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 $S$ ?

• A

  $2 \Omega$
• Correct answer
  B

  $3 \Omega$
• C

  $6 \Omega$
• D

  $1 \Omega$

Solution

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, $\begin{aligned} & & \frac{P}{Q} & =\frac{R}{\left(\frac{6 S}{6+S}\right)} \\ \text { or } & & \frac{2}{2} & =\frac{2(6+S)}{6 S} \\ \text { or } & & 3 S & =6+S \\ \text { or } & & S & =3 \Omega \end{aligned}$

Hence, the correct answer is option: 'B'.