Question

In the figure shown for which values of $R_1$ and $R_2$ the balance point for Jockey is at $40cm$ from $A$. When $R_2$ is shunted by a resistance of $10\Omega$, balance shifts to $50cm$. Find and $R_2$ in ohms $(AB=1m)$:

Answer 1

Given that,

$AB=1m=100cm,$

$l=40cm$

The balancing condition for meter bridge is

$\frac{R1}{R2}=\frac{l}{100-l}$

$R1=R2\left(\frac{l}{100-l}\right)$

$R1=R2\left(\frac{40}{100-40}\right)$

$R1=R2\left(\frac{40}{60}\right)=\frac{4}{6}R2$.......(1)

When R2 is shunted by a resistance of 10 , l = 50 cm hence,

$R1-10=R2\left(\frac{50}{100-50}\right)$

$R1-10=R2$

$\frac{4}{6}R2-10=R2$ ........from(1)

$2R2=10$

$R2=5\Omega$

Using this in (1), we get

$R1=\left(\frac{4}{6}\right)5=\frac{10}{3}\Omega$