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In a typical Wheatstone network, the resistances in cyclic order are $A=10 \Omega, B=5 \Omega, C=4 \Omega$ and $D=4$ $\Omega$ for the bridge to be balanced
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The correct answer is:
$10 \Omega$ should be connected in parallel with $A$
For a balance Wheatstone bridge.
$\begin{aligned}
& \frac{A}{B}=\frac{D}{C} \Rightarrow \frac{10}{5} \neq \frac{4}{4} \text { (Unbalanced) } \\
& \frac{A^{\prime}}{B}=\frac{D}{C} \Rightarrow \frac{A^{\prime}}{5}=\frac{4}{4} \Rightarrow A^{\prime}=5 \Omega
\end{aligned}$
$A^{\prime}(5 \Omega)$ is obtained by connecting a $10 \Omega$ resistance in parallel with $A$.
$\begin{aligned}
& \frac{A}{B}=\frac{D}{C} \Rightarrow \frac{10}{5} \neq \frac{4}{4} \text { (Unbalanced) } \\
& \frac{A^{\prime}}{B}=\frac{D}{C} \Rightarrow \frac{A^{\prime}}{5}=\frac{4}{4} \Rightarrow A^{\prime}=5 \Omega
\end{aligned}$
$A^{\prime}(5 \Omega)$ is obtained by connecting a $10 \Omega$ resistance in parallel with $A$.
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