When the two known resistances R and S are co | vedclass

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

Question

(A) In the first case,the meter bridge balance condition is given by: $\frac{R}{l_1} = \frac{S}{100-l_1} \implies R = S \frac{l_1}{100-l_1}$.In the se

Vedclass · Accepted Answer

$\frac{S l_1(100-l_2)}{100(l_2-l_1)}$

Vedclass · Answer

(A) In the first case,the meter bridge balance condition is given by: $\frac{R}{l_1} = \frac{S}{100-l_1} \implies R = S \frac{l_1}{100-l_1}$.In the second case,$X$ is connected in parallel with $S$. The equivalent resistance $S'$ is given by: $S' = \frac{XS}{X+S}$.The new balance condition is: $\frac{R}{l_2} = \frac{S'}{100-l_2} = \frac{XS}{(X+S)(100-l_2)}$.Substituting $R$ from the first equation: $\frac{S l_1}{(100-l_1) l_2} = \frac{XS}{(X+S)(100-l_2)}$.Simplifying: $\frac{l_1}{l_2(100-l_1)} = \frac{X}{(X+S)(100-l_2)} \implies \frac{X+S}{X} = \frac{l_2(100-l_1)}{l_1(100-l_2)}$.$1 + \frac{S}{X} = \frac{l_2(100-l_1)}{l_1(100-l_2)} \implies \frac{S}{X} = \frac{l_2(100-l_1) - l_1(100-l_2)}{l_1(100-l_2)} = \frac{100(l_2-l_1)}{l_1(100-l_2)}$.Therefore,$X = \frac{S l_1(100-l_2)}{100(l_2-l_1)}$.

Similar Questions

In a meter bridge experiment,the null point is obtained at $20 \ cm$ from one end of the wire when resistance $X$ is balanced against resistance $Y$. If $X < Y$,then the distance of the new null point from the same end when a resistance of $4X$ is balanced against $Y$ will be ............ $cm$.

In a metre bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance point from the left is: (in $cm$)

In a meter bridge experiment,the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $X \Omega$ is added to the $3 \Omega$ resistor to shift the null point by $22.5 \text{ cm}$. The value of $X$ is: (in $Omega$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module