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

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(C) For a balanced Wheatstone bridge,the ratio of resistances is given by $\frac{A}{B} = \frac{C}{D}$.Let $D = x$. Initially,$\frac{A}{B} = \frac{100}

Vedclass · Accepted Answer

$110$

Vedclass · Answer

(C) For a balanced Wheatstone bridge,the ratio of resistances is given by $\frac{A}{B} = \frac{C}{D}$.Let $D = x$. Initially,$\frac{A}{B} = \frac{100}{x}$.When $A$ and $B$ are interchanged,the new condition is $\frac{B}{A} = \frac{121}{x}$.From the first equation,$\frac{B}{A} = \frac{x}{100}$.Equating the two expressions for $\frac{B}{A}$,we get $\frac{x}{100} = \frac{121}{x}$.$x^2 = 121 	imes 100 = 12100$.$x = \sqrt{12100} = 110 \ \Omega$.Thus,the value of $D$ is $110 \ \Omega$.

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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