
For zero galvanometer deflection, the bridge must be balanced. This is a Wheatstone bridge arrangement.
The condition for a balanced Wheatstone bridge is:
$\frac{R_1}{R_2} = \frac{R_3}{R_4}$
Where R1, R2, R3, and R4 represent the resistance in each arm of the Wheatstone bridge. Here, R1 and R2 are the resistances of the wire segments AJ and JB respectively, and R3 = 8 Ω and R4 = 12 Ω.
The length of AB is 40 cm, thus, $\frac{R_1}{R_2} = \frac{8}{12} = \frac{2}{3}$
$\frac{R_1}{R_2} = \frac{x}{40 - x} = \frac{2}{3}$
3x = 80 − 2x 5x = 80 x = 16 cm
Distance of J from B = 40 cm − 16 cm = 24 cm
Figure shows a current carrying square loop ABCD of edge length is $ a $ lying in a plane. If the resistance of the ABC part is $ r $ and that of the ADC part is $ 2r $, then the magnitude of the resultant magnetic field at the center of the square loop is: 
The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.