Question:
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 \(\pm0.01\) V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA ?
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 \(\pm0.01\) 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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Differentiate I with respect to V and use \(e^{1000V/T}=I+1=6\).
Updated On: Nov 11, 2026
- 0.2 mA
- 0.02 mA
- 0.5 mA
- 0.05 mA
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The Correct Option is A
Solution and Explanation
\(I=e^{1000V/T}-1\). For \(I=5\) mA: \(e^{1000V/T}=6\).
Differentiate: \(dI=e^{1000V/T}\cdot\frac{1000}{T}\,dV\).
\(dI=6\times\frac{1000}{300}\times0.01=0.2\) mA.
Answer: (1) 0.2 mA
Differentiate: \(dI=e^{1000V/T}\cdot\frac{1000}{T}\,dV\).
\(dI=6\times\frac{1000}{300}\times0.01=0.2\) mA.
Answer: (1) 0.2 mA
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