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Step-by-Step Solution
Step 1: Recall the Balanced Wheatstone Bridge Condition
A Wheatstone bridge is said to be balanced when the ratio of the resistances in one pair of opposite arms equals the ratio of the resistances in the other pair of opposite arms. Mathematically:
$ \frac{P}{Q} = \frac{R}{S} $
Step 2: Apply the Given Values for the First Balance Condition
We are given the three known resistances: $P = 2\,\Omega$, $Q = 2\,\Omega$, and $R = 2\,\Omega$. Let the unknown resistance be $S$. From the balanced condition:
$ \frac{2}{2} = \frac{2}{S} \quad \Longrightarrow \quad 1 = \frac{2}{S} \quad \Longrightarrow \quad S = 2\,\Omega .$
So, if the bridge were balanced with just $P$, $Q$, $R$, and $S$, then $S$ would have to be $2\,\Omega$.
Step 3: Incorporate the 6 Ω Resistor Connected in Parallel with $S$
However, the problem states that the bridge actually becomes balanced only when a $6\,\Omega$ resistor is placed in parallel with $S$. Let us denote the parallel combination of $S$ and $6\,\Omega$ as $S_{\parallel}$.
For two resistors in parallel, the effective resistance $S_{\parallel}$ is given by:
$ \frac{1}{S_{\parallel}} = \frac{1}{S} + \frac{1}{6} .$
Step 4: Use the Corrected Balanced Condition with the Parallel Combination
Now, for the Wheatstone bridge to balance, the ratio $P : Q$ must still equal the ratio $R : S_{\parallel}$. Since $P = 2\,\Omega$, $Q = 2\,\Omega$, and $R = 2\,\Omega$, we have:
$ \frac{2}{2} = \frac{2}{S_{\parallel}} \quad \Longrightarrow \quad 1 = \frac{2}{S_{\parallel}} \quad \Longrightarrow \quad S_{\parallel} = 2\,\Omega .$
So the parallel combination of $S$ and $6\,\Omega$ must be $2\,\Omega$.
Step 5: Solve for $S$ in the Parallel Combination Equation
We set the parallel combination equal to $2\,\Omega$:
$ S_{\parallel} = 2\,\Omega \quad \Longrightarrow \quad \frac{1}{S_{\parallel}} = \frac{1}{2} .$
But $ \frac{1}{S_{\parallel}} = \frac{1}{S} + \frac{1}{6} . $
Hence:
$ \frac{1}{2} = \frac{1}{S} + \frac{1}{6} .$
Step 6: Simplify to Find $S$
Rearranging the above equation:
$ \frac{1}{S} = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} .$
Therefore:
$ S = 3\,\Omega .$
Final Answer
The value of the unknown resistance $S$ is $3\,\Omega$.
