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JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    2. The current voltage relation of diode is given by \[I=\left( {{e}^{1000V/T}}-1 \right)mA,\]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?   JEE Main  Solved  Paper-2014

    A) 0.5 mA                 

    B) 0.05 mA

    C) 0.2 mA

    D) 0.02 mA

    Correct Answer: C

    Solution :

    \[I=\left( {{e}^{\frac{^{1000V}}{T}}}-1 \right)\] \[\frac{dl}{dV}=\frac{1000}{T}{{e}^{\frac{1000V}{T}}}\] \[dI=\left( \frac{1000}{T}{{e}^{\frac{1000V}{T}}} \right)dV\]                        \[I=5MA\] \[\Rightarrow \]\[e\frac{1000V}{T}=6MA\] \[dI=\left( \frac{1000}{300} \right)(0.01)(6)=\frac{10}{50}=\frac{1}{5}\] \[dI=0.2mA\]                


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