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
- A$\frac{I}{\rho A}$
- B$\frac{I A}{\rho}$
- C$I A \rho$
- D$\frac{I \rho}{A}$
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Similar Questions
$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-
Medium
View Solution$A$ potentiometer wire is $4 \text{ m}$ long and a potential difference of $2 \text{ V}$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $80 \text{ cm}$ of the potentiometer wire is: (in $\text{ V}$)
DifficultMHT CET 2026
View Solution$A$ potentiometer wire is $4 \text{ m}$ long and a potential difference of $3 \text{ V}$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $100 \text{ cm}$ of the potentiometer wire is: (in $\text{ V}$)
DifficultMHT CET 2026
View SolutionFor the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because
$A$ cell can be balanced against $110 \, cm$ and $100 \, cm$ of potentiometer wire,respectively with and without being short-circuited through a resistance of $10 \, \Omega$. Its internal resistance is ............... $\Omega$.
MediumAIPMT 2008
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