In a Wheatstone bridge (see figure),resistanc | vedclass

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(D) For a balanced Wheatstone bridge,the condition is $\frac{P}{Q} = \frac{R}{X}$.Case $1$: When $R = 400 \,\Omega$,the bridge is balanced,so $\frac{P

Vedclass · Accepted Answer

$402$

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(D) For a balanced Wheatstone bridge,the condition is $\frac{P}{Q} = \frac{R}{X}$.Case $1$: When $R = 400 \,\Omega$,the bridge is balanced,so $\frac{P}{Q} = \frac{400}{X} \quad .....(1)$Case $2$: On interchanging $P$ and $Q$,the new balance condition is $\frac{Q}{P} = \frac{405}{X} \quad .....(2)$From equation $(1)$,we have $\frac{Q}{P} = \frac{X}{400}$.Substituting this into equation $(2)$,we get $\frac{X}{400} = \frac{405}{X}$.$X^2 = 400 	imes 405 = 162000$.$X = \sqrt{162000} = \sqrt{400 	imes 405} = 20 	imes \sqrt{405} \approx 20 	imes 20.1246 = 402.492 \,\Omega$.Thus,the value of $X$ is close to $402.5 \,\Omega$.

In a Wheatstone bridge (see figure),resistances $P$ and $Q$ are approximately equal. When $R = 400 \,\Omega$,the bridge is balanced. On interchanging $P$ and $Q$,the value of $R$ for balance is $405 \,\Omega$. The value of $X$ is close to ................. $\Omega$. (in $.5$)

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