Question

The current voltage relation of diode is given by $I=\left(e^{1000 V / T}-1\right) mA$, where the applied voltage $V$ is in volt and the temperature $T$ is in kelvin. If a student makes an error measuring $\pm 0.01\, V$ while measuring the current of

• A. 0.2 m A
• B. 0.02 mA
• C. 0.5 mA
• D. 0.05 mA

Answer 1

Answer: (A)

0.2 m A

Answer 2

Given, I = $( e^{ 1000 V / T } - 1)$ mA, dV = $\pm 0.01$ V
T = 300 K
So, I = $e^{ 1000 V /T } - 1$
I + 1 = $e^{ 1000 V / T}$
Taking log on both sides, we get log (I + 1) = $\frac{1000 \, V}{ T}$
On differentiating. $\frac{ dI}{ I + 1} = \frac{ 1000}{ T}$ dV
dI = $\frac{ 1000}{ T} \times (I + 1)$ dV
$\Rightarrow dI = \frac{ 1000}{ 300} \times ( 5 + 1) \times 0.01$ = 0 . 2 m A
So, error in the value of current is 0.2 mA.