A 2-wire DC distributor cable 800 m long is loaded with 1 A/m. Resistance of each conductor is 0.05 Ω/km. Calculate the maximum voltage drop if the distributor is fed from both ends with equal voltages of 220 V.

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PGCIL DT Electrical 22 Aug 2021 Official Paper (SR I)

Answer 1

Concept:

Total current supplied by the distributor, I = 800 × 1 = 800A

Total resistance of the distributor , R = 2 × 0.05 × 0.8 = 0.08 ohm

For a uniformly loaded DC-distributed wire fed from both sides with equal voltages, the V_min occurs at mid-point( x = l/2).

So the maximum voltage drop is at the mid-point:

Calculation:

\(V_{drop}=\frac{i.r×(l-x)^2}{2}\) = \(\frac{i.r×(l-l/2 )^2}{2}\) = \(\frac{i.l×r.l}{2}\)

Maximum voltage drop = \(\frac{IR}{8}\) = \(\frac{800×0.08}{8}\) = 8 volts