A short circuit occurs in a telephone cable having a resistance of `0.45 Omega m^(-1)`. The circuit is tested with a Wheatstone bridge network have values of `100 Omega` and `1110 Omega` respectively. A balance condition is found when the variable resistor has a value of `400 Omega`. Calculate the distance down the cable, where the short has occured.

Explanation

To solve the problem of determining the distance down the cable where the short circuit has occurred, we can follow these steps:

### Step 1: Understand the Given Information
- Resistance of the cable: \( R_c = 0.45 \, \Omega/m \)
- Resistor values in the Wheatstone bridge: \( R_1 = 100 \, \Omega \) and \( R_2 = 1110 \, \Omega \)
- Variable resistor value at balance: \( R_v = 400 \, \Omega \)

### Step 2: Define the Resistance of the Cable Segment
Let \( L \) be the distance (in meters) from the starting point of the cable to the point where the short circuit occurs. The resistance of this segment of the cable can be expressed as:
\[ R_L = R_c \times L = 0.45 \times L \, \Omega \]

### Step 3: Apply the Wheatstone Bridge Balance Condition
For a Wheatstone bridge to be balanced, the following condition must hold:
\[ \frac{R_1}{R_L} = \frac{R_2}{R_v} \]
Substituting the known values, we have:
\[ \frac{100}{0.45L} = \frac{1110}{400} \]

### Step 4: Cross-Multiply to Solve for \( L \)
Cross-multiplying gives us:
\[ 100 \times 400 = 1110 \times 0.45L \]
This simplifies to:
\[ 40000 = 499.5L \]

### Step 5: Solve for \( L \)
Now, we can solve for \( L \):
\[ L = \frac{40000}{499.5} \approx 80.04 \, m \]

### Step 6: Round the Result
Since we are looking for a practical distance, we can round \( L \) to:
\[ L \approx 80 \, m \]

### Final Answer
The distance down the cable where the short circuit has occurred is approximately **80 meters**.

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