(C) For a balanced Wheatstone bridge,the ratio of resistances in the arms is equal: $\frac{A}{B} = \frac{C}{D}$.Given,in the first case: $\frac{A}{B}

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(C) For a balanced Wheatstone bridge,the ratio of resistances in the arms is equal: $\frac{A}{B} = \frac{C}{D}$.Given,in the first case: $\frac{A}{B} = \frac{100}{D} \quad ... (1)$When $A$ and $B$ are interchanged,the new ratio becomes $\frac{B}{A} = \frac{121}{D} \quad ... (2)$Multiplying equation $(1)$ and $(2)$:$\left(\frac{A}{B}\right) \times \left(\frac{B}{A}\right) = \left(\frac{100}{D}\right) \times \left(\frac{121}{D}\right)$$1 = \frac{12100}{D^2}$$D^2 = 12100$$D = \sqrt{12100} = 110 \ \Omega$.

Class 12 Physics Current Electricity Kirchhoff's Law and Whitestone Bridge Circuit solving

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Four resistors $A, B, C$ and $D$ form a Wheatstone bridge. The bridge is balanced when $C = 100 \ \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \ \Omega$. The value of $D$ is: (in $\Omega$)

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A
$10$
B
$100$
C
$110$
D
$120$

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Four resistors $A, B, C$ and $D$ form a Wheatstone bridge. The bridge is balanced when $C = 100 \ \Omega$. If $A$ and $B$ are interchanged,the bridge balances for $C = 121 \ \Omega$. The value of $D$ is: (in $\Omega$)

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