In a metre bridge experiment (see figure), the positions of the cell, $E$ , and galvanometer, $G$ , are interchanged. We shall observe in the galvanometer :
NEET 2026 (Cancelled)

Question

In a metre bridge experiment (see figure), the positions of the cell, E , and galvanometer, G , are interchanged. We shall observe in the galvanometer : NEET 2026 (Cancelled)

TestHub · Accepted Answer

Let the nodes of the circuit be identified as follows:
Node A: Left L-shaped metallic strip.
Node B: Right L-shaped metallic strip.
Node C: Central metallic strip.
Node D: The point on the wire where the jockey touches.

The resistances in the four arms of the circuit are:
1. Resistance between A and C is $R_{1}$ .
2. Resistance between B and C is $R_{2}$ .
3. Resistance between A and D is $R_{A D}$ (left part of the metre bridge wire).
4. Resistance between B and D is $R_{D B}$ (right part of the metre bridge wire).

In the given figure, the cell $E$ is connected between nodes C and D, and the galvanometer $G$ is connected between nodes A and B. This arrangement forms a Wheatstone bridge.

Let the potential at node D be $0$ and the potential at node C be $V$ .
The current from the cell divides at node C into two parallel branches: C $\to$ A $\to$ D and C $\to$ B $\to$ D.
The potential at node A is given by the voltage divider rule:
$V_{A} = V (\frac{R _{A D}}{R _{1} + R _{A D}})$

Similarly, the potential at node B is:
$V_{B} = V (\frac{R _{D B}}{R _{2} + R _{D B}})$

The galvanometer shows no deflection when the potential difference across it is zero, i.e., $V_{A} = V_{B}$ .
$\frac{R _{A D}}{R _{1} + R _{A D}} = \frac{R _{D B}}{R _{2} + R _{D B}}$
$\Rightarrow \frac{R _{1} + R _{A D}}{R _{A D}} = \frac{R _{2} + R _{D B}}{R _{D B}}$
$\Rightarrow \frac{R _{1}}{R _{A D}} + 1 = \frac{R _{2}}{R _{D B}} + 1$
$\Rightarrow \frac{R _{1}}{R _{A D}} = \frac{R _{2}}{R _{D B}}$

This is the standard balance condition for a Wheatstone bridge. Thus, interchanging the cell and the galvanometer does not affect the balance condition. A null point will still be obtained on the wire.

If the jockey is moved to the left of the balance point, $R_{A D}$ decreases and $R_{D B}$ increases, making $V_{A} < V_{B}$ . Current will flow from B to A through the galvanometer, causing a deflection in one direction.
If the jockey is moved to the right of the balance point, $R_{A D}$ increases and $R_{D B}$ decreases, making $V_{A} > V_{B}$ . Current will flow from A to B, causing a deflection in the opposite direction.

Therefore, the galvanometer will show both right-sided and left-sided deflections depending on the jockey's position, and no deflection at the balance point.

Answer: Both right-sided and left-sided deflection and at balance point, no deflection

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