In a potentiometer experiment,the null point | vedclass

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

Question

(B) The potential gradient $x$ of a potentiometer wire is given by $x = \frac{V}{L} = \frac{E R}{(R + R') L}$,where $E$ is the e.m.f. of the driver ce

Vedclass · Accepted Answer

Increase resistance in main circuit

Vedclass · Answer

(B) The potential gradient $x$ of a potentiometer wire is given by $x = \frac{V}{L} = \frac{E R}{(R + R') L}$,where $E$ is the e.m.f. of the driver cell,$R$ is the resistance of the potentiometer wire,$R'$ is the resistance in the main circuit,and $L$ is the total length of the wire.For a given cell of e.m.f. $\epsilon$,the balancing length $l$ is given by $\epsilon = x \cdot l$,which implies $l = \frac{\epsilon}{x}$.To shift the null point from the $7^{	ext{th}}$ wire to the $9^{	ext{th}}$ wire,the balancing length $l$ must increase.Since $l = \frac{\epsilon}{x}$,increasing $l$ requires a decrease in the potential gradient $x$.Looking at the expression $x = \frac{E R}{(R + R') L}$,to decrease $x$,we must increase the denominator term $(R + R')$.Therefore,we must increase the resistance $R'$ in the main circuit.

In a potentiometer experiment,the null point is obtained on the $7^{\text{th}}$ wire for a given cell. To shift the null point to the $9^{\text{th}}$ wire for the same cell,what should we do?

Similar Questions

When two cells are connected in series in a potentiometer circuit to assist each other,the balancing length is $6 \ m$. When they are connected in series to oppose each other,the balancing length is $2 \ m$. What is the ratio of the $EMF$ of the two cells?

In a potentiometer experiment,the balancing length is $8 \ m$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition,the balancing length is $4 \ m$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is

$A$ $10 \, m$ long wire of $20 \, \Omega$ resistance is connected with a battery of $3 \, V$ $e.m.f.$ (negligible internal resistance) and a $10 \, \Omega$ resistance is joined to it in series. The potential gradient along the wire in $V/m$ is:

$A$ cell in a secondary circuit gives a null deflection for $2.5 \ m$ length of a potentiometer wire having a total length of $10 \ m$. If the length of the potentiometer wire is increased by $1 \ m$ without changing the cell in the primary circuit,the new position of the null point is: (in $m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module