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Three resistances $P, Q, R$ each of $2 \Omega$ and an unknown resistance $S$ form the four arms of a Wheatstone's bridge circuit. When a resistance of $6 \Omega$ is connected in parallel to $S$ the bridge gets balanced. What is the value of $S$ ?
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Verified Answer
The correct answer is:
$3 \Omega$
The situation can be depicted as shown in figure.
As resistances $S$ and $6 \Omega$ are in parallel their effective resistance is
$\frac{6 S}{6+S} \Omega$
As the bridge is balanced, hence it is balanced Wheatstone's bridge.
For balancing condition,
$\begin{aligned}
& & \frac{P}{Q} & =\frac{R}{\left(\frac{6 S}{6+S}\right)} \\
\text { or } & & \frac{2}{2} & =\frac{2(6+S)}{6 S} \\
\text { or } & & 3 S & =6+S \\
\text { or } & & S & =3 \Omega
\end{aligned}$
As resistances $S$ and $6 \Omega$ are in parallel their effective resistance is
$\frac{6 S}{6+S} \Omega$
As the bridge is balanced, hence it is balanced Wheatstone's bridge.
For balancing condition,
$\begin{aligned}
& & \frac{P}{Q} & =\frac{R}{\left(\frac{6 S}{6+S}\right)} \\
\text { or } & & \frac{2}{2} & =\frac{2(6+S)}{6 S} \\
\text { or } & & 3 S & =6+S \\
\text { or } & & S & =3 \Omega
\end{aligned}$
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