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?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

0.5 mA

0.02 mA

0.05 mA

0.2 mA

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE