Answer
372.2k+ views
Hint: Potentiometer is a three-terminal resistor, which produces a voltage divider, through sliding or rolling contact. $r=R\left( \dfrac{{{l}_{1}}}{{{l}_{2}}}-1 \right)$ and $E=\dfrac{\rho L}{A}I=KL$, we can solve the question.
Formula used: $r=R\left( \dfrac{{{l}_{1}}}{{{l}_{2}}}-1 \right)$ and, $E=\dfrac{\rho L}{A}I=KL$
Complete step-by-step answer:
Potentiometer is a three-terminal resistor, which produces a voltage divider, through sliding or rolling contact. It works on the principle that potential across the wire depends on the length of the wire, which has uniform cross-sectional area and constant current flowing through it. It can be used to measure the internal resistance of the primary cell.
Connections are made as given in the diagram.
We know, $V=IR$ Ohm’s law. Where $I$ is current in the circuit, $R$ is the total resistance and $V$ is the voltage drop.
Also $R=\dfrac{\rho L}{A}$. Where $\rho$ is the resistivity, $A$ is the area of cross-section of the wire, $L$ is the length of the wire.
For any potentiometer $\rho\;,A$ are constant.
Then, according to the principle, $E=\dfrac{\rho L}{A}I=KL$, where $E$ is the emf of the cell and $K$ is the potential gradient.
Here $E$ is the emf of the primary cell, and $r$ is its internal resistance. To find $r$, plug in${K_{2}}$ the resistance box $R$, the potentiometer is balanced at say $l_{1}$,
Then$E=Kl_{1}=I(R+r)$.
When the plug out the $K_{2}$, when the potentiometer is balanced at say $l_{2}$, then$V=Kl_{2}=IR$
$\dfrac{E}{V}=\dfrac{ Kl_{1}}{ Kl_{2}}=\dfrac{ I(R+r)}{IR}$
$\dfrac{ l_{1}}{ l_{2}}=\dfrac{ (R+r)}{R}$
Thus $r=R\left( \dfrac{{{l}_{1}}}{{{l}_{2}}}-1 \right)$
Using the given, emf of cell= $E_{1}= 1.25V\; , l_{1}=35cm$ and $E_{2}=E\;, l_{2}=63cm$
$\dfrac{E_{1}}{E_{2}}=\dfrac{l_{1}}{l_{2}}$
$\dfrac{1.25}{E}=\dfrac{35}{63}$
$E=\dfrac{63}{35}\times 1.25=2.25 V$
Hence the emf of the unknown is 2.25 V.
Note: Clearly understand the difference between what happens when $K_{2}$ is plugged in and out. Try to remember the formula used. Remember the circuit diagram and the principle. Observe that the equations are interrelated, in terms of the potential.
Formula used: $r=R\left( \dfrac{{{l}_{1}}}{{{l}_{2}}}-1 \right)$ and, $E=\dfrac{\rho L}{A}I=KL$
Complete step-by-step answer:
Potentiometer is a three-terminal resistor, which produces a voltage divider, through sliding or rolling contact. It works on the principle that potential across the wire depends on the length of the wire, which has uniform cross-sectional area and constant current flowing through it. It can be used to measure the internal resistance of the primary cell.
![seo images](https://www.vedantu.com/question-sets/96b4b583-4514-4d4d-b326-1f9898dbcca01689518970766740531.png)
Connections are made as given in the diagram.
We know, $V=IR$ Ohm’s law. Where $I$ is current in the circuit, $R$ is the total resistance and $V$ is the voltage drop.
Also $R=\dfrac{\rho L}{A}$. Where $\rho$ is the resistivity, $A$ is the area of cross-section of the wire, $L$ is the length of the wire.
For any potentiometer $\rho\;,A$ are constant.
Then, according to the principle, $E=\dfrac{\rho L}{A}I=KL$, where $E$ is the emf of the cell and $K$ is the potential gradient.
Here $E$ is the emf of the primary cell, and $r$ is its internal resistance. To find $r$, plug in${K_{2}}$ the resistance box $R$, the potentiometer is balanced at say $l_{1}$,
Then$E=Kl_{1}=I(R+r)$.
When the plug out the $K_{2}$, when the potentiometer is balanced at say $l_{2}$, then$V=Kl_{2}=IR$
$\dfrac{E}{V}=\dfrac{ Kl_{1}}{ Kl_{2}}=\dfrac{ I(R+r)}{IR}$
$\dfrac{ l_{1}}{ l_{2}}=\dfrac{ (R+r)}{R}$
Thus $r=R\left( \dfrac{{{l}_{1}}}{{{l}_{2}}}-1 \right)$
Using the given, emf of cell= $E_{1}= 1.25V\; , l_{1}=35cm$ and $E_{2}=E\;, l_{2}=63cm$
$\dfrac{E_{1}}{E_{2}}=\dfrac{l_{1}}{l_{2}}$
$\dfrac{1.25}{E}=\dfrac{35}{63}$
$E=\dfrac{63}{35}\times 1.25=2.25 V$
Hence the emf of the unknown is 2.25 V.
Note: Clearly understand the difference between what happens when $K_{2}$ is plugged in and out. Try to remember the formula used. Remember the circuit diagram and the principle. Observe that the equations are interrelated, in terms of the potential.
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