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 ±0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA ?

Question

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 ±0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA ? (A) 0.2 mA (B) 0.02 mA (C) 0.5 mA (D) 0.05 mA

Tardigrade Admin · Accepted Answer

I=(e1000 V / T -1) mA When I=5 mA , e1000 V / T =6 mA Also, d I=(e1000 V / T ) × (1000/ T ) ⋅ d V =(6 mA ) × (1000/300) ×(0.01) =0.2 mA

0.2 mA

0.02 mA

0.5 mA

0.05 mA

I=(e1000V/T−1)mA When I=5mA,e1000V/T=6mA Also, dI=(e1000V/T)×T1000​⋅dV =(6mA)×3001000​×(0.01) =0.2mA

$I = (e^{1000 V / T} - 1) m A$
When $I = 5 m A, e^{1000 V / T} = 6 m A$
Also, $d I = (e^{1000 V / T}) \times \frac{1000}{T} \cdot d V$
$= (6 m A) \times \frac{1000}{300} \times (0.01)$
$= 0.2 m A$