Question

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

Answer 1

Given
$dV=\pm 0.01\text{V}$
$T=300\text{K}$
$I=5\text{mA}$

$I=(e^{\frac{1000V}{T}}-1)\text{mA}$

$\Rightarrow 5+1=e^{\frac{1000V}{T}}$

$\Rightarrow 6=e^{\frac{1000V}{T}}.....\left(1\right)$

Now differentiating equation of $I$, we get

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

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

Using equation(1)

$dI=\frac{1000}{300}\times 6\times 0.01$
$dI=0.2\text{mA}$