In above question if resistance of meter bridge wire is 1 `Omega` /cm then the value of current / is

Options

A. 0.11 A
B. 0.33 A
C. 0.66 A
D. 3.3 A

Answer

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step-by-Step Solution:

1. **Identify the Resistance of the Meter Bridge Wire:**
Given that the resistance of the meter bridge wire is 1 ohm/cm, we can calculate the resistance for different segments of the wire based on their lengths.

2. **Calculate the Resistance for Each Segment:**
- For the first segment (let's say R1), which is 40 cm long:
\[ R_1 = 40 \, \text{cm} \times 1 \, \Omega/\text{cm} = 40 \, \Omega \]
- For the second segment (let's say R2), which is 60 cm long:
\[ R_2 = 60 \, \text{cm} \times 1 \, \Omega/\text{cm} = 60 \, \Omega \]

3. **Set Up the Balance Condition of the Meter Bridge:**
The meter bridge is balanced, which gives us the following relationship:
\[ \frac{AB}{BD} = \frac{AC}{CD} \]
Where:
- \(AB = X\) (the unknown resistance)
- \(BD = 6 \, \Omega\) (the known resistance)
- \(AC = 40 \, \Omega\)
- \(CD = 60 \, \Omega\)

Plugging in the values:
\[ \frac{X}{6} = \frac{40}{60} \]

4. **Solve for X:**
Cross-multiplying gives:
\[ 60X = 240 \implies X = \frac{240}{60} = 4 \, \Omega \]

5. **Determine the Total Resistance in the Circuit:**
After finding \(X\), we can now analyze the circuit. The two resistances \(X\) and \(Y\) (which is 6 ohms) are in series:
\[ R_{\text{total}} = X + Y = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega \]

6. **Calculate the Equivalent Resistance of the Circuit:**
The other two resistances (R1 and R2) are also in series:
\[ R_{\text{parallel}} = R_1 + R_2 = 40 \, \Omega + 60 \, \Omega = 100 \, \Omega \]

Now, we can find the equivalent resistance \(R_{\text{eq}}\) of the two series combinations:
\[ R_{\text{eq}} = \frac{R_{\text{total}} \times R_{\text{parallel}}}{R_{\text{total}} + R_{\text{parallel}}} = \frac{10 \times 100}{10 + 100} = \frac{1000}{110} = \frac{100}{11} \, \Omega \]

7. **Calculate the Current I:**
Using Ohm's law, where \(I = \frac{V}{R}\):
\[ I = \frac{6 \, V}{R_{\text{eq}}} = \frac{6}{\frac{100}{11}} = 6 \times \frac{11}{100} = \frac{66}{100} = 0.66 \, A \]

### Final Answer:
The value of the current \(I\) is \(0.66 \, \text{A}\).

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