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If $P=Q=R=10 \Omega$ and $S=20 \Omega$, then what resistance should be joined with $S$ to balance the Wheatstone's network?
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Verified Answer
The correct answer is:
Join a resistance of $20 \Omega$ in parallel with $S$
The given Wheatstone bridge eircuit is shown belin.
Wheatstune is bulanced only when
$$
\frac{P}{Q}=\frac{k}{\mathrm{~S}}
$$
Thus, the value of resistance $S$ should be 10 $\Omega$. Hut it is given to $20 \Omega$.
Hence, to decresese its vulue upto $10 \Omega$, suppose $x \Omega$ tesistunce is comnected pantlel to it.
$\therefore \quad \frac{1}{10}=\frac{1}{20}+\frac{1}{x}$
$\Rightarrow \quad \frac{1}{x}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$
$\Rightarrow \quad x=20 \Omega$
Wheatstune is bulanced only when
$$
\frac{P}{Q}=\frac{k}{\mathrm{~S}}
$$
Thus, the value of resistance $S$ should be 10 $\Omega$. Hut it is given to $20 \Omega$.
Hence, to decresese its vulue upto $10 \Omega$, suppose $x \Omega$ tesistunce is comnected pantlel to it.
$\therefore \quad \frac{1}{10}=\frac{1}{20}+\frac{1}{x}$
$\Rightarrow \quad \frac{1}{x}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$
$\Rightarrow \quad x=20 \Omega$
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