The current voltage relation of diode is given by I=(e1000 V / T-1) mA, where the applied voltage V is in volt and the temperature T is in 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=(e1000 V / T-1) mA, where the applied voltage V is in volt and the temperature T is in 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 m A (B) 0.02 mA (C) 0.5 mA (D) 0.05 mA

Tardigrade Admin · Accepted Answer

Given, I=(e1000 V / T-1) mA , d V=± 0.01 V T=300 K So, I=e1000 V / T -1 I+1=e1000 V / T Taking log on both sides, we get log (I+1)=(1000 V /T) On differentiating, ⋅ (d I/I+1)=(1000/T) d V d I=(1000/T) ×(I+1) d V ⇒ d I=(1000/300) ×(5+1) × 0.01=0.2 mA So, error in the value of current is 0.2 mA.

Q. The current voltage relation of diode is given by I=(e1000V/T−1)mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ±0.01V while measuring the current of 5mA at 300K, what will be the error in the value of current in mA ?

0.2 m A

0.02 mA

0.5 mA

0.05 mA

Given, I=(e1000V/T−1)mA,dV=±0.01V T=300K So, I=e1000V/T−1 I+1=e1000V/T Taking log on both sides, we get log(I+1)=T1000V​ On differentiating, ⋅I+1dI​=T1000​dV dI=T1000​×(I+1)dV ⇒dI=3001000​×(5+1)×0.01=0.2mA So, error in the value of current is 0.2mA.