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 16Ω resistance is connected in parallel to R₂. The values of resistances R₁ and R₂ are ____ .

Solution

Physics sir · 4 steps

Two balance points give two ratio equations. Solve them together and both resistances come out.

1. First balance. At balance, R₁/R₂ = l/(100 - l), with l measured from P on the R₁ side. With l = 40 cm, R₁/R₂ = 40/60 = 2/3, so R₁ = 2/3R₂.
2. Second balance. With 16 Ω across R₂, the right gap holds 16R₂/(16 + R₂). The null point is now at 50 cm, so the two gaps are equal.

   R₁ = 16R₂/(16 + R₂) ⇒ 2/3R₂ = 16R₂/(16 + R₂)

3. Solve. Cancel R₂: 2(16 + R₂) = 48, so R₂ = 8 Ω. Then R₁ = 2/3 × 8 = 16/3 Ω.
4. So. R₂ = 8 Ω and R₁ = 16/3 Ω, which is option (a). Check: 8 in parallel with 16 gives 16/3 Ω, equal to R₁, so the balance sits at 50 cm.

The idea to keep. In a metre bridge, every balance point gives one ratio equation. Changing a resistance and getting a new null point gives the second equation you need for two unknowns.

Answer from the NTA final answer key. Solution by PrepWiser.