Find the currents

Discussion in 'Homework Help' started by sonutulsiani, Mar 1, 2010.

1. sonutulsiani Thread Starter New Member

Jan 27, 2010
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0
Find currents I1 and I2 by using source transformation theorem

I converted the left 10A into 50V, the right most 100V to 10/3A.

Now what to do? Should I apply KVL or something? But won't that be wrong because it says to solve by source transformation theorem.

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2. Heavydoody Active Member

Jul 31, 2009
140
11
I am no expert myself, and I certainly don't claim to read minds. If it were me, I would do half of what you've done (undo the other half) and follow it up with a healthy dose of nodal analysis. I suspect the idea is to use source transformation to simplify the circuit to pave the way for another technique. But that's just a guess.

Mesh works here just as well.

Last edited: Mar 1, 2010
3. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Yes - it's not clear how many successive source transformations are envisaged in the question.

One approach would be

1. As 'sonutulsiani' suggested in the original post for LHS & RHS sources.

2. Integrate the 10Ω with the above LHS transformation and re-transform to a current source with parallel resistor branch.

3. At this stage you have three resistors in parallel with a current source either side.

4. Reduce this to a single resistor and current source.

5. Calculate the voltage drop across the equivalent resistor.

6. This voltage is effectively the node voltage at the top of the 20Ω

7. Using this node voltage calculate the required currents from the original circuit with only the first LHS reduction done.

In any case it's not possible to do this purely by source transformation alone.

4. Heavydoody Active Member

Jul 31, 2009
140
11
I just toyed with it for a bit and you can use source transformation entirely, although it is not readily apparent. If you reduce the circuit to two voltage sources and three resistors, and then use source transformation on the right loop until the circuit condenses to two sources and one resistor you obtain i1. If you go back and do this to the left loop you obtain i2. Pretty cool, but I doubt I would solve that way intuitively. It may be what you are supposed to do though.

5. Heavydoody Active Member

Jul 31, 2009
140
11
This way looks like the simplest. Pretty groovy.

6. sonutulsiani Thread Starter New Member

Jan 27, 2010
27
0
Anyone worked out the answer? I am getting weird answers the way "HeavyDoody" told in his second post. Because if I condense from LHS I get different answer and from RHS is different. But the method should work same for the same circuits right?

I will try the node method now. Anyone know the answer?

7. Heavydoody Active Member

Jul 31, 2009
140
11
You should be coming up with something resembling this:

Note that in each instance we are solving for the current through the resistor which has remained unchanged, they are different. Furthermore, since current remains the same throughout a series circuit, we can add the resistors into one (in each of the lower circuits) and the current we get from that will be the current through the unchanged resistor in the original circuit.

I personally like t_n_k's solution better. Less steps, less chance for error.

8. sonutulsiani Thread Starter New Member

Jan 27, 2010
27
0
Even i like the other answer. But then the question specifically asks for source transformation theorem. I had done the way you have shown in the figure. I got 0.37 amp flowing right as i1 and i2 as 1.85 flowing left.

9. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
It's a small matter but, given the nominated current directions on the original diagram, one would write I2=-1.85A. You have indicated your recognition of the point anyway by stating that the current is flowing to the left, which is opposite to that indicated on the diagram.

The values look good.

10. sonutulsiani Thread Starter New Member

Jan 27, 2010
27
0
Hey I tried the other way also and the answers match both the ways. I appreciate all of your effort. Thanks a lot for helping me out.