A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at

Question

A meter bridge with two resistances $R_1$ and $R_2$ as shown in figure was balanced (null point) at 40 cm from the point $P$. The null point changed to 50 cm from the point $P$, when $16 \Omega$ resistance is connected in parallel to $R_2$. The values of resistances $R_1$ and $R_2$ are $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

Initially, the balance point is at $40	ext{ cm}$. Thus, $\frac{R_1}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3} \Rightarrow R_1 = \frac{2}{3}R_2$. When a $16\Omega$ resistor is connected in parallel to $R_2$, the new equivalent resistance is $R_2' = \frac{16R_2}{16+R_2}$. The new balance point is at $50	ext{ cm}$, so $\frac{R_1}{R_2'} = \frac{50}{50} = 1 \Rightarrow R_1 = R_2'$. Substituting $R_1$: $\frac{2}{3}R_2 = \frac{16R_2}{16+R_2}$. Solving this yields $32 + 2R_2 = 48 \Rightarrow 2R_2 = 16 \Rightarrow R_2 = 8 \Omega$. Therefore, $R_1 = \frac{2}{3}(8) = \frac{16}{3} \Omega$.