Question

solve this .......☺️

Answer 1

Given :-

I = $\bold{e^{1000V/T}}$ - 1
And,

,5 = e^(1000V/T) - 1

6 = e^(1000V/T)

AND, T = 300k , dv = ±0.01v

Now,

dI/dV = e^(1000V/T) * 1000/T

dI/dv = e^(1000V/T) * 1000/T

dI/dv = 6 * 1000/300

dI/dv = 20

dI = 20 * 0.01

dI = 0.2mA

hence , the error in the value of mA is 0.2mA

Option , (A) is correct

Answer 2

see the attachment given.