The current voltage relation of diode is given by $$I = (e^{1000V/T} - 1)$$ mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring $$\pm 0.01$$ V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?

The current-voltage characteristic of the diode is given as

$$I = \bigl(e^{1000V/T}-1\bigr)\ \text{mA}.$$

Here $$I$$ is in milliampere, $$V$$ is the applied voltage in volts and $$T$$ is the absolute temperature in kelvin.

We are interested in the small error produced in the current when there is a small error in the measured voltage. The standard rule of error propagation for a function of one variable states:

$$\Delta I = \left|\frac{dI}{dV}\right|\;\Delta V,$$

where $$\Delta V$$ is the uncertainty in voltage and $$\Delta I$$ is the resulting uncertainty in current.

First we differentiate the given relation. We have

$$I(V) = e^{1000V/T}-1.$$

Differentiating with respect to $$V$$ gives

$$\frac{dI}{dV} = \frac{1000}{T}\,e^{1000V/T}.$$

Now the measurement is made at temperature $$T = 300\ \text{K}$$, and the reported current is $$I = 5\ \text{mA}.$$ From the defining equation we can determine the exponential term at this operating point:

$$5 = e^{1000V/T} - 1 \quad\Longrightarrow\quad e^{1000V/T} = 5 + 1 = 6.$$

Substituting $$T = 300\ \text{K}$$ and $$e^{1000V/T}=6$$ into the derivative we obtained, we find

$$\frac{dI}{dV} = \frac{1000}{300}\times 6 = \frac{10}{3}\times 6 = \frac{60}{3} = 20\ \text{mA per volt}.$$

The stated voltage uncertainty is $$\Delta V = \pm 0.01\ \text{V}.$$ Applying the error-propagation formula,

$$\Delta I = \left|\frac{dI}{dV}\right|\;\Delta V = 20\ \text{mA V}^{-1}\times 0.01\ \text{V} = 0.20\ \text{mA}.$$

Hence, the correct answer is Option A.