1
\$\begingroup\$

The circuit to be simplified

I have found the equivalent resistance, however, I am finding trouble as to how I should go about finding the equivalent current source.

The answer stated that the 12 ohm resistance was shorted, and that from "the load's perspective", the current through the 20 and 5 ohm resistors were found (each to be I1 = 72/20 = 3.6 A and I2 = 72/5 = 14.4 A respectively). Using that, the answer claimed that the norton current was the algebraic contribution of I1 and I2 - where I (norton) = I2 - I1 = 14.4 - 3.6 = 10.8 A.

While the answer is correct when simulated using software, I do not understand the reasoning behind why each step was done in the answer.

Any help would be greatly appreciated!

EDIT: Initially, I thought of using nodal analysis to find the output voltage, which I could convert into a current source (the Norton current source) to find the answer. However, the method used by the answer seemed much more trivial than mine, yet I do not understand how the method produced the same answer (like why the 72V was considered to be the voltage drop across both R2 and R4 when treated individually, and why the 12 ohm resistor was shorted).

EDIT 2: Modified Circuit: Modified circuit

\$\endgroup\$
  • \$\begingroup\$ allaboutcircuits.com/textbook/direct-current/chpt-10/… \$\endgroup\$ – JIm Dearden Nov 6 '16 at 15:48
  • \$\begingroup\$ I understand how to convert thevenin sources into Norton sources, however, I do not understand why 1. the 12 ohm resistance was shorted, 2. the individual voltage across 5 and 20 ohms was considered to be 72 volts (in relation to the question posed). \$\endgroup\$ – TeraTesla Nov 6 '16 at 15:53
  • \$\begingroup\$ You need to add on your schematic an indication of what I1, I2, etc., are, otherwise it's useless to to try to talk about them. Also adding labels to your nodes will make it easier to discuss your circuit. \$\endgroup\$ – The Photon Nov 6 '16 at 16:24
  • 2
    \$\begingroup\$ Either the answer on your book (or at least what you reported in your question) has something tricky, say untold. R1 resistor MUST somehow be part of solution, just imagine to change its value, let's say to eventually replace it with a short circuit, Norton current will trend up to infinity. Then, for this very set of values it may as well simplify but this is definetively not a rule or a solution, method. \$\endgroup\$ – carloc Nov 6 '16 at 16:25
  • \$\begingroup\$ Given that the load would be R4 (20R) you would replace that by a wire (short circuit) and not R1. In this case R1 and R3 add up to 20R. This would give the 72V across R2 (72/5) and 72V across (R1+R3) = 72/20. The two currents passing through the 'wire'. So I would suggest that the answer was wrong to state a short across the 12R, it should have been across R4 (20R) \$\endgroup\$ – JIm Dearden Nov 6 '16 at 16:31
0
\$\begingroup\$

The Norton source current is just the current that the circuit would deliver to a short-circuit (0-ohm) load.

So if you have the Thevenin voltage, \$V_{th}\$ and the Thevenin equivalent resistance, \$R_{th}\$, then the Norton current is just

$$I_n = \frac{V_{th}}{R_{th}}$$

If you don't already have the Thevenin equivalent circuit, then you could re-draw (at least mentally) your circuit with a short across the outputs, and find the current through that short. It would be equal to the current through \$R_1\$ plus the current through \$R_3\$, both taken from left to right.

\$\endgroup\$
  • \$\begingroup\$ @The Photon I know how to find thevenin voltage and then use source conversion - it was my initial thought. However, it is the understanding of the procedure adopted by the answer sheet that is troubling me, and is what I need help with. \$\endgroup\$ – TeraTesla Nov 6 '16 at 17:16
  • \$\begingroup\$ @Dark_Colossus42, see my comment on your question. You haven't labeled what you mean by I1, I2, etc., so I can't follow your description of that solution. Please make your question more clear if you want an answer that discusses that aspect of it. \$\endgroup\$ – The Photon Nov 6 '16 at 17:17
  • \$\begingroup\$ @The Photon I'm sorry; but the funny thing is, neither the answer sheet does that. It just says that there is 72 volts across some 5 ohm resistor, which gives 14.4 A, and there is 72 volts going across some 20 ohm resistor, that gives 3.6 A. The algebraic sum, which in this case is the difference, would give the result of 10.8 A, which is definitely right. I was wondering if you could help me "decode" what the answer sheet was trying to say. \$\endgroup\$ – TeraTesla Nov 6 '16 at 17:21
  • 1
    \$\begingroup\$ Well, the bit about "12 ohm resistor was shorted" is obvious nonsense, as carloc already pointed out. Beyond that, sometimes bad homework questions are just bad homework questions. \$\endgroup\$ – The Photon Nov 6 '16 at 17:23
  • \$\begingroup\$ @carloc The answer does not mention whether it is R4 or R1 + R3, which is why I need help decoding what answer is saying to me :). \$\endgroup\$ – TeraTesla Nov 6 '16 at 17:27
0
\$\begingroup\$

Find the open circuit voltage and the output resistance into a shorted load.

We find the open circuit voltage by assuming an no load current.

Add R1 = 12 ohms and R3 = 8 ohms together to get Rx = 12 ohms + 8 ohms = 20 ohms.

Rx = 20 ohms is directly in parallel with R2 = 5 ohms. The parallel combination of 5 ohms and 20 ohms is Ry = 1/(1/5 ohms + 1/20 ohms) = 4 ohms.

Next we have a voltage divider between Ry = 4 ohms and R4 = 20 ohms, with 4 ohms on the top and 20 ohms on the bottom. The open circuit voltage is then VOC = 72V * (20 ohms)/(20 ohms + 4 ohms) = 60V.



Next we need to find the output resistance by shorting the load.

With the load shorted. R3 = 8 ohms and R4 = 20 ohms are in parallel. The parallel combination of them Rz = 1/(1/R3+1/R4) = 1/(1/20 ohms + 1/8 ohms) = 5.71 ohms.

Rz = 5.71 ohms is in series with R2 = 5 ohms which makes Rw = R2 + Rz = 5 ohm + 5.71 ohms = 10.71 ohms.

Rw = 10.71 ohms is in parallel with R1 = 12 ohms. The output resistance is therefore ROUT = 1/(1/Rw + 1/R1) = 1/(1/10.71 + 1/12) = 5.66 ohms.

So the whole network is equivalent to a 60 ohms voltage source in series with 5.66 ohms.

\$\endgroup\$
0
\$\begingroup\$

When the answer said that R1 is shorted from the loads point of view, it is saying in a very strange way it is superfluous. Shorting the output puts the resistor in parallel with what is being considered an ideal voltage source. This means it has no effect on the Isc analysis. Now you can sequentially take equivalents when solving circuits. So if you took the Norton equivalent across R4, you would get an Req of 4 and a short circuit current of 72/5. At this point I would just solve the circuit. Apparently the book prefers to go back and take the Norton equivalent again. This time across R2. This is the 72/20 Isc. It looks like this works but I can't really justify it. Maybe I'll think about it some more, but probably not.

This is a strange way to look at the problem to me. I would have summed the currents through R1 and R3. I would have used the Thevenin equivalent V1, R2, and R4 to do it more quickly on a test.

It looks to me like the book is using Norton twice to sum the currents into node 3 as a way to find Is. I can't really justify the third step. On a test, node equations would take a long time. What was done here seems fast but I don't know if it work in general. Like I said if you take R1's current 72/12 and add the R3 current this can be fast using Thevenin at nodes 1 and 3. The voltage is just 72*20/25. You can find the equivalent resistance if 4 is a couple of seconds. So the current through R3 is 72*20/25*1/(4+8). This added to 72/12 is 10.8. If you find out what they were doing please let us know.

Using Norton to sum currents at node 3?

\$\endgroup\$
  • \$\begingroup\$ Of course, \$R_1\$ can be separated. Just look at the circuit as having two 72 V sources (break the V-source into two identical sources, one that is behind \$R_1\$ and one that is behind the rest.) The other aspect, if it helps, is that once you Nortonize \$V_1\$ and \$R_2\$ after breaking it that way, you have two parallel values, \$5\:\Omega\$ and \$20\:\Omega\$, supplied by the current source. The current in the \$5\:\Omega\$ piece is reduced by the current in the \$20\:\Omega\$, hence the subtraction. \$\endgroup\$ – jonk Nov 6 '16 at 19:43
  • \$\begingroup\$ But the p.d. across the 5 ohm and 20 ohm resistor (once you Nortonize the V1 and R2 aspect) would not be 72 volts in that case, which is unlike the solution provided by my answer book, where they used 72/5 and 72/20, and subtracted both - the reason for which I do not understand. \$\endgroup\$ – TeraTesla Nov 6 '16 at 21:45
0
\$\begingroup\$

The book's solution that you paraphrased seems to be a lucky coincidence, because I2 is not actually 14.4A but 6.72A if you solve this circuit (either manually or in simulator).

In some quarters, this circuit is used an example of circuit that cannot be solved by using source transformations, so this brings at least part of owg60's solution in question. Furthermore the numeric answer in user96037's solution is wrong, so I didn't read that much further.

There is, IMHO, one way to simplify some calculations here and not apply the full mesh method for determining Isc. First, observe that the current through R1 is immediately determined by R1 and the 72V source since R1 is connected across it. So I1 = 72V/12ohm = 6A.

Now if we knew the current through R3 (I3) then Isc = I1 + I3. We can indeed ignore R1 for the purpose of calculating the current through R3 (but not for the purpose of calculating Isc, since I1 is included in that).

Ignoring R1 (which is essential, otherwise V1 and R2 don't form a 1-port network), calculate Ieq1 (Norton equivalent of V1 and R2) as 72/5A = 14.4A. Note however that once you've performed a source transformation that's no longer the current through the original R2 but the current through the current source in parallel with the new R2. I think this is what your book is doing, but the wording is muddy. The actual I2 through the original R2 (or even through the new/transformed R2) is not this 14.4A, as you'll see in a moment.

After we've done this source transformation of V1 and R2, we have a current source of 72/5A that's splitting its current three ways: through (the new) R2, through R3, and through R4. Borrowing owg60's 2nd diagram here: enter image description here

This is basically a 3-way current divider. I3 is then: (1/8 / (1/5 + 1/8 + 1/20)) * 72/5 = 4.8A. So Isc = I1 + I3 = 6A + 4.8A = 10.8A indeed.

But what about I2? Which I2?? We have to be very careful, because the I2 through the transformed R2 is different than the one though the original R2! Indeed the former is (1/5 / (1/5 + 1/8 + 1/20)) * 72/5 = 7.68A while the latter is (1/8 + 1/20) / (1/5 + 1/20 + 1/8) * 72/5 = 6.72A since it's the sum of I3 and I4. In either case, the book is obviously wrong about I2.

Also, you cannot make a 1-port network out of V1 and (just) R4, even after removing R1. So that (3.6A) transformation doesn't make any sense to me... But you can make a 1-port network of V1, R2 and R4 (again after removing R1). The latter is 14.4A (or 57.6V) source with a 20 || 5 = 4 ohm resistor. Now 57.6V / (4 + 8) = 4.8A again since R3 is now in series with this 4 ohm resistor in the Thevenin form. And this calculation is faster than the current divider I did above, but not very illuminating as to the currents through the actual R2 etc.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.