The given figure represents an arrangement of | vedclass

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

Question

(A) The formula for the internal resistance $(r)$ of a cell using a potentiometer is given by:$r = R \left( \frac{l_1}{l_2} - 1 \right)$where $l_1$ is

Vedclass · Accepted Answer

$22.1$

Vedclass · Answer

(A) The formula for the internal resistance $(r)$ of a cell using a potentiometer is given by:$r = R \left( \frac{l_1}{l_2} - 1 ight)$where $l_1$ is the balancing length when the key is open (measuring $EMF$ $E$) and $l_2$ is the balancing length when the key is closed (measuring terminal voltage $V$).Given:$l_1 = 70.0 \, cm$$l_2 = 60.0 \, cm$$R = 132.40 \, \Omega$Substituting the values into the formula:$r = 132.40 \left( \frac{70.0}{60.0} - 1 ight)$$r = 132.40 \left( \frac{70.0 - 60.0}{60.0} ight)$$r = 132.40 \left( \frac{10.0}{60.0} ight)$$r = 132.40 	imes \frac{1}{6}$$r = 22.066 \, \Omega \approx 22.1 \, \Omega$Thus,the correct option is $A$.

Similar Questions

In a potentiometer circuit, a cell of $EMF$ $1.5\, V$ gives a balance point at $36\, cm$ length of wire. If another cell of $EMF$ $2.5\, V$ replaces the first cell, then at what length of the wire will the balance point occur? (in $cm$)

$AB$ is a potentiometer wire as shown in the figure. If the value of $R$ is increased,in which direction will the balance point $J$ shift?

In a potentiometer experiment,the balancing length with a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is ................. $\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module