Question

Given is
( 5=mathrm{e}^{1000 mathrm{V} / mathrm{T}}-1 )
This implies, ( 6=mathrm{e}^{1000 mathrm{V} / mathrm{T}} )
Again, ( I=e^{1000 V / T}-1 )
( mathrm{dI} / mathrm{d} mathrm{V}=mathrm{e}^{1000 mathrm{V} / mathrm{T}} * 1000 / mathrm{T} )
( mathrm{dI}=mathrm{e}^{1000 mathrm{V} / mathrm{T}} * 1000 / mathrm{T} mathrm{d} mathrm{V} )
( mathrm{dI}=0.2 mathrm{mA} )

# 12. The current voltage relation of diode is given by I = (e1000V/T - 1) mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error measuring +0.01V while measuring the current of 5 mA at 300 K, what will be error in the value of current in ma? (JEE(Main)-2014] (1) 0.5 mA (2) 0.05 mA (3) 0.2 mA (4) 0.02 mA

Solution