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 at $\mathrm{S}$ the bridge gets balanced. What is the value of $\mathrm{S}$ ?

Question

Aicharya · Accepted Answer

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 parallel, 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 $3 \mathrm{~S}=6+\mathrm{S} \Rightarrow \mathrm{S}=3 \Omega$

Three resistances $P, Q, R$ each of $2 \Omega$ and an unknown resistances $\mathrm{S}$ form the four arms of…

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