An ideal p - n junction diode has a current-voltage characteristic...

The dynamic resistance $r_d$ of a diode is given by the reciprocal of the derivative of current with respect to voltage:

$r_d = \dfrac{dV}{di}$

Given the current-voltage characteristic:

$i = i_0 \left(e^{eV/kT} - 1\right)$

Differentiating with respect to $V$:

\dfrac{di}{dV} =$i_0 \left(\dfrac{e}{kT}\right)$e^{eV/kT}$<br><br>Therefore, the dynamic resistance is:<br><br>$r_d = $\dfrac{kT}{e i_0 e^{eV/kT}}<br><br>At$T = 300 \text{ K}$, the thermal voltage \dfrac{kT}{e} is approximately:<br><br>\dfrac{kT}{e} = \dfrac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} \approx$0.026 \text{ V}$<br><br>Substituting the given values$r_d = 0.2 \Omega and $i_0 = 10 \mu\text{A} = 10^{-5} \text{ A}$:

$0.2 =$\dfrac{0.026}{10^{-5} e^{V/0.026}}

$e^{V/0.026} =$\dfrac{0.026}{0.2 \times 10^{-5}} =$13000$

Taking the natural logarithm on both sides:

\dfrac{V}{0.026} = \ln($13000) \approx 9.47$

$V = 9.47 \times 0.026 \approx 0.246 \text{ V}$

The closest value among the given options is $0.25 \text{ V}$.

Answer: $0.25 \text{ V}$