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ElectricalPower System
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A 3-phase, 50 Hz transmission line has a capacitance of line to neutral Cₙ = 0.01 μF/km. The voltage of the line is 100 kV. The charging current per kilometer of line is

A

181 A/km

B

181×10–³ A/km

C

51 A/km

D

73×10³ A/km

Correct Answer

⚙️ TE • Technical Direct FormulaElectricalPower System
Option B

181×10–³ A/km

Quick Summary:

Given: System voltage VLV_{L}VL​ = 100 kV = 10⁵ V, Frequency f = 50 Hz, Capacitance CnC_{n}Cn​ = 0.01 μF/\mu F/μF/km = 0.01 ×10−6F/\times 10^{-6} F/×10−6F/km.

📐MAMath SolutionDirect Formula
📋 Given

System voltage VLV_{L}VL​ = 100 kV = 10⁵ V, Frequency f = 50 Hz, Capacitance CnC_{n}Cn​ = 0.01 μF/\mu F/μF/km = 0.01 ×10−6F/\times 10^{-6} F/×10−6F/km.

🔢 Formula Used

Ich=VL3×2πfCnI_{ch} = \frac{V_L}{\sqrt{3}} \times 2 \pi f C_nIch​=3​VL​​×2πfCn​

🔢 Step-by-Step Solution
1

Calculate phase voltage

Since the line voltage is given as 100 kV, the phase-to-neutral voltage is Vph=VL3=1053≈57735VV_{ph} = \frac{V_L}{\sqrt{3}} = \frac{10⁵}{\sqrt{3}} \approx 57735 VVph​=3​VL​​=3​105​≈57735V.

Vph=1053V_{ph} = \frac{10⁵}{\sqrt{3}}Vph​=3​105​

2

Determine angular frequency

The angular frequency ω\omegaω for 50 Hz is calculated as ω=2πf\omega = 2 \pi fω=2πf.

ω=2×π×50=314.16 rad/s\omega = 2 \times \pi \times 50 = 314.16 \text{ rad/s}ω=2×π×50=314.16 rad/s

3

Calculate charging current

Substitute the values into the charging current formula: Ich=Vph×ω×CnI_{ch} = V_{ph} \times \omega \times C_nIch​=Vph​×ω×Cn​.

Ich=1053×314.16×0.01×10−6I_{ch} = \frac{10⁵}{\sqrt{3}} \times 314.16 \times 0.01 \times 10^{-6}Ich​=3​105​×314.16×0.01×10−6

4

Final Result

Performing the arithmetic calculation: Ich≈57735×3.1416×10−6≈0.1814A/kmI_{ch} \approx 57735 \times 3.1416 \times 10^{-6} \approx 0.1814 A/kmIch​≈57735×3.1416×10−6≈0.1814A/km.

Ich≈181×10−3 A/kmI_{ch} \approx 181 \times 10^{-3} \text{ A/km}Ich​≈181×10−3 A/km

✅

B is correct because the calculated charging current per kilometer is approximately 181 mA/km, which is equivalent to 181 × 10⁻³ A/km.

Core Concepts Used
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Charging Current Phase Voltage Shunt Capacitance
💡 EXAM TIP

Understanding charging current is vital for calculating Ferranti effect and voltage regulation in long transmission lines where shunt capacitance becomes significant.

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