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
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
Updated On: Apr 28, 2025
- 0.2 m A
- 0.02 mA
- 0.5 mA
- 0.05 mA
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The Correct Option is A
Solution and Explanation
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.
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.
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