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Asked in AIPMT 2002 · Circuits with diodes
Idea: the diode and R are in series across the source, so V divides between them in proportion to their resistances.
Forward biased, the junction's resistance is very small compared with R.
The share taken by the diode is then negligible, and almost the whole of V appears across R.
So in forward bias the voltage across R is V.
Reverse biased the position reverses: the junction becomes the large resistance, takes nearly all of V, and leaves R with almost none.
In neither case can any element carry 2V, since the loop has only V to share out.
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