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

(D) Both right-sided and left-sided deflection and at balance point, no deflection
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: \(C \rightarrow A \rightarrow D\) and \(C \rightarrow B \rightarrow D\).
The potential at node A is given by the voltage divider rule:
\(V_A=V\left(\frac{R_{A D}}{R_1+R_{A D}}\right)\)
Similarly, the potential at node B is:
\(V_B=V\left(\frac{R_{D B}}{R_2+R_{D B}}\right)\)
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.