011) 
#3




Hello, thank you kindly for helping.
I have attached another picture of two circuits including the one you advised on. I have included loop formulas as I understand them, however I dont think I am quite correct. Can you please advise if and where I am going wrong ? Again, thank you very much. Last edited by fran1942; 04272011 at 02:09 AM. 
#4




Check out the attachment.
Only one of your loop equations was correct. I suggest placing polarity signs on the resistors and voltage sources based on indicated current flows (for the resistors)  the voltage sources have fixed polarity. As you go around the loop (you seem to use anticlockwise convention  which is fine) look at the sign you encounter at each successive loop element, whether it be a resistor or voltage source  that sign then becomes the sign [+ or ] for the voltage potential difference you assign to that part of the loop. Once you complete the full loop traversal you then place the "=0" part to complete the equation. Note that this particular approach for your examples requires you to formulate another equation which relates the individual currents assigned in the various branches  this is based on Kirchoff's current rule. It happens to work out that in both examples I1+I2=I3 You'll also note in the second example the current in the 3Ω across the 5V source is immediately known from I=V/R. Also note that I3 "splits" into two parts as it passes through the 5V source in parallel with the 3Ω. The current in the 3Ω will have no bearing on what happens in the rest of the circuit  the 5V source fixes the potential difference for that part of the circuit. 
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fran1942 (04272011) 
#6




t_n_k, the loop equations he has written in the circuits is correct, however he is mixing the loop mesh currents from my equations (lower case i's) with the upper case I's illustrated in the schematic. At least in the second schematic. By moving the equation totally to the left side of the equal sign, it appears that the loop was set up counter clock wise.
With the equations I listed in my post you can completely solve the problem without the KCL node. However, to relate the uppercase I's to the loop i's, it is necessary to use this fact. I should have posted the relationship in my first post So: I1 = i1 I2 = i2  i1 I3 = i2 Took me a few seconds to identify what was happening. But confusing the loop currents where KVL is applied to 0 the loop with the listed currents on the diagram is the main problem. You should see that the answer is the same between both my and t_n_k's responses. Just make sure you know the difference between the loop currents (lower case i's) and the currents identified on the schematic (upper case I's). I must say that the labeling of the I3 on the diagram (second circuit) is contributing to the confusion. That is a very poorly designed problem, in my humble opinion. Last edited by mjhilger; 04272011 at 05:37 AM. 
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fran1942 (04272011) 
#7




The second circuit equations look OK  if we are referring to mesh currents  rather than the designated currents on the schematic.
The top circuit equations are still incorrect.... both of them. Let's take the first circuit with the two loops. I1 will be the clockwise loop current in the lefthand mesh and I2 the clockwise current in the righthand mesh. Assume I1 & I2 aren't the same as those shown on the original diagram. The two equations would be 11+5*I1I2=0 or (with sign reversal throughout) 115*I1+I2=0 8I1+6*I2=0 or 8+I16*I2=0 To fran1942: You would do well to clearly show the intended current paths and directions on your diagrams. There is likely to be less confusion as has arisen here. Several members will often respond to the same post. So clearly stating your problem & solution is important to avoid this confusion. 
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fran1942 (04272011) 
#8




thanks guys for your great help. I understand what you have discussed.
However, I have one more, more complicated circuit that I cannot figure out. If I can just see how you derive the formulas for this one, then I will be totally confident. I have attached the picture of the circuit. If you have time, it would be much appreciated. I think there will be 5 loops comprising I1, I2, I3, I4 and I5 ? nb. t_n_k  the way you solved the first set of circuits that I posted, is how we interpret them. I followed that method well. 
#9




This problem has 3 different resistor branches where the current is solely determined by a single voltage source.
In such cases you can work out the current in those branches and then remove them from the circuit to analyze for the current in other branches which are dependent on more than one source. This allows a simplified analysis  notionally from 5 loop currents to 2 loop currents in this case. I've attempted to illustrate this in the attachment. NB If you use this method  you must remember that if required to calculate the total current drawn from the various voltage sources, you would then need to add the currents (magnitude & direction) flowing in the branches removed in the circuit reduction applied. That's only important if you were asked to calculate the current drawn from the sources. 
Tags 
analysis, basic, circuits, loop 
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