Question

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

• A. $\frac{10}{3}\Omega ,5\Omega$
• B. $20\Omega ,30\Omega$
• C. $10\Omega ,15\Omega$
• D. $5\Omega ,\frac{15}{2}\Omega$

Answer 1

Answer: (A)

$\frac{10}{3}\Omega ,5\Omega$

For balancing condition we can write,

$R_1=R_2\frac{l}{100-l}$

in first case,

$R_1=R_2\frac{40}{100-40}=R_2\frac{4}{6}=R_2\frac{2}{3}$ .....(1)

in second case,

$R_1=\left(\frac{10R_2}{10+R_2}\right)\frac{l}{100-l}=\left(\frac{10R_2}{10+R_2}\right)\left(\frac{50}{100-50}\right)$

$R_1=\frac{10R_2}{10+R_2}$

from(1)

$R_2\frac{2}{3}=\frac{10R_2}{10+R_2}$

$-10R_2+2R_2^2=0$

$2R_2=10$

$R_2=5\Omega$

hence, $R_1=R_2\frac{2}{3}=\left(5\right)\frac{2}{3}=\frac{10}{3}\Omega$