The resistance R = frac{V}{I} where V = (200 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given,$R = \frac{V}{I}$.According to the rules of error analysis for division,the relative error in $R$ is given by:$\frac{\Delta R}{R} = \frac{\D

Vedclass · Accepted Answer

$3.5$

Vedclass · Answer

(A) Given,$R = \frac{V}{I}$.According to the rules of error analysis for division,the relative error in $R$ is given by:$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$Substituting the given values $V = 200 \ V$,$\Delta V = 5 \ V$,$I = 20 \ A$,and $\Delta I = 0.2 \ A$:$\frac{\Delta R}{R} = \frac{5}{200} + \frac{0.2}{20}$$\frac{\Delta R}{R} = 0.025 + 0.01 = 0.035$To find the percentage error,multiply by $100$:$	ext{Percentage error} = \frac{\Delta R}{R} 	imes 100 = 0.035 	imes 100 = 3.5 \%$

The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) \ V$ and $I = (20 \pm 0.2) \ A$. The percentage error in the measurement of $R$ is: (in $\%$)

Similar Questions

The relative error in the determination of the surface area of a sphere is $\alpha$. Then the relative error in the determination of its volume is

The maximum percentage errors in the measurement of mass $(M)$,radius $(R)$,and angular velocity $(\omega)$ of a ring are $2 \%$,$1 \%$,and $1 \%$ respectively. Find the maximum percentage error in the measurement of its angular momentum $(J = I \omega)$ about its geometrical axis. (in $\%$)

If $\theta_1 = 25.5 \pm 0.1 \, ^\circ C$ and $\theta_2 = 35.3 \pm 0.1 \, ^\circ C$,find $\theta_1 - \theta_2$.

Error in the measurement of radius of the sphere is $2 \%$. The error in the calculated value of its volume is (in $\%$)

What is called as relative error? Define fractional error.

Vedclass Test Series

Exam Paper Generator

Online Exam Module