In the given figure,the value of resistance X | vedclass

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(C) The potential difference between $B$ and $D$ is zero,which means the Wheatstone bridge is in a balanced condition.For a balanced Wheatstone bridge

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$8$

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(C) The potential difference between $B$ and $D$ is zero,which means the Wheatstone bridge is in a balanced condition.For a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal:$\frac{R_{AB}}{R_{BC}} = \frac{R_{AD}}{R_{DC}}$First,calculate the equivalent resistances of the arms:$R_{AB} = 15 + 6 = 21 \, \Omega$$R_{BC} = 3 + \frac{8X}{8 + X} \, \Omega$$R_{AD} = 15 + (6 || 6) = 15 + 3 = 18 \, \Omega$$R_{DC} = 4 + (4 || 4) = 4 + 2 = 6 \, \Omega$Now,apply the balanced condition:$\frac{21}{3 + \frac{8X}{8 + X}} = \frac{18}{6}$$\frac{21}{3 + \frac{8X}{8 + X}} = 3$$7 = 3 + \frac{8X}{8 + X}$$4 = \frac{8X}{8 + X}$$4(8 + X) = 8X$$32 + 4X = 8X$$4X = 32$$X = 8 \, \Omega$

In the given figure,the value of resistance $X$ will be,when the potential difference between $B$ and $D$ is zero. (in $\Omega$)

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