# Superposition theorem wrong values ?

Discussion in 'Homework Help' started by mauz243, Apr 11, 2012.

1. ### mauz243 Thread Starter New Member

Apr 7, 2012
3
0
I think my values are wrong, since the final answer doesn't seem to match. This is the first part where I am getting values without removing any supplies. This is the circuit:

Values I got:

 I1 Ω I2 Ω I3 Ω E V I kontuur 25 15 0 65 II kontuur 0 -15 30 -90 III kontuur 1 -1 -1 0

Some of the text on the circuit is in estonian, but I think it's pretty clear.

Anyway, I'd really need some help. Thanks

2. ### WBahn Moderator

Mar 31, 2012
18,089
4,917

1) The diagram has two loops (I'm assuming that's what 'kontuur' means, which sems reasonable given the similarity to our 'contour') but your table has three.

2) I also don't see an 'E' marked in the diagram, so I don't know what the 'E' column in the table is refering to.

3) You also have the units on I1, I2, and I3 indicated as ohms, not amps.

4) At all times, the relation I1 = I2 + I3 must hold, and it doesn't in any of your table rows.

5) Taking the last two points into consideration, are the first three columns in the table supposed to be currents, or equivalent resistances (and, if so, where)?

Unless you aren't allowed to do so, I would recommend first simplifying the circuit by combining the resistances in the outer two branches into a single resistance (in each branch). And, again if you allowed, also combine the two supplies in the center branch into a single equivalent source. Now you have three parallel branches each having one source and one resistance. Apply superposition to get the branch currents with each supply turned on in exactly one analysis and then combine them to get the total branch currents. You can then apply those back to the original diagram to get the any specific current or voltage you need very easily.

Assuming I didn't make any mistakes, the total voltage (with all supplies turned on) at your red dot (relative to the node directly below it at the bottom of the middle branch) is 25V. With just the supply in the left branch turned on it is 14.29V, with just the two center supplies turned on it is -7.14V, and with just the right supply turned on it is 17.86V.

Those should give you some useful check points (again, assuming I didn't make any mistakes, but I did check that my total branch currents sum correctly, so I think these are correct).

Good luck!

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3. ### rac1 New Member

Apr 8, 2012
5
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if i were you, i would find the equivalent resistor values in every current way.And after i would make E2+ E3=15V
then there will be 3 sources, you can kill all and leave 1 source alone, after do this 3 times for every voltage source.Find the algebrically sum of the values. The summation of the currents should be carefully added
for example.
when you kill e2+e3=15v and e4 =75 volt,
i assume you found I1=2A-->
after when you take only e4 and find I1=3A<--
than when you take only e2+e3=15 V ( i assume it as one voltage source)
you found current 1A<---

I1=2A--> + 3A<-- + 1A <-- = 2A <-- Be carefull in here.

4. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
If we turn off e2, e3, and e4 and assume that I1 is 2A left to right, then the voltage at the red dot (relative to the bottom of the center branch) is going to be 0V, meaning that I2 and I3 must also be zero, which violates KCL at the red dot.

You can't just take any voltage and divide it by any resistance to get any current you are interested is. Ohm's law describes the relationship between the voltage across a specific resistance to the current through that resistance and the value of that resistance.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi rac1,

As WBahn indicates, I1 can't be 2A. Re-check and you'll find I1=1A.

6. ### mauz243 Thread Starter New Member

Apr 7, 2012
3
0
Thanks a lot for your help everyone. Hopefully I can fix it now.