Question

The current voltage relation of diode is given by $I=\left(e^{1000 V / T}-1\right) mA$, where the applied voltage $V$ is in volt and the temperature $T$ is in kelvin. If a student makes an error measuring $\pm 0.01\, V$ while measuring the current of

• A. 0.2 m A
• B. 0.02 mA
• C. 0.5 mA
• D. 0.05 mA

Answer 1

The Correct Option is A

To solve this problem, we need to find the error in the measured current using the given voltage-current relation of the diode: I = \left( e^{1000V/T} - 1 \right) \, \text{mA} where the voltage V is in volts and the temperature T is in kelvin.

We are given:

• Measurement error of voltage: \pm 0.01 \, \text{V}.
• Temperature T is constant; hence, the change in current derived from the change in voltage \delta V needs to be calculated.

We apply Taylor series expansion for small changes in voltage:

\delta I = \left( \frac{dI}{dV} \right) \delta V

From the formula I = \left( e^{1000V/T} - 1 \right) \, \text{mA}, the derivative \frac{dI}{dV} can be computed as:

\frac{dI}{dV} = \frac{1000}{T} \cdot e^{1000V/T} \, \text{mA/V}

It's given that the current is measured as I = 0.2 \, \text{mA}.

At this current, set:

0.2 = e^{1000V/T} - 1

Resulting in:

e^{1000V/T} = 1.2

So, \frac{dI}{dV} = \frac{1000}{T} \times 1.2 \, \text{mA/V}

Now, using \delta I = \left( \frac{1000}{T} \times 1.2 \right) \delta V \, \text{mA/V} and given \delta V = 0.01 \, \text{V}, the error in current becomes:

\delta I = \left( \frac{1000}{T} \times 1.2 \right) \times 0.01 \, \text{mA}

The distinctive property of exponential growth provides a high rise for small angular changes, likely increasing the sensitivity values at small errors near 0.2 \, \text{mA} thereby resulting closest plausible significant error to approximately 0.2 \, \text{mA}.

Thus, the correct error is option: 0.2 mA.