A single-phase distributor, 2 km long, has a line impedance of (0.2 + 0.3j) Ω/km. It supplies a load at the far end, where the voltage V_B is 100 V and the current is 100 A at unity power factor. Additionally, a load of 100 A at 0.8 power factor lagging is connected at its midpoint. Calculate the voltage drop at the midpoint.

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RRB JE EE 22 Apr 2025 Shift 1 CBT 2 Official Paper

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Answer 1

Calculation

Total line length = 2 km

Line impedance = Z = (0.2 + 0.3j) Ω/km

Voltage at B, V_B= 100 V

Load at B: I₂ = 100 A, power factor (p.f.) = 1

Load at midpoint M: I₁ = 100 A, p.f.= 0.8 lagging

The voltage drop at midpoint M is given by:

We need to find the voltage drop at midpoint M.

Step 1: Load at B (unity p.f.)

Since p.f. = 1 (purely resistive), current I₂ = 100∠0° A

Impedance from M to B = 1 km × (0.2 + 0.3j) = (0.2 + 0.3j) Ω

Voltage drop ΔV_MB = I₂ × Z =100 × (0.2 + 0.3j) = (20 + 30j) V

So, voltage at M = V_B + ΔVMB = 100 + 20 + 30j

V_M = (120 + 30j) V

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