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Q.
In a meter bridge the point $D$ is a neutral point as shown in figure.
Current Electricity
Solution:
When there is a neutral point at $D$ in meter-bridge, then
$\frac{R}{S}=\frac{l_{1}}{(100-l_{1})}$
For the given values of $R$ and $S$, there will be only one value of $l_{1}$ for which we shall get the neutral point on bridge wire.
In this case $V_{A}-V_{B}=V_{A}-V_{D}$ or $V_{B}=V_{D}$.
Therefore the galvanometer shows no deflection when jockey contacts a point at $D$. There is no current in galvanometer arm.
When jockey contacts a point $D_{1}$ on meter-bridge wire left of $D$, the resistance of arm $A D_{1}$ becomes smaller than previous value.
Due to it, $V_{A}-V_{B}>V_{A}-V_{D_{1}}$ or $V_{D_{1}}>V_{B}$.
Therefore current flows to $B$ from the wire through galvanometer.
When jockey contacts a point $D_{2}$ on meter bridge wire right of $D$, the resistance of arm $A D_{2}$ becomes more than first value.
Due to it, $V_{A}-V_{D_{2}}>V_{A}-V_{B}$ or $V_{B}>V_{D_{2}}$ Therefore the current will flow from $B$ to the wire through galvanometer.
When $R$ is increased, the neutral point will shift to the right instead of left on bridge wire.
