In a metre bridge experiment the balance point is obtained if the gaps are closed by 2Ω and 3Ω. A shunt of X Ω is added to 3Ω resistor to shift the balancing point by 22.5 cm. The value of X is ___________.

Solution

Physics sir · 4 steps

A shunt can only lower the 3 Ω side, so the balance point must move away from the 2 Ω side.

1. First balance. With 2 Ω in the left gap and 3 Ω in the right, 2/3 = l/(100 - l). This gives l = 40 cm from the left end.
2. Which way it shifts. A shunt in parallel makes the right-gap resistance smaller than 3 Ω. That raises the ratio 2/S, so l must grow: l = 40 + 22.5 = 62.5 cm.
3. New right resistance. 2/S = 62.5/37.5 = 5/3, so S = 1.2 Ω. This S is the parallel combination of 3 Ω and X.

   3X/(3 + X) = 1.2 ⇒ 3X = 3.6 + 1.2X ⇒ X = 3.6/1.8

4. So. X = 2 Ω, so the answer is 2.

The common slip: 1.2. Stopping at the combined resistance of the right gap gives 1.2, not 2. That 1.2 Ω is 3 Ω and X in parallel, so you still have to solve for X.

The idea to keep. In a metre bridge, first decide the direction of the shift from physics: a shunt lowers a resistance, a series resistor raises it. Then solve the new balance and undo the parallel combination.

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