Two cells having unknown e.m.f.s E_{1} and E_ | vedclass

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

Question

(D) In a potentiometer circuit,when two cells are connected to assist each other,the balancing length $\ell_{1}$ is proportional to $(E_{1} + E_{2})$.

Vedclass · Accepted Answer

$1.45$

Vedclass · Answer

(D) In a potentiometer circuit,when two cells are connected to assist each other,the balancing length $\ell_{1}$ is proportional to $(E_{1} + E_{2})$. Thus,$E_{1} + E_{2} = k \ell_{1}$,where $k$ is the potential gradient.When they are connected to oppose each other,the balancing length $\ell_{2}$ is proportional to $(E_{1} - E_{2})$. Thus,$E_{1} - E_{2} = k \ell_{2}$.Dividing the two equations,we get: $\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{\ell_{1}}{\ell_{2}}$.Using the given values $\ell_{1} = 490 \ cm$ and $\ell_{2} = 90 \ cm$:$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{490}{90} = \frac{49}{9}$.Applying componendo and dividendo:$\frac{E_{1}}{E_{2}} = \frac{49 + 9}{49 - 9} = \frac{58}{40} = 1.45$.

Similar Questions

In a potentiometer experiment,the balancing length with a cell $E_{1}$ of unknown e.m.f. is $\ell_{1} \ cm$. By shunting the cell with a resistance $R \ \Omega$,the balancing length becomes $\frac{\ell_{1}}{2} \ cm$. The internal resistance $(r)$ of the cell is:

In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is

Let $A$ be the cross-sectional area and $\rho$ be the specific resistance (resistivity) of a potentiometer wire. If $I$ is the current passing through the wire,then the potential gradient along the length of the wire is

Sensitivity of a potentiometer can be increased by:

$A$ cell in the 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