A meter bridge with two resistances R₁ and R₂ 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 Ω resistance is connected in parallel to R₂. The values of resistances R₁ and R₂ ar $e______.$

Question

G1 Physics · Accepted Answer

Initially the meter bridge balances at 40 cm from P. Therefore, R₁/R₂ = 40/60 = 2/3. (1) When a 16 Ω resistor is connected in parallel with R₂, the balance point becomes 50 cm. At 50 cm, the two bridge-wire lengths are equal, so the two gap resistances must be equal: R₁ = R₂(parallel). The parallel combination is R₂(parallel) = 16R₂/(16 + R₂). Hence, R₁ = 16R₂/(16 + R₂). (2) Using equation (1), R₁ = 2R₂/3. Therefore, 2R₂/3 = 16R₂/(16 + R₂). Cancel R₂: 2/3 = 16/(16 + R₂). Thus, 2(16 + R₂) = 48,…

A meter bridge with two resistances R₁ and R₂ 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 Ω resistance is connected in parallel to R₂. The values of resistances R₁ and R₂ ar e______.