Four resistances 10, 10 , 10 and `15 Omega` are connected so as to form Wheatstone's bridge. The resistance connected across `15 Omega` resistance to balance the bridge is

Options

A. `10Omega`
B. `20Omega`
C. `30Omega`
D. `40Omega`

Explanation

To solve the problem of finding the resistance connected across the 15-ohm resistor to balance the Wheatstone bridge, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Resistor Configuration**:
We have four resistors: three of them are 10 ohms (let's call them R1, R2, and R3) and one is 15 ohms (let's call it R4). The Wheatstone bridge is set up as follows:
- R1 = 10 ohms
- R2 = 10 ohms
- R3 = 10 ohms
- R4 = 15 ohms

2. **Understand the Condition for Balance**:
The Wheatstone bridge is balanced when the ratio of the resistances in one branch is equal to the ratio of the resistances in the other branch. Mathematically, this can be expressed as:
\[ \frac{R1}{R2} = \frac{R3}{R4'} \]
where R4' is the equivalent resistance of the 15-ohm resistor (R4) and the unknown resistance (R) connected in parallel.

3. **Set Up the Equation**:
To balance the bridge, we want:
\[ \frac{10}{10} = \frac{10}{R4'} \]
This simplifies to:
\[ 1 = \frac{10}{R4'} \]
Therefore, we need:
\[ R4' = 10 \text{ ohms} \]

4. **Calculate the Equivalent Resistance (R4')**:
The equivalent resistance of R4 (15 ohms) and the unknown resistance (R) in parallel is given by:
\[ \frac{1}{R4'} = \frac{1}{R4} + \frac{1}{R} \]
Substituting R4' = 10 ohms and R4 = 15 ohms, we have:
\[ \frac{1}{10} = \frac{1}{15} + \frac{1}{R} \]

5. **Solve for R**:
Rearranging the equation gives:
\[ \frac{1}{R} = \frac{1}{10} - \frac{1}{15} \]
Finding a common denominator (30):
\[ \frac{1}{R} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30} \]
Therefore:
\[ R = 30 \text{ ohms} \]

6. **Conclusion**:
The resistance that needs to be connected across the 15-ohm resistor to balance the Wheatstone bridge is **30 ohms**.