Three resistances P, Q, R each of 2Ω and an unknown resistance S form the four arms of a Wheatstone's bridge circuit. When a resistance of 6Ω is connected in parallel to S the bridge gets balanced. What is the value of S?

2 Ω

3 Ω

6 Ω

1 Ω

Solution

3 Ω

The situation can be depicted as shown in figure

As resistances S and 6Ω are in parallel their effective resistance is:

$\frac{6 S}{6 + S} Ω$

As the bridge is balanced, hence it is balanced Wheatstone's bridge.

For balancing condition

$\Rightarrow \frac{P}{Q} = \frac{R}{(\frac{6 S}{6 + S})} \Rightarrow \frac{2}{2} = \frac{2 (6 + S)}{6 S} \Rightarrow 3 S = 6 + S$

⇒S = 3 Ω

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