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Assume the voltages sources(InducedV1/V2/V3) are created by an inductor/or a strong magnetic field changing rapidly(or in anyway) the PS would power the load, however, there are induced-EMF's V1 in support while as V2/V3 in opposition. Since they are parallel and connected in series, I believe they all cancel out, the remaining voltage is 20V and the load current drawn is 0.2A from the power supply.

I was wondering, is the diode(D1) blocking any current? In the parallel sub-circuit of InducedV1 and the diode and the wire?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ D1 has a lower voltage on its anode than its cathode, so if it was conducting then current would be flowing backwards through it. So it's not conducting. \$\endgroup\$ – user253751 Apr 25 '15 at 22:01
  • \$\begingroup\$ D1 is clearly reverse biased hence off. From zero reference on the lower rail: up through PS;up through V1; down through V2 and/or V3, giving +20+10-10 = 20V on the load. Just go on a tour of the circuit and do what you're told by every voltage source you encounter on your journey. \$\endgroup\$ – Chu Apr 26 '15 at 0:14
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At the cathode of D1 relative to ground (the bottom line on your diagram) there is +30V. V2 and V3 are equal and in parallel and subtract 10V from the 30V leaving 20 volts relative to ground i.e. the load sees 20 volts and diode D1 does not conduct.

If the induced voltage was produced by an alternating field it will change direction after some time and the diode will become forward biased and may indeed destroy itself with that current.

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  • \$\begingroup\$ However, the current is divide in that sub-circuit isn't it? Assume the resistance is equal and it's 100ohms, that's 0.2A of current flow. At the parallel sub-circuit of D1 and IV1, the current divides 0.10A throughout that parallel circuit? Or is it 0.20A passing through IV1 since D1 does not conduct? And since All the induced-Voltages are DC and cancel out, the diode should conduct because of it's voltage at the anode is 20V? And cathode is 20V? \$\endgroup\$ – Pupil Apr 25 '15 at 21:31
  • \$\begingroup\$ The diode is reverse biased by V1 it's as simple as that - it's got nothing to do with the 20V nor the other induced voltage sources. The current flow equally divides between V2 and V3. \$\endgroup\$ – Andy aka Apr 25 '15 at 21:41
  • \$\begingroup\$ Thanks Andy, what if the diode was removed? Will this somehow short the circuit? \$\endgroup\$ – Pupil Apr 26 '15 at 0:57
  • \$\begingroup\$ The diode is reverse biased therefore removing it will have no effect. \$\endgroup\$ – Andy aka Apr 26 '15 at 8:42
  • \$\begingroup\$ On a conceptual level, wouldn't the cathode of the diode be 20V total? After canceling out with the V2&V3? Prior to the conduction of any current the circuit? \$\endgroup\$ – Pupil May 20 '15 at 9:19
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many points are unclear in this circuit. like impedance or V1, V2 and V3.

regarding your question, if D1 is blocking current or not...

notice one thing that, V1's polarity is supporting the main voltage source. if we consider the voltage sources to be ideal then, for load current, the path with V1 is best as it offers no resistance and diode will never conduct.

(or else provide some more information or description for this circuit)

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