1. Given diode relation

$I = \left(e^{100V/T}-1\right)\,\text{mA}$

We need the error in current when the voltage measurement has an error of

$\Delta V = \pm 0.01\,\text{V}$

at:

• $I = 5\,\text{mA}$
• $T = 300\,\text{K}$

2. Use differential error relation

For small errors,

$\Delta I \approx \left|\frac{dI}{dV}\right|\Delta V$

Now,

$I = \left(e^{100V/T}-1\right)\text{ mA}$

Differentiate with respect to $V$:

$\frac{dI}{dV} = \frac{100}{T}e^{100V/T}\,\text{mA/V}$

At $T=300\,\text{K}$,

$\frac{dI}{dV} = \frac{100}{300}e^{100V/300} = \frac{1}{3}e^{100V/300}\,\text{mA/V}$

3. Find $e^{100V/T}$ using given current

Given current is $5\,\text{mA}$:

$5 = e^{100V/T}-1$

so,

$e^{100V/T} = 6$

4. Compute slope at this operating point

$\frac{dI}{dV} = \frac{100}{300}\times 6 = 2\,\text{mA/V}$

5. Calculate error in current

$\Delta I \approx \frac{dI}{dV}\Delta V = 2 \times 0.01 = 0.02\,\text{mA}$

Thus the error in current is

$\boxed{0.02\,\text{mA}}$

6. Check options

• A: $0.2\,\text{mA}$
• B: $0.02\,\text{mA}$
• C: $0.5\,\text{mA}$
• D: $0.05\,\text{mA}$

So the correct option is:

$\boxed{\text{B}}$

7. Comparison with stored answer

Stored correct answer: A

My derived answer is B. The likely mistake in the stored answer is a decimal-place error, since the correct propagation gives $0.02\,\text{mA}$, not $0.2\,\text{mA}$.