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In a circuit the voltage drop is equal on all main "branches".

In this diagram the Zener diode has 3.6 break voltage:

schematic

simulate this circuit – Schematic created using CircuitLab

In the example above I can calculate that the voltage drop on R1 will be 5-3.6=1.4 and current 1.4/100 A. On the second branch the voltage drop on R2 will be 5V. This would imply that if I lower R1's value, the voltage drop on R2 should still be 5V. But if I reach a short-circuit state(resistance very small) then of course I am wrong about R2.

Where am I wrong? I think some infinity case will arise about the current in the second case which would allow an exceptional case about the voltage...

Note: I am most certain that a simple Kirchhoff rule would solve the problem but I am more interested in solving this with Ohm's law and common intuition(like voltage splitting in an intersection and currents adding to zero in nodes)

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    \$\begingroup\$ Ideal circuits stop working in some extreme situations. Think of two ideal voltage sources in parallel with a different voltage. A "short circuit" is another situation where ideal circuits often fail. Practically if R1 is zero D1 would blow up (unless it is an ideal zener etc.) \$\endgroup\$
    – Oldfart
    Mar 16 '20 at 20:27
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It's easy.

  • You can't have short-circuits on ideal voltage sources. Infinite current would have to flow.
  • You can't have open circuits on constant-current sources. Infinite voltage would be generated.

In the real world the voltage source's internal resistance would cause the voltage to sag and the voltage drop would occur in the voltage source. Alternatively the electronic current limiter would cut-in and limit the current which in turn would cause the voltage to drop.

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  • \$\begingroup\$ but if R1 were to be for ex. 0.001 ohm then R2 would still have a 5v drop on it right? \$\endgroup\$ Mar 16 '20 at 20:34
  • \$\begingroup\$ Yes, if the voltage source could supply 1.4 / 0.001 = 1400 A with no voltage sag and your Zener could dissipate 3.6 \$ \times \$ 1400 = 5 kW. \$\endgroup\$
    – Transistor
    Mar 16 '20 at 20:38
  • \$\begingroup\$ No, your zener would blow up. (specked to max ~900mA) \$\endgroup\$
    – Oldfart
    Mar 16 '20 at 20:38
  • \$\begingroup\$ what if we put a resistor just before the GND? Then there's no shortcircuit (only for R2..and forget the zener). Thus the circuit would not blow but the voltage should still "split"...or not? Of course now we need to calculate the new current on the whole circuit..and the voltage drop on the new resistor... \$\endgroup\$ Mar 16 '20 at 20:39
  • \$\begingroup\$ With Real components, if R1 was .001 Ohm, there would be a large current flowing through it and the Zener diode. If the Zener could handle more current than the supply can produce, the supply voltage and the voltage on R2 will drop. If the supply can produce more current than the Zener can handle, the Zener will fail, and the supply voltage will remain at 5 volts. In any case, you are in an area where smoke, flame, and other exciting things will occur. \$\endgroup\$ Mar 16 '20 at 20:41
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Ideal components don't exist. For example wires are not superconductors, 5V supply would have source impedance and it would not source infinite currents. The zener would also have internal impedance, the voltage over it would depend on current through it. Real world components also blow up when dissipating too much power as heat. So ideal models don't work when you have some edge case like R1 as zero ohms, it is not withing boundary of sane solutions, as the current won't be infinite or more than few hundred milliamps, and the voltage can't be 3.6V and 5V simultaneously.

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  • \$\begingroup\$ Giving the very low value of the resistors, the zener is likely to burn and the source to see a short circuit situation. Values should be 10x higher to have a usability in the real world. As a theorical problem, it's anti-pedagogical. \$\endgroup\$
    – Fredled
    Mar 16 '20 at 20:58

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