Answer:
Option A
Explanation:
Given, $I=(e^{1000V/T}-1)mA$
$dV=\pm0.01V $
$T=300K\Rightarrow mA$
So, $I= e^{1000V/T}-1$
$I+1= e^{1000V/T}$
Taking log onboth sides , we get
log(I+1)= $\frac{1000V}{T}$
On differentiating, $\frac{dI}{I+1}=\frac{1000}{T}dV$
$dI= \frac{1000}{T}\times (I+1)dV$
$\Rightarrow dI= \frac{1000}{300}\times (5+1)0.01=0.2mA$
So, error in the value of current is 0.2 mA