Mathematics in Physics

JEE Main level, single correct, medium

115 of 121

Question 115 of 121

The current voltage relation of diode is given by I = (e1000 V/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\pm \text{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 ?

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Answer: (A) 0.2 mA

I=(e1000V/T−1)mA\text{I}=\left({\text{e}}^{\text{1000} \text{V}/\text{T}}-1\right)\text{mA}

WhenI=5mA,e1000V/T=6mA\text{When} \text{I}=5 \text{mA}\text{,} {\text{e}}^{\text{1000} \text{V}/\text{T}}=6 \text{mA}

Also, dI=(e1000V/T)×1000T⋅dV\text{dI} =\left({\text{e}}^{\text{1000} \text{V}/\text{T}}\right)\times \dfrac{\text{1000}}{\text{T}}\cdot \text{dV}

=(6mA)×1000300×(0.01)=\left(6 \text{mA}\right)\times \dfrac{\text{1000}}{\text{300}}\times \left(\text{0.01}\right)

=0.2mA= \text{0.2} \text{mA}

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