A telephone wire of length `200 km` has a capacitance of `0.01 4 muF per km`. If it carries an `AC` of frequency `5kHz` what should be the value of an inductor required to be connected in series so that impedence of the circuit is minimum ?

Options

A. (a)0.35 mH
B. (b)3.5 mH
C. (c)2.5 mH
D. (d)zero

Answer

To solve the problem step by step, we need to find the value of the inductor required to minimize the impedance of the circuit. Let's break down the solution:

### Step 1: Calculate the total capacitance of the telephone wire
The capacitance per kilometer is given as \(0.014 \, \mu F/km\) and the length of the wire is \(200 \, km\).

\[ C_{total} = \text{Capacitance per km} \times \text{Length of the wire} \]
\[ C_{total} = 0.014 \, \mu F/km \times 200 \, km = 2.8 \, \mu F \]

### Step 2: Convert capacitance to farads
Since \(1 \, \mu F = 10^{-6} \, F\), we convert \(C_{total}\) to farads:
\[ C_{total} = 2.8 \, \mu F = 2.8 \times 10^{-6} \, F \]

### Step 3: Calculate the angular frequency
The frequency \(f\) is given as \(5 \, kHz\). We need to convert this to hertz:
\[ f = 5 \, kHz = 5 \times 10^3 \, Hz \]
Now, calculate the angular frequency \(\omega\):
\[ \omega = 2 \pi f = 2 \pi (5 \times 10^3) \approx 31415.93 \, rad/s \]

### Step 4: Set the condition for minimum impedance
For minimum impedance, the inductive reactance \(X_L\) must equal 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 5: Solve for inductance \(L\)
Rearranging the equation gives:
\[ L = \frac{1}{\omega^2 C} \]

### Step 6: Substitute the values into the equation
Substituting \(\omega\) and \(C_{total}\):
\[ L = \frac{1}{(31415.93)^2 \times (2.8 \times 10^{-6})} \]

### Step 7: Calculate \(L\)
Calculating the denominator:
\[ (31415.93)^2 \approx 9.8696 \times 10^8 \]
Now substituting:
\[ L = \frac{1}{9.8696 \times 10^8 \times 2.8 \times 10^{-6}} \approx \frac{1}{2765.888} \approx 0.0003624 \, H \]

### Step 8: Convert to milliHenries
\[ L \approx 0.3624 \, mH \approx 0.35 \, mH \]

### Final Answer
The required inductance \(L\) is approximately \(0.35 \, mH\).

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