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A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers?
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A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is sup...
Given:
- The voltage source is a balanced, Y-connected, three-phase system with a voltage of 400 V and a frequency of 50 Hz.
- Load 1 is ∆ connected and draws 25 kVA at a power factor of 0.8 leading.
- Load 2 is Y connected and draws 10 kVA at a power factor of 0.75 lagging.
To Find:
- Line current
- Phase current
- Power factor of the system
- All powers (active, reactive, apparent)
Solution:
Step 1: Line Voltage
The line voltage in a Y-connected system is equal to the phase voltage. Therefore, the line voltage is 400 V.
Step 2: Line Current of Load 1
The apparent power of Load 1 can be calculated using the formula:
S1 = √3 * V * I1
25 kVA = √3 * 400 V * I1
I1 = 25,000 VA / (1.732 * 400 V) = 36.13 A
Step 3: Phase Current of Load 1
The phase current in a ∆-connected system is equal to the line current. Therefore, the phase current of Load 1 is 36.13 A.
Step 4: Power Factor of Load 1
The power factor of Load 1 is given as 0.8 leading.
Step 5: Line Current of Load 2
The apparent power of Load 2 can be calculated using the formula:
S2 = √3 * V * I2
10 kVA = √3 * 400 V * I2
I2 = 10,000 VA / (1.732 * 400 V) = 14.42 A
Step 6: Phase Current of Load 2
The line current in a Y-connected system is equal to the square root of 3 times the phase current. Therefore, the phase current of Load 2 is:
I2_phase = I2 / √3 = 14.42 A / 1.732 = 8.33 A
Step 7: Power Factor of Load 2
The power factor of Load 2 is given as 0.75 lagging.
Step 8: Total Line Current
The total line current can be calculated by summing the line currents of Load 1 and Load 2. Therefore, the total line current is:
I_line = I1 + I2 = 36.13 A + 14.42 A = 50.55 A
Step 9: Power Factor of the System
The power factor of the system can be calculated using the formula:
PF_system = (Total active power) / (Total apparent power)
The total active power is the sum of the active powers of Load 1 and Load 2, and the total apparent power is the sum of the apparent powers of Load 1 and Load 2. Therefore, the power factor of the system is:
PF_system = (0.8 * 25 kVA + 0.75 * 10 kVA) / (25 kVA + 10 k
- The voltage source is a balanced, Y-connected, three-phase system with a voltage of 400 V and a frequency of 50 Hz.
- Load 1 is ∆ connected and draws 25 kVA at a power factor of 0.8 leading.
- Load 2 is Y connected and draws 10 kVA at a power factor of 0.75 lagging.
To Find:
- Line current
- Phase current
- Power factor of the system
- All powers (active, reactive, apparent)
Solution:
Step 1: Line Voltage
The line voltage in a Y-connected system is equal to the phase voltage. Therefore, the line voltage is 400 V.
Step 2: Line Current of Load 1
The apparent power of Load 1 can be calculated using the formula:
S1 = √3 * V * I1
25 kVA = √3 * 400 V * I1
I1 = 25,000 VA / (1.732 * 400 V) = 36.13 A
Step 3: Phase Current of Load 1
The phase current in a ∆-connected system is equal to the line current. Therefore, the phase current of Load 1 is 36.13 A.
Step 4: Power Factor of Load 1
The power factor of Load 1 is given as 0.8 leading.
Step 5: Line Current of Load 2
The apparent power of Load 2 can be calculated using the formula:
S2 = √3 * V * I2
10 kVA = √3 * 400 V * I2
I2 = 10,000 VA / (1.732 * 400 V) = 14.42 A
Step 6: Phase Current of Load 2
The line current in a Y-connected system is equal to the square root of 3 times the phase current. Therefore, the phase current of Load 2 is:
I2_phase = I2 / √3 = 14.42 A / 1.732 = 8.33 A
Step 7: Power Factor of Load 2
The power factor of Load 2 is given as 0.75 lagging.
Step 8: Total Line Current
The total line current can be calculated by summing the line currents of Load 1 and Load 2. Therefore, the total line current is:
I_line = I1 + I2 = 36.13 A + 14.42 A = 50.55 A
Step 9: Power Factor of the System
The power factor of the system can be calculated using the formula:
PF_system = (Total active power) / (Total apparent power)
The total active power is the sum of the active powers of Load 1 and Load 2, and the total apparent power is the sum of the apparent powers of Load 1 and Load 2. Therefore, the power factor of the system is:
PF_system = (0.8 * 25 kVA + 0.75 * 10 kVA) / (25 kVA + 10 k
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A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers?
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A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers?.
A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers?.
Solutions for A balanced ,Y-connected ,three phase,400 V,50 Hz voltage source is supplying power to two balanced three phase loads .Load 1 is∆ connected and draws 25kVA at 0.8 leading PF, while load 2 is Y connected and draws 10kVA at 0.75lagging PF.find everything. Line current,phase current,pf all powers? in English & in Hindi are available as part of our courses for GATE.
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