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Q.

The current -voltage relation of diode is given by $I = (e^{1000 \frac{V}{T}} - 1) mA,$ 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 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?

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a

0.5 mA

b

0.02 mA

c

0.05 mA

d

0.2 mA

answer is C.

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Detailed Solution

Given 5 = $e^{1000 \frac{V}{T}} - 1$

$\Rightarrow e^{1000} \frac{V}{T} = 6$ -----(i)

Again, I = $e^{1000 \frac{V}{T}} - 1$

$\frac{dI}{dV} = e^{\frac{1000}{T} V} \frac{1000}{T}$

$dl = \frac{1000}{T} e^{\frac{1000}{T} V} dV$

Using (i) $∆ I = \frac{1000}{T} \times 6 \times 0.01 = \frac{60}{T} = \frac{60}{300} = 0.2 mA$

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The current -voltage relation of diode is given by I = (e1000VT-1)mA, where the applied voltage V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ±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?