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I'm studying recently Electric Circuits (9th edition) by Nilsson. In Chapter 2, an example bothers me.

schematic

simulate this circuit – Schematic created using CircuitLab

I don't understand why these above figures can't be permissible. (I know KVL and KCL, but I want to analyse these circuits with another physical meaning.)

In addition, I want to know what situations will appear in these circuits.

Peter Mortensen
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Junu
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    What "physical meaning" do you expect from `5 = 10`? – Dmitry Grigoryev Mar 20 '18 at 12:48
  • There has to be the assumption that the wires are 'ideal' and have no resistance value that applies here. –  Mar 21 '18 at 03:35
  • "physical meaning" which I mentioned is what situation will appear in these circuits. For example, which source will explode or which source will be gone. – Junu Mar 22 '18 at 08:21

4 Answers4

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Ask yourself:

For the left circuit, what will be the voltage across V1 ?

  • 10 V ? No because V2 puts 5 V directly across V1.

  • 5 V ? No because V1 puts 10 V directly across V1.

Do you spot the contradiction?

For the right circuit, what will be the current flowing ?

  • 5 A ? No because I2 forces 2 A to flow.

  • 2 A ? No because I1 forces 5 A to flow.

Do you spot the contradiction?

In these circuits there is no solution as the sources are ideal so their voltage or current must be obeyed (for example: a 5 V voltage source must have 5 V across its terminals).

For the left circuit the KVL states that V1 = V2 as the sources are in parallel, fill that in and you get: 10 = 5

We all know that 10 is not equal to 5.

For the right circuit KCL states that I1 = I2 as the sources are in series, fill that in and you get: 5 = 2

We all know that 5 is not equal to 2.

Bimpelrekkie
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    The next logical question is, what would happen if we did this in real life? The answer is our current/voltage sources are non-ideal, so one or both would "break their contract" and not provide the voltage/current specified. Depending on what type of source it is (chemical, physical, etc), other bad things could happen, like the source permanently breaking, or even exploding. – BlueRaja - Danny Pflughoeft Mar 20 '18 at 15:26
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    @BlueRaja-DannyPflughoeft I'm tempted to say (sarcastically) that you have just re-invented a battery charger :-) – yo' Mar 21 '18 at 07:45
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I don't understand why these above figures can't be permissable.

Because you are literally saying that 10 = 5 in the left figure, and that 5 = 2 in the right figure. Both of which are false and doesn't help you solving the problem at hand.

Circuit theory is using math to solve the equations, if you are going to break math, then you are also going to break the circuit theory and end up with nonsense.

Harry Svensson
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Such circuits cannot exist in theory because the math does not sum up, as other answers state. However, I think it's worth noting what would happen if you try to build this in practice. The voltage source example would actually become:

schematic

simulate this circuit – Schematic created using CircuitLab

That's because there is no cable or voltage source in real life with 0 resistance. R1 and R2 would be the combination of the resistance of the cables and the internal resistance of the voltage sources. Their values would probably be very low. The cables would heat up until possibly burning due to the potentially high current.

In parallel (no pun intented), the current source example would become:

schematic

simulate this circuit

Where R1 and R2 would be the internal resistance of the current sources, and R3 and R4 the resistance of the cables. R1 and R2 would be very high, and R3 and R4 would be very low, so again, the cables would probably heat up and burn.

Daniel García Rubio
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    I don't think this is true, since the problem lies at an even 'rawer' level. The problem Isn't that there is no load, its that 5 != 10. – James T Mar 20 '18 at 13:01
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    Yeah, that is the theoretical reason for those circuits being wrong. I wanted to give a more practical approach, by explaning why you can't "build" this, because it would actually become something different than you intended. – Daniel García Rubio Mar 20 '18 at 13:04
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    James, what Daniel says is true. Just try to put 2 batteries in parallel and see what happens. The ideal circuit however cannot work due to the output impedance of the voltage sources being 0 ohms. Same with the current sources, real current sources always have a very high output impedance, but not infinite! – lucas92 Mar 20 '18 at 13:05
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    @lucas92 thanks for pointing out the internal impedance of the current sources, i've updated the answer to include that example. – Daniel García Rubio Mar 20 '18 at 13:28
  • I agree that it would be nice to see more of an explanation as to why the circuit is impossible. However, when paired with the other answers which cover that question, I like this answer because it approaches the intuitive misunderstanding. Most likely the OP is thinking "I can connect together any sources I like in real life," and this answer explains why that works but the ideal case does not. – Cort Ammon Mar 20 '18 at 19:55
  • Note that even with superconductors, the circuit is impossible in real life: any superconductor has a limit to how much current it can carry, and once you exceed that limit, it stops superconducting, possibly with explosive results. – Mark Mar 20 '18 at 22:03
  • @Mark if you use superconducting wires, you'll just shift the "bottleneck" to be inside the sources themselves. Even if the sources are made of superconducting wire, the internal structure of a boost converter (for example) limits the amount of charge transferred on each cycle. (In the specific case of a boost converter, the lower voltage one would probably just shut off, leaving the voltage at 10V) – user253751 Mar 21 '18 at 21:40
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Figure 1 says that there is 10 V between the top and bottom wires at the same time as it is 5 V. This is impossible. With real power supplies (not ideal ones) a very high current would flow and this would be limited by the internal resistance of the power supplies.

Figure 2 says that there is 5 A flowing in the circuit loop at the same time there is 2 A flowing in the loop. This is also impossible. The 5 A supply will increase its output voltage to infinity to try to drive 5 A. In a real PSU the 5 A supply will increase its output voltage to the maximum it can.

Neil_UK
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Transistor
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