Solution:
Solution:
Given, I = $$ ( e^{ 1000 V / T } - 1) $$ mA, dV = $$\pm 0.01$$ V
T = 300 K
So, I = $$ e^{ 1000 V /T } - 1 $$
I + 1 = $$ e^{ 1000 V / T} $$
Taking log on both sides, we get log (I + 1) = $$ \frac{1000 \, V}{ T} $$
On differentiating. $$ \frac{ dI}{ I + 1} = \frac{ 1000}{ T} $$ dV
dI = $$ \frac{ 1000}{ T} \times (I + 1) $$ dV
$$\Rightarrow dI = \frac{ 1000}{ 300} \times ( 5 + 1) \times 0.01 $$ = 0 . 2 m A
So, error in the value of current is 0.2 mA.
