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 $5 \,mA$ at $300 \,K$, what will be the error in the value of current in $mA$ ?

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

Answer 1

Answer: (A)

Given, $ I=\left(e^{1000 V / T}-1\right) mA , d V=\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, $\cdot \frac{d I}{I+1}=\frac{1000}{T} d V$
$d I=\frac{1000}{T} \times(I+1) d V $
$\Rightarrow d I=\frac{1000}{300} \times(5+1) \times 0.01=0.2 \,mA$
So, error in the value of current is $0.2\, mA$.