(B) In a potentiometer,the null point length $l$ is directly proportional to the potential difference $V$ across the points being measured,i.e.,$V = k

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(B) In a potentiometer,the null point length $l$ is directly proportional to the potential difference $V$ across the points being measured,i.e.,$V = kl$,where $k$ is the potential gradient.For the potential difference between $A$ and $B$,the $EMF$ is $E_1$. Thus,$E_1 = k(0.9)$.For the potential difference between $A$ and $C$,the net $EMF$ is $E_1 - E_2$. Thus,$E_1 - E_2 = k(0.3)$.Dividing the two equations:$\frac{E_1}{E_1 - E_2} = \frac{0.9}{0.3} = 3$$E_1 = 3(E_1 - E_2)$$E_1 = 3E_1 - 3E_2$$3E_2 = 2E_1$$\frac{E_2}{E_1} = \frac{2}{3}$

Class 12 Physics Current Electricity Potentiometer

Medium

$A$ potentiometer is used to measure the potential difference between $A$ and $B$. The null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. The ratio $\frac{E_2}{E_1}$ is $\left(E_1 > E_2\right)$.

MHT CET 2022 All MHT CET

A
$3:1$
B
$2:3$
C
$1:3$
D
$1:2$

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$A$ 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 $m$)

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In a potentiometer experiment, a cell of emf $1.25 \,V$ gives a balancing length of $30 \,cm$. If the cell is replaced by another cell, the balancing length is found to be $40 \,cm$. What is the emf of the second cell?

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$A$ potentiometer is used to measure the potential difference between $A$ and $B$. The null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. The ratio $\frac{E_2}{E_1}$ is $\left(E_1 > E_2\right)$.

Similar Questions

In a potentiometer, the area of cross-section of the wire is $4 \, cm^2$, the current flowing in the circuit is $1 \, A$ and the potential gradient is $7.5 \, V/m$, then the resistivity of the potentiometer wire is

The current in the primary circuit of a potentiometer is $0.2 \, A$. The specific resistance and cross-section of the potentiometer wire are $4 \times 10^{-7} \, \Omega \cdot m$ and $8 \times 10^{-7} \, m^2$ respectively. The potential gradient will be equal to .............. $V/m$.

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

$A$ 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 $m$)

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