Four resistances of `3Omega, 3Omega, 3Omega and 4Omega` respectively are used to form a Wheatstone bridge. The `4Omega` resistance is short circuited with a resistance R in order to get bridge balanced. The value of R will be

Options

A. `10Omega`
B. `11Omega`
C. `12Omega`
D. `13Omega`

Explanation

To solve the problem of finding the resistance \( R \) that balances the Wheatstone bridge, we can follow these steps:

### Step 1: Understand the Wheatstone Bridge Configuration
In a Wheatstone bridge, we have four resistances arranged in a diamond shape. Here, we have three resistances of \( 3 \Omega \) each and one resistance of \( 4 \Omega \). The \( 4 \Omega \) resistance is short-circuited with an unknown resistance \( R \).

### Step 2: Identify the Resistors
Let:
- \( R_1 = 3 \Omega \)
- \( R_2 = 3 \Omega \)
- \( R_3 = 3 \Omega \)
- \( R_4 = 4 \Omega \) (which is short-circuited with \( R \))

### Step 3: Calculate the Equivalent Resistance of \( R_4 \)
When the \( 4 \Omega \) resistor is short-circuited with \( R \), the equivalent resistance \( R_{eq} \) of \( R_4 \) can be calculated using the formula for resistors in parallel:
\[ R_{eq} = \frac{R \cdot 4}{R + 4} \]

### Step 4: Set Up the Balance Condition
For the Wheatstone bridge to be balanced, the following condition must hold:
\[ \frac{R_1}{R_2} = \frac{R_{eq}}{R_3} \]
Substituting the known values:
\[ \frac{3}{3} = \frac{R_{eq}}{3} \]
This simplifies to:
\[ 1 = \frac{R_{eq}}{3} \]

### Step 5: Solve for \( R_{eq} \)
From the balance condition, we can express \( R_{eq} \):
\[ R_{eq} = 3 \Omega \]

### Step 6: Substitute \( R_{eq} \) into the Equivalent Resistance Formula
Now we can substitute \( R_{eq} \) back into the equivalent resistance formula:
\[ 3 = \frac{R \cdot 4}{R + 4} \]

### Step 7: Cross-Multiply and Solve for \( R \)
Cross-multiplying gives:
\[ 3(R + 4) = 4R \]
Expanding this:
\[ 3R + 12 = 4R \]
Rearranging gives:
\[ 12 = 4R - 3R \]
Thus:
\[ R = 12 \Omega \]

### Final Answer
The value of the resistance \( R \) that balances the Wheatstone bridge is:
\[ \boxed{12 \Omega} \]