In the case of a Wheatstone bridge,the balanced condition is.....

For the given circuit,the currents $I$ and $I_2$ are respectively:

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:

In the circuit shown below,the ammeter reading is zero. Then the value of the resistance $R$ is (in $Omega$)

For the circuit shown in the figure,the current through the $6 \Omega$ resistor connected between the junctions $A$ and $B$ is (in $A$)

In the following circuit,the current $I_3$ is