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
Hide Solution
Verified By Collegedunia
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.
Was this answer helpful?
0
0
Top Questions on Uncertainty in Measurement
If the uncertainty in velocity and position of a minute particle in space are, \(2.4 × 10^{–26}\) \((m s^{–1)}\) and \(10^{–7} (m)\), respectively. The mass of the particle in g is _____ . (Nearest integer)
(Given : \(h = 6.626 × 10^{–34} Js\))- JEE Main - 2022
- Chemistry
- Uncertainty in Measurement
- The density of a material in $CGS$ system of units is $4\,g\,cm^3$. In a system of units in which unit of length is $10\,cm$ and unit of mass is $100\,g$. the value of density of material will be
- NEET (UG) - 2011
- Chemistry
- Uncertainty in Measurement
- The magnitude of any physical quantity
- Punjab PMET - 2006
- Chemistry
- Uncertainty in Measurement
- Match the following prefixes with their multiples:
S.No Prefixes Multiples (i) micro 106 (ii) deca 109 (iii) mega 10–6 (iv) giga 10–15 (v) femto 10 - CBSE Class XI
- Chemistry
- Uncertainty in Measurement
- How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034- CBSE Class XI
- Chemistry
- Uncertainty in Measurement
View More Questions
Questions Asked in JEE Main exam
- If the equation of the parabola with vertex $\left(\frac{3}{2},\, 3\right)$ and the directrix $x + 2y = 0$ is
$$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, $$ then $\alpha + \beta + \gamma$ is equal to:- JEE Main - 2025
- Calculus
- In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $ t_1 $ and $ t_2 $ (s), respectively. The ratio $ t_1 / t_2 $ will be:
- JEE Main - 2025
- Chemical Kinetics
- HA $ (aq) \rightleftharpoons H^+ (aq) + A^- (aq) $ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given: $ K_f(H_2O) = 1.8 \, \text{K kg mol}^{-1} $, molality ≡ molarity
- JEE Main - 2025
- Colligative Properties
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)
- JEE Main - 2025
- Stoichiometry and Stoichiometric Calculations
- Which of the following binary mixture does not show the behavior of minimum boiling azeotropes?
- JEE Main - 2025
- Solutions
View More Questions
Concepts Used:
Uncertainty in Measurement
- Atoms and molecules consist of very low mass. But they are present in huge numbers.
- Chemists have to deal with numbers as large as 602,200,000,000,000,000,000,000, which is the number of molecules of 2g of hydrogen.
- There are even other constants like the speed of light, charges on particles, Avogadro’s number and etc.
- To handle these large or small numbers, we use the following notation: m × 10n, which is, m times ten raised to the power of n. In this expression, n is an exponent having positive and negative values and m is that number that varies from 1.000… and 9.999…
- The scientific notation 573.672 can be written as 5.73672 × 102 and in the same way, 0.000089 can also be written as 8.9 × 10-5.
- All of these help us to attain easier handling, better precision, and accuracy while performing operations on numbers with high magnitudes.
Read More: Uncertainty in Measurement