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

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(D) Given the relation: $I = (e^{1000V/T} - 1) \; mA$.At $I = 5 \; mA$,we have $5 = e^{1000V/T} - 1$,which implies $e^{1000V/T} = 6$.Differentiating t

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$0.2$

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(D) Given the relation: $I = (e^{1000V/T} - 1) \; mA$.At $I = 5 \; mA$,we have $5 = e^{1000V/T} - 1$,which implies $e^{1000V/T} = 6$.Differentiating the expression for $I$ with respect to $V$:$\frac{dI}{dV} = e^{1000V/T} 	imes \frac{1000}{T}$.For small errors,we can write $\Delta I = \frac{dI}{dV} \Delta V$.Substituting the known values: $\Delta I = (e^{1000V/T}) 	imes \frac{1000}{T} 	imes \Delta V$.Since $e^{1000V/T} = 6$,$T = 300 \; K$,and $\Delta V = 0.01 \; V$:$\Delta I = 6 	imes \frac{1000}{300} 	imes 0.01$.$\Delta I = 6 	imes \frac{10}{3} 	imes 0.01 = 20 	imes 0.01 = 0.2 \; mA$.

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