Question

The Current voltage relation of diode is given by $I=(e^{1000V/T}-1)\text{mA}$, where the applied voltage $V$ is in volts and the temperature $T$ is in degree Kelvin. If a student makes an error measuring $\pm 0.01\text{V}$ while measuring the current of $5\text{mA}$ at $300\text{K}$, what will be the error in the value of current in $\text{mA}$ ?

• A. $0.05\text{mA}$
• B. $0.02\text{mA}$
• C. $0.2\text{mA}$
• D. $0.5\text{mA}$

Answer 1

The correct option is C $0.2\text{mA}$

Given
$I=e^{\frac{1000V}{T}}-1$
$I+1=e^{\frac{1000V}{T}}$
On applying logarithm,
$\log (I+1)=\frac{1000V}{T}$

On differentiating, above equation

$\frac{d(I+1)}{I+1}=\frac{1000}{T}dV$

$\frac{dI}{I+1}=\frac{1000}{T}dV$

$\frac{dI}{(5+1)\text{mA}}=\frac{1000}{300}\left(0.01\right)$

$dI=0.2\text{mA}$