I=(e1000V/T−1)mA
WhenI=5mA,e1000V/T=6mA
Also, dI=(e1000V/T)×T1000⋅dV
=(6mA)×3001000×(0.01)
=0.2mA
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 ?
Held on 6 Apr 2014 · Verified 6 Jul 2026.
0.2 mA
0.02 mA
0.5 mA
0.05 mA
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