Three resistance `P, Q, R` each of `2 Omega` and an unknown resistance `S` from the four amrs 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` ?

Options

A. `2Omega`
B. `3Omega`
C. `6Omega`
D. `1Omega`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Wheatstone Bridge Condition
In a Wheatstone bridge, the bridge is balanced when the ratio of the resistances in one arm is equal to the ratio of the resistances in the other arm. The condition for balance is given by:

\[ \frac{P}{Q} = \frac{R}{S'} \]

Where \( S' \) is the equivalent resistance of \( S \) when a \( 6 \, \Omega \) resistor is connected in parallel with it.

### Step 2: Assign Values to the Known Resistors
Given that:
- \( P = 2 \, \Omega \)
- \( Q = 2 \, \Omega \)
- \( R = 2 \, \Omega \)

### Step 3: Set Up the Equation for Balance
Substituting the known values into the balance condition:

\[ \frac{2}{2} = \frac{2}{S'} \]

This simplifies to:

\[ 1 = \frac{2}{S'} \]

From this, we can find \( S' \):

\[ S' = 2 \, \Omega \]

### Step 4: Find the Equivalent Resistance \( S' \)
The equivalent resistance \( S' \) of the resistors \( S \) and \( 6 \, \Omega \) in parallel is given by the formula:

\[ S' = \frac{S \cdot 6}{S + 6} \]

### Step 5: Set Up the Equation
Now we can set the equation we derived for \( S' \) equal to \( 2 \, \Omega \):

\[ \frac{S \cdot 6}{S + 6} = 2 \]

### Step 6: Cross Multiply to Solve for \( S \)
Cross multiplying gives us:

\[ S \cdot 6 = 2(S + 6) \]

Expanding this:

\[ 6S = 2S + 12 \]

### Step 7: Rearranging the Equation
Rearranging the equation to isolate \( S \):

\[ 6S - 2S = 12 \]

This simplifies to:

\[ 4S = 12 \]

### Step 8: Solve for \( S \)
Dividing both sides by 4 gives:

\[ S = 3 \, \Omega \]

### Final Answer
The value of the unknown resistance \( S \) is:

\[ \boxed{3 \, \Omega} \]

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