Recently on a book i came across a problem-
Here when calculating the current flow through the diode they didn't consider the voltage drop across the diode.If it is correct then what am i missing?
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2\$\begingroup\$ You are missing "assume the diode to be ideal". Ideal diodes don't have any voltage drop. \$\endgroup\$– Peter BennettCommented Dec 12, 2017 at 18:40
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2\$\begingroup\$ What a silly problem, If diode is ideal ... Id = V/R1 => 10/50... \$\endgroup\$– Trevor_GCommented Dec 12, 2017 at 18:52
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1\$\begingroup\$ @Trevor I think it's a reasonable "trick question" to make sure that the student understands that no current will flow through R2 if there is a short-circuit. \$\endgroup\$– pipeCommented Dec 12, 2017 at 19:30
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1\$\begingroup\$ @pipe Indeed. If nothing else it proves Thevenin's theorem :) \$\endgroup\$– Trevor_GCommented Dec 12, 2017 at 19:37
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2 Answers
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From the question:
Assume the diode to be ideal
There is no voltage drop across ideal diodes.
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1\$\begingroup\$ Which makes the whole problem rather silly. \$\endgroup\$– Trevor_GCommented Dec 12, 2017 at 18:46
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4\$\begingroup\$ Well the title of the question says "Current flow across a diode", so I believe we've entered the silly valley already at the title. \$\endgroup\$ Commented Dec 12, 2017 at 18:47
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\$\begingroup\$ @HarrySvensson ya I was talking about the exercise in the book though, not the OPs question.. \$\endgroup\$– Trevor_GCommented Dec 12, 2017 at 18:59
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Since ideal diode has no voltage drop and it is forward biased, it can be considered as a short. Thus R2 is shorted and the current flowing will be just : I = V/R1.
