A cell is connected to a potentiometer, and t | vedclass

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

Question

(D) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 \right)$, where $l_1$ is

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) The formula for the internal resistance $r$ of a cell using a potentiometer is given by $r = R \left( \frac{l_1}{l_2} - 1 \right)$, where $l_1$ is the balancing length without the external resistor and $l_2$ is the balancing length with the external resistor $R$.Given: $l_1 = 2 \, m$, $l_2 = 3 \, m$, and $R = 5 \, \Omega$.Substituting the values into the formula:$r = 5 \left( \frac{2}{3} - 1 \right)$ is incorrect based on the standard potentiometer setup where $l_1$ corresponds to $EMF$ $(E)$ and $l_2$ corresponds to terminal voltage $(V)$.Correct approach: $E = k l_1$ and $V = k l_2$.Since $V = E - Ir = E - \left( \frac{E}{R+r} \right)r = E \left( \frac{R}{R+r} \right)$, we have $\frac{V}{E} = \frac{l_2}{l_1} = \frac{R}{R+r}$.Substituting the values: $\frac{3}{2} = \frac{5}{5+r}$.$3(5+r) = 2(5) \Rightarrow 15 + 3r = 10 \Rightarrow 3r = -5$. This implies the values provided in the prompt lead to a negative resistance, which is physically impossible. Re-evaluating the prompt values: If $l_1 = 3 \, m$ and $l_2 = 2 \, m$:$\frac{2}{3} = \frac{5}{5+r} \Rightarrow 10 + 2r = 15 \Rightarrow 2r = 5 \Rightarrow r = 2.5 \, \Omega$. Given the options provided, if we assume the question intended $l_1 = 3 \, m$ and $l_2 = 2 \, m$ with $R = 5 \, \Omega$, the result is $2.5 \, \Omega$. If the question intended $l_1 = 2 \, m$ and $l_2 = 1.5 \, m$ with $R = 5 \, \Omega$, then $r = 5(2/1.5 - 1) = 5(0.5/1.5) = 1.66 \, \Omega$. Given the options, there is a discrepancy. Assuming the standard calculation $r = R(l_1/l_2 - 1)$, if $l_1=3, l_2=2, R=2$, then $r=1$. If $l_1=3, l_2=2, R=5$, then $r=2.5$. Based on the provided options, $D$ is the closest logical fit if parameters were adjusted.

$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?

Similar Questions

$A$ potentiometer wire of length $4 \, m$ and resistance $5 \, \Omega$ is connected in series with a resistance of $992 \, \Omega$ and a cell of e.m.f. $4 \, V$ with internal resistance $3 \, \Omega$. The length of $0.75 \, m$ on the potentiometer wire balances the e.m.f. of (in $ \, mV$)

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is:

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?

In a potentiometer circuit, there is a cell of $e.m.f.$ $2\, V$, a resistance of $5\, \Omega$ and a wire of uniform thickness of length $1000\, cm$ and resistance $15\, \Omega$. The potential gradient in the wire is:

$A$ wire of length $100\, cm$ is connected to a cell of emf $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential difference of $1\, mV/cm$ is ............. $\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module