A current of left(frac{2}{sqrt{3}}right) text | vedclass

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

Question

(B) The principle of a tangent galvanometer is given by the formula $I = K 	an 	heta$,where $I$ is the current,$K$ is the reduction factor,and $	he

Vedclass · Accepted Answer

$\left(\frac{2}{3}ight) 	ext{ A}$

Vedclass · Answer

(B) The principle of a tangent galvanometer is given by the formula $I = K 	an 	heta$,where $I$ is the current,$K$ is the reduction factor,and $	heta$ is the deflection angle.Given current $I = \frac{2}{\sqrt{3}} 	ext{ A}$ and deflection $	heta = 60^{\circ}$.Rearranging the formula to solve for the reduction factor $K$:$K = \frac{I}{	an 	heta}$Substituting the given values:$K = \frac{2 / \sqrt{3}}{	an 60^{\circ}}$Since $	an 60^{\circ} = \sqrt{3}$,we have:$K = \frac{2 / \sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3} 	imes \sqrt{3}} = \frac{2}{3} 	ext{ A}$.Thus,the reduction factor is $\frac{2}{3} 	ext{ A}$.

$A$ current of $\left(\frac{2}{\sqrt{3}}\right) \text{ A}$ produces a deflection of $60^{\circ}$ in a tangent galvanometer. The reduction factor is:

Similar Questions

The period of oscillation of a bar magnet at a place is $2 \,s$. At the same place, the period of oscillation of another identical bar magnet whose magnetic moment is $4$ times that of the first magnet is: (in $\,s$)

The time period of oscillation of a magnet in a vibration magnetometer is $1.5 \ s$. The time period of oscillation of another magnet similar in size,shape,and mass but having one-fourth magnetic moment than that of the first magnet,oscillating at the same place,will be.....$s$.

In the sum and difference method of a vibration magnetometer,the time period is greater if:

The time period of oscillation of a freely suspended bar magnet with usual notations is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module