In a wheatstone bridge resistance of each of the four sides is `10Omega`. If the resistance of the galvanometer is also `10Omega` , then effective resistance of the bridge will be

Options

A. `10Omega`
B. `5Omega`
C. `20Omega`
D. `40Omega`

Answer

To solve the problem of finding the effective resistance of a Wheatstone bridge with given resistances, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Resistors**:
In a Wheatstone bridge, we have four resistors (R1, R2, R3, R4) and a galvanometer. Given that each resistor has a resistance of 10Ω, we can denote:
- R1 = 10Ω
- R2 = 10Ω
- R3 = 10Ω
- R4 = 10Ω
- Resistance of the galvanometer (Rg) = 10Ω

2. **Check for Balance Condition**:
The Wheatstone bridge is balanced when the ratio of the resistances is equal:
\[ \frac{R1}{R2} = \frac{R3}{R4} \]
Substituting the values:
\[ \frac{10}{10} = \frac{10}{10} \implies 1 = 1 \]
Since the bridge is balanced, there will be no current flowing through the galvanometer.

3. **Combine Resistors in Series**:
Since there is no current through the galvanometer, we can treat R1 and R3 as being in series, and R2 and R4 as being in series:
- R_series1 = R1 + R3 = 10Ω + 10Ω = 20Ω
- R_series2 = R2 + R4 = 10Ω + 10Ω = 20Ω

4. **Combine Series Resistances in Parallel**:
Now, we have two resistances (R_series1 and R_series2) in parallel:
\[ R_{eq} = \frac{R_{series1} \times R_{series2}}{R_{series1} + R_{series2}} \]
Substituting the values:
\[ R_{eq} = \frac{20Ω \times 20Ω}{20Ω + 20Ω} = \frac{400Ω^2}{40Ω} = 10Ω \]

5. **Final Result**:
The effective resistance of the Wheatstone bridge is:
\[ R_{eq} = 10Ω \]

### Conclusion:
The effective resistance of the Wheatstone bridge is **10Ω**.

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