The given diode current-voltage relation is: $I=(e^{\frac{1000V}{T}}-1)$ mA, where V is in volts and T is in Kelvin. We are told that at T=300K, the measured current is I=5mA, and there is an error in voltage measurement of $\mathrm{\Delta }V=\pm 0.01$ V. We need to find the resulting error in current, $\mathrm{\Delta }I$.

Step 1: Find the operating voltage V.
We know I=5 mA and T=300K. Let's plug these values into the diode equation to find V. $5=e^{\frac{1000V}{300}}-1$
$e^{\frac{10V}{3}}=6$ (since 5 + 1 = 6)
Taking natural logarithm (ln) on both sides:
$\frac{10V}{3}=\ln \left(6\right)$
$V=\frac{3}{10}\ln \left(6\right)$
We will use this expression for V later. The exact numerical value is not needed for the error calculation.

Step 2: Relate the error in current to the error in voltage.
The error $\mathrm{\Delta }I$ is approximately given by the derivative of I with respect to V, multiplied by the error $\mathrm{\Delta }V$.
$\mathrm{\Delta }I\approx \left|\frac{dI}{dV}\right|\times \mathrm{\Delta }V$

Step 3: Calculate the derivative dI/dV.
The current equation is $I=e^{\frac{1000V}{T}}-1$.
Differentiating with respect to V:
$\frac{dI}{dV}=\frac{1000}{T}\times e^{\frac{1000V}{T}}$
From Step 1, we found that at our operating point (I=5mA, T=300K), $e^{\frac{1000V}{T}}=6$.
Substituting T=300 and this value into the derivative:
$\frac{dI}{dV}=\frac{1000}{300}\times 6=\frac{10}{3}\times 6=20$ mA/V

Step 4: Calculate the error in current.
Now we can find $\mathrm{\Delta }I$:
$\mathrm{\Delta }I\approx \left|\frac{dI}{dV}\right|\times \mathrm{\Delta }V=20\frac{\mathrm{mA}}{V}\times 0.01V=0.2$ mA

Final Answer: The error in the value of the current is $0.2$ mA.

Related Concepts and Formulae

Error Propagation: When a quantity y depends on another quantity x (y = f(x)), the error in y ($\mathrm{\Delta }y$) due to a small error in x ($\mathrm{\Delta }x$) is given by $\mathrm{\Delta }y\approx \left|\frac{dy}{dx}\right|\times \mathrm{\Delta }x$. This formula is derived from the first-order Taylor series expansion and is valid for small errors.

Diode Equation: The given equation is a simplified version of the Shockley diode equation, $I=I_s(e^{\frac{V}{\eta V_T}}-1)$, where $I_s$ is the saturation current, $\eta$ is the ideality factor, and $V_T=\frac{kT}{q}$ is the thermal voltage (~26 mV at 300K). The constant 1000 in the exponent suggests it's a normalized form of this standard equation.