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Q.

The current voltage relation of diode is given by I=(e1000V/T1)mA, where the applied voltage V is in volts and the temperature T is in degree Kelvin. If a student makes an error of ±0.01V while measuring the current of 5mA at 300K, what will be the error in the value of current in  mA?

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a

0.5mA

b

0.2mA

c

0.02mA

d

0.05mA

answer is B.

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Detailed Solution

As, I=(e1000V/T1)mA     ….(i)

Here, I=5mA at T=300K

dV=0.01V, 5=(e1000V/T1)e(1000V/T)=6mA

Differentiating eqn. (i), we get dI=(1000T)e(1000V/T)dV

dI=1000300(6)(0.01)=0.2mA

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