Units and Dimensions — JEE Main PYQ Quiz

Units and Dimensions — JEE Main Practice Quiz

Question:
The current–voltage relation of a diode is given by \( I = \left(e^{\frac{1000V}{T}} - 1\right)\,\text{mA} \), where the applied voltage \(V\) is in volts and the temperature \(T\) is in Kelvin. A student makes an error of ±0.01 V while measuring the current of 5 mA at 300 K. What will be the resulting error in the current (in mA)?

Choose the correct option:

A) 0.2 mA
B) 0.02 mA
C) 0.5 mA
D) 0.05 mA

Correct Answer: A) 0.2 mA

Solution:
\( I = \left(e^{\frac{1000V}{T}} - 1\right)\,\text{mA} \)

Given: \( I = 5 \text{ mA}, \; T = 300 \text{ K} \)

\( 5 = e^{\frac{1000V}{300}} - 1 \Rightarrow 6 = e^{\frac{1000V}{300}} \)

Differentiate: \( \frac{dI}{dV} = e^{\frac{1000V}{T}} \cdot \frac{1000}{T} = 6 \cdot \frac{1000}{300} = 20 \, \text{mA/V} \)

Error: \( \Delta I = 20 \times 0.01 = 0.2 \text{ mA} \)

Difficulty: Medium
Tags: Units and Dimensions, Error Analysis, JEE Physics