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Three resistances $P, Q, R$ each of $2 \Omega$ and an unknown resistances $\mathrm{S}$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 \Omega$ is connected in parallel ot $\mathrm{S}$ the bridge gets balanced. What is the value of $\mathrm{S}$ ?
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The correct answer is:
$3 \Omega$
The bridge formed with given resistances is a balanced Wheatstone's bridge
The situation can be depicted as shown in the figure.
As resistances \(S\) and \(6 \Omega\) are in paralle, their effective resistance is \(\frac{6 S}{6+S} \Omega\).
For balancing condition, \(\frac{P}{Q}=\frac{R}{\left(\frac{6 S}{6+S}\right)}\)
or \(\frac{2}{2}=\frac{2(6+\mathrm{S})}{6 \mathrm{~S}}\)
or \(3 \mathrm{~S}=6+\mathrm{S} \Rightarrow \mathrm{S}=3 \Omega\)
The situation can be depicted as shown in the figure.
As resistances \(S\) and \(6 \Omega\) are in paralle, their effective resistance is \(\frac{6 S}{6+S} \Omega\).
For balancing condition, \(\frac{P}{Q}=\frac{R}{\left(\frac{6 S}{6+S}\right)}\)
or \(\frac{2}{2}=\frac{2(6+\mathrm{S})}{6 \mathrm{~S}}\)
or \(3 \mathrm{~S}=6+\mathrm{S} \Rightarrow \mathrm{S}=3 \Omega\)
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