A telegraph line of length 100 km has a capacity of 0.01 μF/km and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is _____________ mH. (if π = √10)

Solution

Physics sir · 4 steps

Minimum impedance means resonance: pick L so that X_L exactly cancels the line's X_C.

1. Total capacitance. The line has 0.01 μF for every km, so 100 km gives C = 100 × 0.01 = 1 μF = 10⁻⁶ F.
2. Why resonance. In series, Z = √(R² + (X_L - X_C)²). It is smallest when X_L = X_C, that is, when ω L = 1/(ω C).

   L = 1/(ω² C) = 1/(4π² f² C)

3. Put in numbers. With π² = 10, f = 500 Hz and C = 10⁻⁶ F: L = 1/(4 × 10 × 2.5 × 10⁵ × 10⁻⁶) = 1/10 = 0.1 H.
4. So. L = 0.1 H = 100 mH.

The common slip: 10000. Using 0.01 μF as the whole capacitance and forgetting to multiply by 100 km gives L = 10 H, which is 10000 mH, not 100.

The idea to keep. Whenever a question asks for minimum impedance or maximum current in a series circuit, set X_L = X_C. Then L = 1/(4π² f² C) in SI units, and convert at the end.

Answer from the NTA final answer key. Solution by PrepWiser.