A cell can be balanced against 110 , cm and 1 | vedclass

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

Question

(C) Let $\varepsilon$ be the $EMF$ of the cell and $r$ be its internal resistance. Let $V/L$ be the potential gradient of the potentiometer wire.When

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $\varepsilon$ be the $EMF$ of the cell and $r$ be its internal resistance. Let $V/L$ be the potential gradient of the potentiometer wire.When the cell is not short-circuited,the balancing length is $l_1 = 110 \, cm$. The $EMF$ is balanced as:$\varepsilon = (V/L) 	imes 110$  ....$(i)$When the cell is short-circuited through an external resistance $R = 10 \, \Omega$,the terminal voltage $V_t$ is balanced at $l_2 = 100 \, cm$. The terminal voltage is given by $V_t = \frac{\varepsilon R}{R + r}$.$V_t = (V/L) 	imes 100$  ....$(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{\varepsilon}{V_t} = \frac{110}{100}$$\frac{\varepsilon}{\varepsilon R / (R + r)} = \frac{11}{10}$$\frac{R + r}{R} = 1.1$$1 + \frac{r}{R} = 1.1$$\frac{r}{R} = 0.1$$r = 0.1 	imes R = 0.1 	imes 10 \, \Omega = 1 \, \Omega$.

$A$ 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 ............... $\Omega$.

Similar Questions

In the given potentiometer circuit,the balancing length for points $B$ and $C$ is $40 \, cm$. What is the balancing length for points $C$ and $D$ in $cm$?

In a potentiometer,a balance point is obtained when:

$A$ 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 $\text{ V}$)

$A$ potentiometer wire has length $4 \, m$ and resistance $5 \, \Omega$. It is connected in series with $495 \, \Omega$ resistance and a cell of e.m.f. $4 \, V$. The potential gradient along the wire is (in $V/m$)

The reading of the voltmeter in the following circuit is ................ $V$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module