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Assuming the diode is ideal but with a forward voltage drop of. 65 volts, what is the voltage vr1, if v1=6v, and r1=921ω?

Sagot :

The ideal diode's current is 0.064 mA at a forward biased voltage drop of 65 V.

An ideal diode is a particular kind of electronic component that has two terminals, and only permits current to flow in one direction, with less zero resistance in one direction and infinite resistance in another.

The most often used type of diode is a semiconductor diode.

It is a crystalline semiconductor part with a PN junction that is connected to two electrical terminals.

We have provided values here.

[tex]V_{2}[/tex]=65 volts

[tex]v_{1} =6v[/tex]

[tex]r_{1} =921w[/tex]

[tex]I=V_{2 - V_{1} /I[/tex]=  65-6/921=0.064

The ideal diode's current is 0.064 mA at a forward biased voltage drop of 65 V.

Here is further information about the ideal diode:

brainly.com/question/14988926

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