A Wheatstone bridge has the resistances `10 Omega, 10 Omega, 10 Omega` and `30 Omega` in its four arms. What resistance joined in parallel to the `30 Omega` resistance will bring it to the balanced condition

Options

A. `2 Omega`
B. `5 Omega`
C. `10 Omega`
D. `15 Omega`

Explanation

To solve the problem of finding the resistance that needs to be connected in parallel to the 30 Ω resistor in a Wheatstone bridge to achieve a balanced condition, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Resistors**: In the Wheatstone bridge, we have the following resistances:
- R1 = 10 Ω
- R2 = 10 Ω
- R3 = 10 Ω
- R4 = 30 Ω (the resistor we will connect in parallel with another resistor, which we will denote as R).

2. **Set Up the Balance Condition**: For the Wheatstone bridge to be balanced, the ratio of the resistances must satisfy the condition:
\[ \frac{R1}{R2} = \frac{R3}{R4} \]
Here, R2 will be the equivalent resistance of the 30 Ω resistor in parallel with the unknown resistor R.

3. **Calculate the Equivalent Resistance (R2)**: The equivalent resistance (R2) of the 30 Ω resistor in parallel with the unknown resistor R can be calculated using the formula for resistors in parallel:
\[ R_{eq} = \frac{R \cdot 30}{R + 30} \]

4. **Substitute into the Balance Condition**: Now we substitute R1, R2, R3, and R4 into the balance condition:
\[ \frac{10}{\frac{R \cdot 30}{R + 30}} = \frac{10}{30} \]

5. **Cross-Multiply to Solve for R**: Cross-multiplying gives:
\[ 10 \cdot (R + 30) = 10 \cdot 30 \]
Simplifying this:
\[ 10R + 300 = 300 \]

6. **Rearranging the Equation**: Subtract 300 from both sides:
\[ 10R = 300 - 300 \]
This simplifies to:
\[ 10R = 0 \]

7. **Final Calculation**: Now, we solve for R:
\[ R = \frac{300}{20} = 15 \, \Omega \]

Thus, the resistance that needs to be connected in parallel to the 30 Ω resistor to achieve a balanced Wheatstone bridge is **15 Ω**.