Two students X and Y perform a potentiometer | vedclass

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

Question

(D) The potential drop across the potentiometer wire $AB$ is given by $V_{AB} = E \cdot \frac{R_{AB}}{R + R_{AB} + r}$,where $E$ is the $EMF$ of the d

Vedclass · Accepted Answer

would shift towards point $B$,would shift towards point $A$

Vedclass · Answer

(D) The potential drop across the potentiometer wire $AB$ is given by $V_{AB} = E \cdot \frac{R_{AB}}{R + R_{AB} + r}$,where $E$ is the $EMF$ of the driver cell,$R$ is the external resistance,$R_{AB}$ is the resistance of the wire,and $r$ is the internal resistance of the driver cell.$(i)$ When student $X$ increases $R$,the total resistance of the primary circuit increases,so the current $I = \frac{E}{R + R_{AB} + r}$ decreases. Consequently,the potential gradient $k = \frac{V_{AB}}{L}$ decreases. Since the balancing condition is $E_1 = k \cdot l$,where $l$ is the balancing length,if $k$ decreases,$l$ must increase to maintain the same $E_1$. Thus,the null point shifts towards $B$.(ii) When student $Y$ decreases $S$,the terminal potential difference across the cell $E_1$ is $V = E_1 - I_1 r_1$,where $I_1 = \frac{E_1}{S + r_1}$. Decreasing $S$ increases the current $I_1$ drawn from the cell $E_1$,which increases the voltage drop $I_1 r_1$ across the internal resistance $r_1$. Therefore,the terminal voltage $V$ across the cell $E_1$ decreases. Since $V = k \cdot l$,a decrease in $V$ requires a smaller balancing length $l$. Thus,the null point shifts towards $A$.

Similar Questions

$A$ battery of $6\,V$ is connected to the terminals of a $3\,m$ long wire of uniform thickness and resistance of the order of $100\,\Omega$. The difference of potential between two points separated by $50\,cm$ on the wire will be .......... $V$.

In a potentiometer experiment for the determination of the internal resistance of a cell,when an external resistance of $R$ is connected in parallel to the cell,the balancing length decreases by $10 \%$. The internal resistance of the cell is

To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module