A 10 ,m long potentiometer wire is connected | vedclass

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

Question

(B) The potential gradient $k$ of a potentiometer wire is given by $k = \frac{V}{L} = \frac{I \cdot R}{L} = \frac{I \cdot \rho \cdot L}{A \cdot L} = \

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The potential gradient $k$ of a potentiometer wire is given by $k = \frac{V}{L} = \frac{I \cdot R}{L} = \frac{I \cdot \rho \cdot L}{A \cdot L} = \frac{I \cdot \rho}{A}$,where $I$ is the current,$\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.However,in a standard potentiometer circuit,the potential drop across the wire is $V_{wire} = I \cdot R_{wire} = I \cdot \frac{\rho \cdot L}{A}$.When the cell is balanced,the emf of the cell $E = k \cdot l$,where $l$ is the balancing length.If the length $L$ of the potentiometer wire is kept constant and the cross-section $A$ is doubled,the resistance of the wire $R = \rho \frac{L}{A}$ becomes half.Since the battery voltage is steady,the current $I$ through the wire increases,but the potential drop across the wire remains equal to the battery voltage.Thus,the potential gradient $k = \frac{V_{battery}}{L}$ remains unchanged because both $V_{battery}$ and $L$ are constant.Since $E = k \cdot l$ and both $E$ and $k$ are constant,the balancing length $l$ remains $4 \,m$.

$A$ $10 \,m$ long potentiometer wire is connected to a battery having a steady voltage. $A$ Leclanche cell is balanced at $4 \,m$ length of the wire. If the length is kept the same,but its cross-section is doubled,the null point will be obtained at ........... $m$.

Similar Questions

If the resistivity of a potentiometer wire is $\rho$ and the area of cross-section is $A$,then what will be the potential gradient along the wire,given that a current $I$ flows through it?

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:

$A$ $10 \, m$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module