The current-voltage relation of a diode is gi | vedclass

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(D) The given current-voltage relation is $I = (e^{1000V/T} - 1) 	ext{ mA}$.Given $I = 5 	ext{ mA}$,we have $5 = e^{1000V/T} - 1$,which implies $e^{

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$0.2$

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(D) The given current-voltage relation is $I = (e^{1000V/T} - 1) 	ext{ mA}$.Given $I = 5 	ext{ mA}$,we have $5 = e^{1000V/T} - 1$,which implies $e^{1000V/T} = 6$.To find the error in current $(dI)$,we differentiate the expression with respect to $V$:$dI = \frac{d}{dV} (e^{1000V/T} - 1) \cdot dV = \left( e^{1000V/T} \cdot \frac{1000}{T} ight) dV$.Given $dV = 0.01 	ext{ V}$ and $T = 300 	ext{ K}$,substitute the values:$dI = (6) \cdot \left( \frac{1000}{300} ight) \cdot (0.01)$.$dI = 6 \cdot \left( \frac{10}{3} ight) \cdot 0.01 = 2 \cdot 10 \cdot 0.01 = 0.2 	ext{ mA}$.Thus,the error in the value of current is $0.2 	ext{ mA}$.

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