In a Wheatstone bridge circuit where the four arms have resistances $P, Q, R,$ and $S$ arranged in order, if the bridge is balanced, the potential difference between the two mid-points $A$ and $B$ (where the galvanometer is connected) is .............. $V$.

In the cube shown,current $i$ enters at $H$ and leaves at $C$. If $i_{AB} = \frac{i}{6}$,$i_{DC} = \frac{2i}{3}$,$i_{HA} = \frac{i}{2}$,$i_{GF} = \frac{i}{6}$,and $i_{HE} = \frac{i}{6}$,choose the branch in which the current is zero.

In the given circuit,it is observed that the current $I$ is independent of the value of the resistance $R_6$. Then the resistance values must satisfy

If the internal resistance of the battery is $1\, \Omega$, then what is the reading of the ammeter?

The Wheatstone bridge principle is used to measure the specific resistance $(S_1)$ of a given wire,having length $L$ and radius $r$. If $X$ is the resistance of the wire,then the specific resistance is: $S_1 = X \left( \frac{\pi r^2}{L} \right)$. If the length of the wire is doubled,then the value of the specific resistance will be:

The equivalent resistance of the following circuit when no current flows in the resistance of $5 \Omega$ is nearly (in $\Omega$)