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Regarding the following circuit, SPICE calculates the current I3 through R3 as 3.333A.

enter image description here

I try to solve that myself and can only reach that result if I first set V1 as zero(short) and solve for the current and then set V2 as zero(short) and solve for the current. Then the total sum through R3 becomes 3.333A.

But there are some parts which are not yet clear to me.

How can we set V1 or V2 as short? How can the current produced by V1 can pass through V2 at all? As far as I know they can do that in AC small signal analysis but here we have DC currents. How come a theoretical voltage source or a power supply in real pass DC?

If I was asked this question I would first remove V1 and solve for the current through R3 and then remove V2 and solve for the current through R3 and add these currents. But that would give 5A which is not the case.

So my thinking is somehow not matching the results, because I cannot imagine how current produced by V2 can pass trough V1 or vice versa. I would think the power supply would block such current. And even if not the case, here V1 = V2 how come current flows through the voltage sources?

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  • \$\begingroup\$ This can be solved easily using the nodal equation: Uk/(R1|R2|R3) = 5/R1 + 5/R2 + 0/R3, Uk being the voltage in the middle node. \$\endgroup\$ – Bart Jan 28 at 15:09
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    \$\begingroup\$ An ideal voltage source can have a current flowing through it in either direction and of any magnitude. The ideal voltage source constrains the voltage between its terminals but has no control over the current flowing through it. \$\endgroup\$ – Elliot Alderson Jan 28 at 16:21
  • \$\begingroup\$ Replace R3 with the parallel of two 2 ohm resistors. Split the circuit in two circuits (you can do it, why?) and compute the current in the series of 5V generator and 3 ohms. The current in the original R3 is double that (why?). \$\endgroup\$ – Sredni Vashtar Jan 28 at 22:28
  • \$\begingroup\$ @SredniVashtar I didnt say V1 and V2 are utopic ideal sources. They can be any voltage real source with output impedance neglected. In that case, if V1 and V2 are LDO your theorem will not hold. Read the accepted answer and comments. \$\endgroup\$ – user1999 Jan 29 at 9:36
  • \$\begingroup\$ So, the symmetric circuit is not symmetric? My bad. \$\endgroup\$ – Sredni Vashtar Jan 30 at 17:07
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How can we set V1 or V2 as short?

That's the rule when using superposition theorem.

If I was asked this question I would first remove V1 and solve for the current through R3 and then remove V2 and solve for the current through R3 and add these currents. But that would give 5A which is not the case.

No, you have misused the superposition theorem.

With V1 removed (and a short put in its place), the current from V2 flows into R2 + R1 || R3 and that is 5 volts / 1.5 ohms = 3.333 amps (split equally between R1 and R3).

If you do the same for V2 you get the same but for R2 and R3 and, this ultimately means that 3.333 amps flows through R3. But you would have probably seen that if you get rid of V2 and connected the right node of R2 directly to V1 - you can do this because the voltage sources (V1 and V2) are equal in value.

How come a theoretical voltage source or a power supply in real pass DC?

That's pretty much what is normally expected of them. They do it with ease.

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  • \$\begingroup\$ By removing I meant open circuiting the V1 or V2 so R2 or R1 becomes floating. \$\endgroup\$ – user1999 Jan 28 at 15:01
  • \$\begingroup\$ Can an SMPS(voltage source) sink current like in this theoretical example? \$\endgroup\$ – user1999 Jan 28 at 15:04
  • \$\begingroup\$ That's the wrong way to use the superposition theorem; voltage sources get replaced with shorts. \$\endgroup\$ – Andy aka Jan 28 at 15:04
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    \$\begingroup\$ It's the same with (series linear) voltage regulators - they can supply lots of current (within reason) but if you drag the output upwards to a higher rail they offer no resistance to this action and can easily be destroyed. \$\endgroup\$ – Andy aka Jan 28 at 15:43
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    \$\begingroup\$ " I thought voltage source can not sink DC current. " IS WRONG e.g. Battery current can be bi-directional. But an LDO cannot, by design \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 28 at 22:35
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You asked a good question. How to solve the problem seems trivia, but how to get the intuition about the superposition theorem seems not that intuitive.

Referring to this site, superposition Principle states that

When there exist multiple energy sources in the circuit, any voltage and current in the circuit can be found as the algebraic sum of the corresponding values obtained by assuming only one source at a time, with all other sources turned off:

A voltage source is turned off if treated as short-circuit so that the voltage across it is guaranteed to be zero. A current source is turned off if treated as open-circuit so that the current through it is guaranteed to be zero.

This may still not answer your puzzle. Let's try to get a sense of it.

  1. Using KCL or superposition method, we can get the voltage at the mid point,

enter image description here

So the current I_R3 = V_NV / R3 = (R1V2+R2V1)/(R1R2+R1R3+R2R3).

(Here I just used an online symbolic circuit analysis tool, CircuitNav, to get the solution quickly)

The thing to notice is that V_NV and I_R3 are just linear combination of V1 and V2. At this point, V1 and V2 are not only 5V but variables.

  1. If we set V2 = 0, we will get I_R3_V2zero = R2V1/(R1R2+R1R3+R2R3). Setting V1 = 0, we get I_R3_V1zero = R1V2/(R1R2+R1R3+R2R3). Summing this two currents, I_R3 = (R1V2+R2V1)/(R1R2+R1R3+R2R3). So far so good?

  2. When we set V1 or V2 equal to zero, what does this mean? As a voltage source, the terminal voltage is fixed but the current through it can be anything. If the voltage source voltage is zero, it means it is a short circuit. Can this become your intuition now? By the same token, when a current source is set to zero current, it is equivalent to open circuit, current being set to zero with any possible voltage depending on other part of the circuit.

  3. Turning back your example circuit, when one source, i.e. V2 is set to zero or say shorted, IR1 = IV1 = 10/3 A (out of V1), IR3 = 5/3 A, IR2 = IV2 = - 5/3 A (into V2). When V1 is set to zero, IR1 = IV1 = -5/3 A (into V1), IR3 = 5/3 A and IR2 = IV2 = 10/3 A (out of V2). Summing each up, we get, IR1 = IV1 = 5/3 A, IR3 = 10/3 A, and IR2 = IV2 = 5/3 A.

Hope this makes some more sense to you.

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    \$\begingroup\$ You directly copied some of the content in this answer from another website. You need to provide a citation for the website where you got the text, and make it very clear which text was copied. \$\endgroup\$ – Elliot Alderson Jan 28 at 21:31
  • \$\begingroup\$ Elliot, if you mean the definition part of the superposition principle, I added the link. If not, let me know. Thanks for reading through. \$\endgroup\$ – X J Jan 28 at 21:45
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Try to imagine that the right side (R2, V2) is "mirrored" on the left, co we can assume that we have on the left: - V1 in parallel to V2 - R1 in parallel to R2

Next, both V1 and V2 are equal, so we can think about one voltage source, V12, with voltage 5V. Similar, the R1 becomes in parallel to R2, so we can replace them using R12 with value R12 = R1 || R2 = 0.5Ohm.

It means, the voltage on R3 will be: V12 * R3 / (R12 + R3), and current of R3 will be: V12 / (R12 + R3) = 5V / 1.5Ohm = 3.3(3)A

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The voltage at the node R1-R2-R3 (call V0) is :

V0 = (V1/R1 + V2/R2 + 0/R3)/(1/R1 + 1/R2 + 1/R3) = 10/3 V.

This formula is easily obtained by summing the currents entering the node :

(V1-V0)/R1 + (V2-V0)/R2 + (gnd-V0)/R3 = 0.

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Another way to look at it is to use a thevenin equivalent. Open circuit voltage would be 5V and Rth equivalent resistance 0.5 Ohm. Then 5/1.5 = 3.33 A.

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