The magnitude of current in the diode is - 4.19 mA in the reverse situation.
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<u>Given data</u>
forward-bias situation
current, I = 200 mA
potential difference V = 100mV
temperature = 300 K
reversed potential difference = - 100mV
<h3>How to find the magnitude of current in the diode in reverse bias situation</h3>
The current voltage relationship in an ideal diode is given by
where
k = Boltzmann constant in J/k
T = temperature in Kelvin, K
forward bias situation
Reverse bias situation
V = -100 v
I = 4.19 mA = 4.19*10^-3 A
Therefore the magnitude of current in the diode is 4.19 mA
Read more about diode current here: brainly.com/question/26540960
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