In a Wheatstone's brigde all the four arms have equal resistance `R`. If the resistance of the galvanometer arm is also `R`, the equivalent resistance of the combination as seen b the battery is

Options

A. R
B. 2R
C. `(R)/(4)`
D. `(R)/(2)`

Answer

To find the equivalent resistance of a Wheatstone bridge with all four arms having equal resistance \( R \) and the galvanometer arm also having resistance \( R \), we can follow these steps:

### Step 1: Understand the Wheatstone Bridge Configuration
In a Wheatstone bridge, there are four resistors arranged in a diamond shape. Let's label them as follows:
- Resistor \( P = R \)
- Resistor \( Q = R \)
- Resistor \( R = R \)
- Resistor \( S = R \)

The galvanometer is connected between the midpoints of the two pairs of resistors.

### Step 2: Analyze the Balanced Condition
For a balanced Wheatstone bridge, the ratio of the resistances in one branch is equal to the ratio in the other branch:
\[ \frac{P}{Q} = \frac{R}{S} \]
Since all resistors are equal, this condition is satisfied.

### Step 3: Remove the Galvanometer Branch
In a balanced Wheatstone bridge, there is no current flowing through the galvanometer. Therefore, we can ignore the galvanometer branch for the purpose of calculating the equivalent resistance.

### Step 4: Combine the Resistors in Series
The two pairs of resistors (P and Q, R and S) can be considered as two series combinations:
- The first series combination (P and Q) gives:
\[ R_{1} = R + R = 2R \]
- The second series combination (R and S) gives:
\[ R_{2} = R + R = 2R \]

### Step 5: Combine the Series Combinations in Parallel
Now, we have two resistances \( R_{1} \) and \( R_{2} \) in parallel:
\[ R_{eq} = \frac{1}{\frac{1}{R_{1}} + \frac{1}{R_{2}}} = \frac{1}{\frac{1}{2R} + \frac{1}{2R}} = \frac{1}{\frac{2}{2R}} = \frac{R}{2} \]

### Step 6: Conclusion
Thus, the equivalent resistance of the combination as seen by the battery is:
\[ R_{eq} = R \]

### Final Answer
The equivalent resistance of the combination as seen by the battery is \( R \).

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