A `100 km` telegraph wire hasd capacity of `0.02 mu F//km`, if it carries an alternating current of frequency `5 kHZ`. The value of an inductrance required to be connected in series so that the impedance is minimum.

Options

A. `50.7 mH`
B. `5.07 mH`
C. `0.507 mH`
D. `507 mH`

Explanation

To find the value of inductance \( L \) required to be connected in series with a telegraph wire to minimize the impedance when carrying an alternating current, we can follow these steps:

### Step 1: Calculate the total capacitance \( C \)
The capacitance per kilometer is given as \( 0.02 \, \mu F/km \). For a length of \( 100 \, km \):
\[ C = 0.02 \, \mu F/km \times 100 \, km = 2.0 \, \mu F = 2.0 \times 10^{-6} \, F \]

### Step 2: Determine the angular frequency \( \omega \)
The frequency \( f \) is given as \( 5 \, kHz = 5000 \, Hz \). The angular frequency \( \omega \) is calculated as:
\[ \omega = 2\pi f = 2\pi \times 5000 \, Hz \approx 31415.93 \, rad/s \]

### Step 3: Set up the resonance condition
At resonance, the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \):
\[ X_L = X_C \]
Where:
\[ X_L = \omega L \quad \text{and} \quad X_C = \frac{1}{\omega C} \]
Thus, we have:
\[ \omega L = \frac{1}{\omega C} \]

### Step 4: Solve for inductance \( L \)
Rearranging the equation gives:
\[ L = \frac{1}{\omega^2 C} \]
Substituting the values of \( \omega \) and \( C \):
\[ L = \frac{1}{(31415.93)^2 \times (2.0 \times 10^{-6})} \]

### Step 5: Calculate \( L \)
Calculating \( \omega^2 \):
\[ \omega^2 \approx (31415.93)^2 \approx 9.8696 \times 10^8 \]
Now substituting back:
\[ L = \frac{1}{9.8696 \times 10^8 \times 2.0 \times 10^{-6}} \approx \frac{1}{1.97392 \times 10^3} \approx 0.5066 \times 10^{-3} \, H \]
Thus, rounding gives:
\[ L \approx 0.507 \, mH = 0.507 \times 10^{-3} \, H \]

### Final Answer
The value of inductance required is approximately:
\[ L \approx 0.507 \, mH \]

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