A three-phase line with an impedance of (0.2 1 j1.0) Ω /phase feeds three balanced three-phase loads connected in parallel. Load 1: Absorbs a total of 150 kW and 120 kvar. Load 2: Delta connected with an impedance of (150 2 j48) Ω /phase. Load 3: 120 kVA at 0.6 PF leading. If the line-to-neutral voltage at the load end of the line is 2000 v (rms), determine the magnitude of the line-to-line voltage at the source end of the line.

Source voltage (L-L) = 3479.50<2.13 V (in polar form)

Source voltage (L-L) = 3477.1 + j129.57 V (rectangular form)

Given Information:

A 3 phase source is feeding three loads connected in parallel.

Load voltage (L-N) = VLoad = 2000 V

Impedance of line = ZLine = 0.2 + j1.0 Ω

Load 1 = S1 = P + jQ = 150 + j120 kVA

Load 2 = S2 = Delta connected with Z2 = 150 - j48 Ω

Load 3 = S3 = 120 KVA at PF = 0.6 leading

Source voltage (L-L) = ?

Explanation:

The source voltage is = VLoad + total current*(ZLine)

Where total current is = I1 + I2 + I3

Lets first find current flowing in each of the loads

Load 1:

3 phase apparent power is given S1 = 150 + j120 kVA

Convert into per phase by diving by 3

S1 = (150 + j120)/3

S1 = 50 + j40 kVA

As we know, S = V1* (where * is the conjugate)

I1 = S1*/VLoad

I1 = (50,000 - j40,000)/2000  (notice minus sign due to conjugate)

I1 = 25 - j20 A

Load 2:

first convert delta impedance into wye by the relation

Zy = Zdelta/3

Zy = (150 - j48)/3

Zy = 50 - j16 Ω

I2 = V/Zy

I2 = 2000/(50 - j16)

I2 = 36.29 + j11.61 A

Load 3:

apparent power = 120 KVA

PF = cos(θ) = 0.6 leading

As we know, P = S*cos(θ) and Q = S*sin(θ)

θ = cos⁻¹(PF) = cos⁻¹(0.6) = 53.13°

P3 = 120*cos(53.13°) = 72 kW

Q3 = 120*sin(53.13°) = 96 kVAR

S3 = 72 + j96 kVA

Convert into per phase by diving by 3

S3 = (72 - 96)/3  (minus sign due to leading PF)

S3 = 24 - j32 kVA

I3 = S3*/VLoad

I3 = (24,000 + j32,000)/2000

I3 = 12 + j16 A

Total current = I1 + I2 + I3

Total current = (25 - j20) + (36.29) + j11.61) + (12 + j16)

Total current = 73.29 + j7.61

Source voltage (L-N) = VLoad + total current*(ZLine)

Source voltage (L-N) = 2000 + (73.29 + j7.61)*(0.2 + j1.0)

Source voltage (L-N) = 2007.05 + j74.81

Source voltage (L-L) = *Source voltage (L-N)

Source voltage (L-L) =   (2007.05 + j74.81)

Source voltage (L-L) = 3477.1 + j129.57 V

Source voltage (L-L) = 3479.50<2.13 V (in polar form)

fanswer:

explanation:

f

answer:

on fri feb the green land, the only one who can