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For a diode, the current–voltage relation is I = (e^(1000V/T) - 1) mA, where V is the applied voltage in volts and T is the temperature in kelvin. If the current is 5 mA at 300 K and the voltage measurement has an uncertainty of ±0.01 V, what is the resulting error in the current (in mA)?

1. 0.2 mA
2. 0.02 mA
3. 0.5 mA
4. 0.05 mA

Correct answer: 0.2 mA

Solution

I = e^(1000V/T)-1 = 5 mA -> e^(1000V/T) = 6. Then dI/dV = (1000/T)e^(1000V/T) = (1000/300)(6) = 20 mA/V. Error dI = 20 x 0.01 = 0.2 mA.

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