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**Capacitive Current And Reactance**

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**Question: The following phase schematic diagram (FIGURE 4) shows an 11 kV, 50 Hz, 3-phase, short line feeding a load. By calculation or constructing the phasor diagram (use a scale of 1 mm = 2 A) for the load current with VR as reference, determine the capacitive current and**

**1. Calculate the capacitive reactance/ph such that the load power factor is increased to 0.98 lag**

**2. Calculate the percentage reduction in line current with this value of capacitive reactance in circuit.**

**Answer:** Conside the figure given below

When voltage and length of the transmission line increases the capacitance has greater importance.

The phasor diagram can be constructed as follows

Receiving End

V_{R} = 11/(√3) KV

Apparent Power = V_{R} I_{R} cos θ

I_{R} = (Apparent Power)/(V_{R} cos θ)

= (3.2 × 10^{6})/[(11 × (1/√3) × 10^{3} × cosθ)

Now Cos θ = KW/(KW + KVAR) = 3.2/(3.2 + 1.9) = 0.6

Now I_{R} = (3.2 × 10^{6})/(11 × (1/√3) × 10^{3} × 0.6) = 629A

As we know Cos θ = I_{C}/I_{R}

So I_{C} = I_{R} × cosθ = 0.6 × 629 = 377.4A

C = (-jI_{C})/2πf = (-j × 377.4)/(2 × 3.14 × 50) = -j1.20Ω

When load power factor is increased to 0.98 the capacitive reactance can be calculated as follows

I_{R} = (3.2 × 10^{6})/(11 × (1/√3) × 10^{3} ×0.98) = 514A

I_{C} = I_{R} × cosθ = 0.98 × 514 = 503.72A

C = (-jI_{C})/2πf = (-j × 503.72)/(2 × 3.14 × 50) = -j1.60Ω

Percentage reduction in line current = (624 - 514)/624= 0.17%

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