Question

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

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

Answer 1

Answer: (B)

$0.02mA$

$\begin{array}{c}I=\text{current}\\ V=\text{Voltage}\\ T=\text{Temperature}\end{array}$

$\begin{array}{c}I=(e^{1000V/T}-1)mA\\ \text{Taking log both side}\\ \log I=\log (e^{\log \theta /T}-1)\end{array}$

$\begin{array}{cc}\log I& =\frac{1000V}{T}\\ \log I& =\frac{1000V}{308}\\ \log I& =\frac{10}{3}\end{array}$

$\text{Differentiating both side}$

$\begin{array}{cc}\frac{\delta I}{I}& =\frac{10dv}{3}\\ \frac{\delta I}{5}& =\frac{10}{3}\times 0.01\\ \delta I& =\frac{0.1}{3}\times 5\\ & =\frac{0.5}{3}=0.166\end{array}$

$\approx 0.2mA$