A single-phase distributor, 2 km long, has a line impedance of (0.2 + 0.3j) ?/km. It supplies a load at the far end, where the voltage V 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

Q: What is the total impedance of the distributor up to the midpoint?

(0.2 + 0.3j) ? — The distributor is 2 km long, thus midpoint is 1 km from the source. The impedance per km is given so the impedance up to the midpoint is 1km * (0.2+0.3j) ?/km.

Q: What is the current flowing through the section of the distributor from the source to the midpoint?

200 A at an unknown power factor — The current flowing from source to midpoint is the vector sum of currents at midpoint and far end load: 100 A at 1.0 p.f. + 100 A at 0.8 p.f. lagging.

Q: Calculate the current phasor of the far-end load.

100 + j0 A — The far end load is 100 A at unity power factor; therefore it is purely real and the current phasor is (100 + j0) A.

Q: Calculate the current phasor of the load at midpoint (100 A at 0.8 power factor lagging).

80 - j60 A — Power factor is 0.8 lagging so the current phasor has positive real and negative imaginary components: 100*(0.8 - j0.6) = 80 - j60 A.

Q: What is the approximate voltage drop from the source to the midpoint? (Hint: consider the combined load current)

35 V — Find total current from source (approximately 200 A), and find the voltage drop using I*Z, where Z is impedance to midpoint. Voltage drop is appoximately 200 * (0.2^2 + 0.3^2)^0.5 = 200 * 0.36 = 72. Then consider the phase angle, so voltage drop is approximately 35V.

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Question 1 of 5

What is the total impedance of the distributor up to the midpoint?

The distributor is 2 km long, thus midpoint is 1 km from the source. The impedance per km is given so the impedance up to the midpoint is 1km * (0.2+0.3j) ?/km.

Question 2 of 5

What is the current flowing through the section of the distributor from the source to the midpoint?

The current flowing from source to midpoint is the vector sum of currents at midpoint and far end load: 100 A at 1.0 p.f. + 100 A at 0.8 p.f. lagging.

Question 3 of 5

Calculate the current phasor of the far-end load.

The far end load is 100 A at unity power factor; therefore it is purely real and the current phasor is (100 + j0) A.

Question 4 of 5

Calculate the current phasor of the load at midpoint (100 A at 0.8 power factor lagging).

Power factor is 0.8 lagging so the current phasor has positive real and negative imaginary components: 100*(0.8 - j0.6) = 80 - j60 A.

Question 5 of 5

What is the approximate voltage drop from the source to the midpoint? (Hint: consider the combined load current)

Find total current from source (approximately 200 A), and find the voltage drop using I*Z, where Z is impedance to midpoint. Voltage drop is appoximately 200 * (0.2^2 + 0.3^2)^0.5 = 200 * 0.36 = 72. Then consider the phase angle, so voltage drop is approximately 35V.

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