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Assuming the diode is ideal but with a forward voltage drop of. 65 volts, what is the current in ma if v1=0v, and r1=490ω?

Sagot :

tanisharawat014

The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.

To find the answer, we have to know more about the ideal diode.

What is an ideal diode?

  • A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
  • A semiconductor diode is the kind of diode that is used the most commonly.
  • It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.

How to find the current in ideal diode?

  • Here we have given with the values,

                       [tex]V_2=65V\\V_1=0V\\R_1=490Ohm.[/tex]

  • We have the expression for current in mA of the ideal diode with forward biased voltage drop as,

                [tex]I=\frac{V_2-V_1}{R_1} =\frac{65}{490} =132.6mA[/tex]

Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.

Learn more about the ideal diode here:

https://brainly.com/question/14988926

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