Four resistors A, B, C and D form a Wheatston | vedclass

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(C) For a balanced Wheatstone bridge,the condition is $\frac{A}{B} = \frac{C}{D}$.Case $1$: When $C = 100 \Omega$,the bridge is balanced,so $\frac{A}{

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$110$

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(C) For a balanced Wheatstone bridge,the condition is $\frac{A}{B} = \frac{C}{D}$.Case $1$: When $C = 100 \Omega$,the bridge is balanced,so $\frac{A}{B} = \frac{100}{D}$ ... $(i)$Case $2$: When $A$ and $B$ are interchanged,the new resistances in the arms are $B$ and $A$ respectively. The bridge balances for $C = 121 \Omega$,so $\frac{B}{A} = \frac{121}{D}$ ... (ii)From equation $(i)$,we have $\frac{A}{B} = \frac{100}{D}$. Therefore,$\frac{B}{A} = \frac{D}{100}$.Substituting this into equation (ii),we get $\frac{D}{100} = \frac{121}{D}$.$D^2 = 100 	imes 121 = 12100$.$D = \sqrt{12100} = 110 \Omega$.

Four resistors $A, B, C$ and $D$ form a Wheatstone bridge as shown in the figure. 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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