Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length l₁ is 40 cm. Now an unknown resistance x is connected in series with P and new balancing length is found to be 80 cm measured from the same end. Then the value of x will be ____________ Ω.

Solution

Physics sir · 3 steps

Use the first balance to find Q, then use the second balance to find what P has grown into.

1. First balance. At balance, P/Q = l₁/(100 - l₁). With P = 4 Ω and l₁ = 40 cm, this gives 4/Q = 40/60, so Q = 6 Ω.
2. Second balance. Now x is in series with P, so the left gap holds 4 + x. The new balancing length is 80 cm from the same end A, which leaves 20 cm on the Q side.

   (4 + x)/6 = 80/20 = 4

3. So. 4 + x = 24, which gives x = 20 Ω.

The common slip: 24. Reporting the new left-gap resistance 4 + x gives 24, not 20. Subtract the original 4 Ω to get x alone.

The idea to keep. In a metre bridge, each balance point gives the ratio of the two gap resistances. A series addition changes only one side, so first find the fixed side, then solve for the new total.

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