When the value of R in the balanced Wheatston | vedclass

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(D) According to the balanced condition of a Wheatstone bridge,the ratio of resistances in opposite arms is equal: $\frac{P}{Q} = \frac{R}{S}$.In the

Vedclass · Accepted Answer

$7.5$

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(D) According to the balanced condition of a Wheatstone bridge,the ratio of resistances in opposite arms is equal: $\frac{P}{Q} = \frac{R}{S}$.In the first case:$\frac{P}{Q} = \frac{5}{S} \quad \dots(i)$In the second case,$R$ is increased to $7 \Omega$ and $S$ is increased by $3 \Omega$ (i.e.,$S + 3 \Omega$):$\frac{P}{Q} = \frac{7}{S + 3} \quad \dots(ii)$Equating $(i)$ and $(ii)$:$\frac{5}{S} = \frac{7}{S + 3}$$5(S + 3) = 7S$$5S + 15 = 7S$$2S = 15$$S = 7.5 \Omega$Thus,the initial value of $S$ is $7.5 \Omega$.

When the value of $R$ in the balanced Wheatstone bridge,shown in the figure,is increased from $5 \Omega$ to $7 \Omega$,the value of $S$ has to be increased by $3 \Omega$ in order to maintain the balance. What is the initial value of $S$ (in $Omega$)?

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