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For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is
- a)0.325∠-100°
- b)0.325∠ 80°
- c)0.433∠-100°
- d)0.433∠80°
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
For a power system network with n nodes, Z33 of its bus impedance matr...
Understanding the Problem
To find the current drawn by the capacitor connected at node 3, we need to analyze the voltage and reactance involved.
Given Data
- Reactance of capacitor, Xc = -j3.5 per unit
- Voltage at node 3, V3 = 1.3∠-10° per unit
Calculating Capacitor Current
The current drawn by a capacitor can be calculated using the formula:
I = V / Xc
Where:
- I is the current drawn by the capacitor
- V is the voltage across the capacitor
- Xc is the reactance of the capacitor
Applying the Values
1. Convert Voltage to Rectangular Form:
V3 = 1.3∠-10°
V3 = 1.3 * (cos(-10°) + j*sin(-10°))
V3 ≈ 1.3 * (0.9848 - j*0.1736) ≈ 1.278 - j0.225
2. Using the Reactance:
Xc = -j3.5
Therefore, to find the current:
I = V3 / Xc = (1.278 - j0.225) / (-j3.5)
I = (1.278 - j0.225) * (j / 3.5)
I = (1.278*j/3.5 + 0.225/3.5)
I ≈ (0.365 + j0.065)
3. Finding the Magnitude and Angle:
To convert this back to polar form:
- Magnitude: √(0.365^2 + 0.065^2) ≈ 0.433
- Angle: arctan(0.065/0.365) ≈ 10°
Thus, the final current in polar form is approximately 0.433∠80°.
Conclusion
The current drawn by the capacitor is 0.433∠80°, confirming that the correct answer is option 'D'.
To find the current drawn by the capacitor connected at node 3, we need to analyze the voltage and reactance involved.
Given Data
- Reactance of capacitor, Xc = -j3.5 per unit
- Voltage at node 3, V3 = 1.3∠-10° per unit
Calculating Capacitor Current
The current drawn by a capacitor can be calculated using the formula:
I = V / Xc
Where:
- I is the current drawn by the capacitor
- V is the voltage across the capacitor
- Xc is the reactance of the capacitor
Applying the Values
1. Convert Voltage to Rectangular Form:
V3 = 1.3∠-10°
V3 = 1.3 * (cos(-10°) + j*sin(-10°))
V3 ≈ 1.3 * (0.9848 - j*0.1736) ≈ 1.278 - j0.225
2. Using the Reactance:
Xc = -j3.5
Therefore, to find the current:
I = V3 / Xc = (1.278 - j0.225) / (-j3.5)
I = (1.278 - j0.225) * (j / 3.5)
I = (1.278*j/3.5 + 0.225/3.5)
I ≈ (0.365 + j0.065)
3. Finding the Magnitude and Angle:
To convert this back to polar form:
- Magnitude: √(0.365^2 + 0.065^2) ≈ 0.433
- Angle: arctan(0.065/0.365) ≈ 10°
Thus, the final current in polar form is approximately 0.433∠80°.
Conclusion
The current drawn by the capacitor is 0.433∠80°, confirming that the correct answer is option 'D'.
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Community Answer
For a power system network with n nodes, Z33 of its bus impedance matr...
Power system having n nodes.
Z33 = j 0.5 PU
V3 = 1.3∠ -10° PU
A capacitor having a reactance of -j 3.5 PU is now added to the network
IC = 0.433∠ 80°
Z33 = j 0.5 PU
V3 = 1.3∠ -10° PU
A capacitor having a reactance of -j 3.5 PU is now added to the network
IC = 0.433∠ 80°
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Question Description
For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit isa)0.325∠-100°b)0.325∠ 80°c)0.433∠-100°d)0.433∠80°Correct answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit isa)0.325∠-100°b)0.325∠ 80°c)0.433∠-100°d)0.433∠80°Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit isa)0.325∠-100°b)0.325∠ 80°c)0.433∠-100°d)0.433∠80°Correct answer is option 'D'. Can you explain this answer?.
For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit isa)0.325∠-100°b)0.325∠ 80°c)0.433∠-100°d)0.433∠80°Correct answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit isa)0.325∠-100°b)0.325∠ 80°c)0.433∠-100°d)0.433∠80°Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3∠-10° per unit. If a capacitor having reactance of –j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit isa)0.325∠-100°b)0.325∠ 80°c)0.433∠-100°d)0.433∠80°Correct answer is option 'D'. Can you explain this answer?.
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