In Wheatstone bridge, four resistors of resistances `15 Omega, 12 Omega, 4Omega` and `10 Omega` respectively, are connected in cyclic order. The resistance ( in `Omega `) that is to be connected in parallel with the resistance of `10 Omega` to balance the network is

Answer

To solve the problem of finding the resistance that needs to be connected in parallel with the 10 Ω resistor in a Wheatstone bridge to achieve balance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Resistors**:
The resistors in the Wheatstone bridge are given as:
- R1 = 15 Ω
- R2 = 12 Ω
- R3 = 4 Ω
- R4 = 10 Ω (the resistor with which we need to find the parallel resistance R)

2. **Set Up the Balance Condition**:
For a balanced Wheatstone bridge, the following condition must hold:
\[ \frac{R1}{R4} = \frac{R2}{R3} \]

3. **Substitute the Values**:
Plugging in the known values into the balance condition:
\[ \frac{15}{10} = \frac{12}{4} \]

4. **Simplify the Ratios**:
Simplifying both sides:
- Left side: \(\frac{15}{10} = 1.5\)
- Right side: \(\frac{12}{4} = 3\)

Since \(1.5 \neq 3\), we need to find the equivalent resistance \(R4\) when the 10 Ω resistor is in parallel with the unknown resistor \(R\).

5. **Calculate the Equivalent Resistance \(R4\)**:
The equivalent resistance \(R4\) when \(R\) is in parallel with 10 Ω is given by:
\[ \frac{1}{R4} = \frac{1}{10} + \frac{1}{R} \]

6. **Set the Equivalent Resistance Equal to the Required Value**:
From the balance condition, we found that:
\[ \frac{15}{R4} = \frac{12}{4} \]
This simplifies to:
\[ R4 = 5 Ω \]

7. **Substitute Back to Find \(R\)**:
Now substituting \(R4 = 5\) into the parallel resistance equation:
\[ \frac{1}{5} = \frac{1}{10} + \frac{1}{R} \]

8. **Solve for \(R\)**:
Rearranging gives:
\[ \frac{1}{R} = \frac{1}{5} - \frac{1}{10} \]
Finding a common denominator:
\[ \frac{1}{R} = \frac{2}{10} - \frac{1}{10} = \frac{1}{10} \]
Therefore:
\[ R = 10 Ω \]

### Final Answer:
The resistance that needs to be connected in parallel with the 10 Ω resistor to balance the Wheatstone bridge is **10 Ω**.