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V2 is blocking any current passing diode(1) because of the voltage being higher,however, V3 is equal and opposing V2, therefore canceling out and allowing current flow through the diode?

schematic

simulate this circuit – Schematic created using CircuitLab

Assume V2/V3 are induced-EMF's going beyond the PS.

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  • \$\begingroup\$ @Samuel adjusted and edited, V1/V2 are induced-EMF's from an exterior source or they could be powerful inductors and so on... just wanted to see the effect on an experiment. \$\endgroup\$ – Pupil Apr 25 '15 at 17:21
  • \$\begingroup\$ Absorbing PS into V1 and V2 would make the same arrangement. This means V2 is effectively 30V and V1 is effectively 15V. There still won't be any current flowing through the diode, it will be flowing out of V2 and into V3. \$\endgroup\$ – Samuel Apr 25 '15 at 17:30
  • \$\begingroup\$ Why wouldn't it be the other way around as well? V2/V3 cancel out leaving 10V throughout the circuit and 15V for V1? \$\endgroup\$ – Pupil Apr 25 '15 at 17:45
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    \$\begingroup\$ @Key, please clarify the last statement about the EMF's - it doesn't make sense to my childish brain :) \$\endgroup\$ – Dzarda Apr 25 '15 at 17:47
  • \$\begingroup\$ Assume V2/V3 being induced-EMF sources via a changing magnetic field, or an inductor, if they cancel out since they are equal and opposing what's left is V1 and the PS? \$\endgroup\$ – Pupil Apr 25 '15 at 17:49
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Let me redraw the central part of the circuit. Without the diode being present, current will flow clockwise. When D1 is inserted, it is reverse-biased, because V2 is higher than V1 (they are in parallel). Reverse-biased means no current flows.

From then on, you can treat this entire thing as its Thevenin equivalent - no matter what you do the path V1-D1 will always be inert, because V2 makes sure D1 stays reverse-biased.

schematic

simulate this circuit – Schematic created using CircuitLab


Just to clarify this issue, let's make a thought experiment.

  1. The circuit of V1, V2, D1 is clearly a sub-circuit of the circuit in question.
  2. Voltage between NODE1 and NODE2 is guaranteed to be 20V, since V2 is connected directly between the two. (V2 being an ideal voltage source, voltage across it cannot change by definition.)
  3. Since V1-D1 and V2 are in parallel, they share the same voltage across them.
  4. When you subtract 5V from 20V (this is perfectly legal with voltage sources), you're left with 15V between D1's anode and cathode.
  5. Diode will act as an open circuit, since the potential at its cathode is higher than its anode = it is reverse biased.
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  • \$\begingroup\$ However, V3 cancels out with V2,allowing current to flow from V1 and from the PS to the load. \$\endgroup\$ – Pupil Apr 25 '15 at 17:23
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    \$\begingroup\$ @Key V3 as no effect whatsoever on V2. \$\endgroup\$ – Dzarda Apr 25 '15 at 17:24
  • \$\begingroup\$ How? They are equal and opposing? \$\endgroup\$ – Pupil Apr 25 '15 at 17:26
  • \$\begingroup\$ See edit.654321 \$\endgroup\$ – Dzarda Apr 25 '15 at 17:44
  • \$\begingroup\$ I'm still trying to connect V3 to V2(or to the circuit generally), which you said had no effect, if they are induced simultaneously, being equal and opposite don't they cancel out since they are connected in series to oppose on another? \$\endgroup\$ – Pupil Apr 25 '15 at 17:51
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Take the bottom rail as reference 0V. The junction V1/V2 is 10V (going up through PS), hence the anode is at 15V, the cathode is at 30V, therefore the diode is off. Drop the 30V down through 20V (through V3), giving 10V at the top of the load, current through load = 0.1A

i.e. o.1A flows clockwise through PS, V2, V3, load resistor

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  • \$\begingroup\$ That makes sense, I'm assuming the voltage is 10V because V2/V3 cancel out, and because the remaining voltage is 10V in reverse bias, the V1 is useless and blocked out by D1. Am I right here? \$\endgroup\$ – Pupil Apr 25 '15 at 20:17
  • \$\begingroup\$ Start at reference zero and travel clockwise through each component. For a voltage source, if you go from - to +, you go up by the appropriate voltage and if you go from + to - you go down. So up by 10V through PS, and that makes the junction V1/V2 = 10V. Then up by 5V through V2, that makes the anode 15V. From jn V1/V2 up by 20V through V3, making the cathode 30V, so the diode must be off. Thus there's 30V at the jn of V2/V3/D1. Finally, down by 20V through V3 which makes the voltage at the top of R = 10V. \$\endgroup\$ – Chu Apr 26 '15 at 0:03

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