Question

​in the circuit shown in the figure the value of resistance X when a potential difference between the points Band D is zero will be

Answer 1

HERE IS YOU ANSWER

HOPE IT HELPS

Answer 2

Answer: 8 Ω​

Explanation: The given circuit is equivalent to the balanced Wheatstone's network.  

Let, resistance in arm AB is P, P=15+6=21Ω

So now the resistance in arm BC is Q,$Q=(\frac{8X}{8+X})+3$Ω​

And now the resistance in arm DA is R, $R=(\frac{(6)(6)}{6+6} )+15=3+15=18$Ω​

Resistance in arm CD is S, $S=(\frac{(4)(4)}{4+4} )+(4)=2+4=6$Ω​

So now by using balancing conditions for Wheatstone's network.

$\frac{P}{Q} =\frac{R}{S}\\ \frac{21}{(\frac{8X}{8+X} +3)}=\frac{18}{6}\\ \frac{21}{(\frac{8X}{8+X} +3)}=3\\(\frac{8X}{8+X} )+3=\frac{21}{3}\\ (\frac{8X}{8+X} )+3=7\\(\frac{8X}{8+X} )=4\\8X=32+4X\\4X=32\\X=\frac{32}{4}\\ X=8$

As the value of resistance X in order that the potential difference between the points B and D is zero, will be 8 Ω​.

#SPJ3