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At an Industrial substation with a 4 MW load a capacitor of 2.5 MVAR is installed to maintain the load power factor at 0.97 lag. If the capacitor goes out of service, the load power factor becomes – 
  • a)
    1.00 lag
  • b)
    1.00 lead
  • c)
    0.75 lag
  • d)
    0.65 lag
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
At an Industrial substation with a 4 MW load a capacitor of 2.5 MVAR i...
PL = 4MW
QC = 2.5MV AR
Pf = 0.97 lag
⇒ Power factor =cosϕ=0.75lag
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Most Upvoted Answer
At an Industrial substation with a 4 MW load a capacitor of 2.5 MVAR i...
0.88 lag.

To solve this problem, we can use the formula:

Q = P * tan(arccos(PF))

where Q is the reactive power in MVAR, P is the real power in MW, and PF is the power factor.

Before the capacitor is out of service, the load has a power factor of 0.97 lag, so:

Q = 4 * tan(arccos(0.97)) = 0.794 MVAR

The capacitor is providing 2.5 MVAR of reactive power, so the total reactive power is:

Qtotal = Q + 2.5 = 3.294 MVAR

When the capacitor goes out of service, the load power factor drops to 0.88 lag. Using the same formula:

Qnew = 4 * tan(arccos(0.88)) = 1.816 MVAR

The total reactive power is now:

Qtotal = Qnew = 1.816 MVAR

The difference in reactive power is:

ΔQ = Qtotal without capacitor - Qtotal with capacitor
ΔQ = 3.294 - 1.816
ΔQ = 1.478 MVAR

This means that without the capacitor, the reactive power is lower by 1.478 MVAR. To maintain the same power factor as before, the load would need to compensate for this with an equivalent amount of reactive power. This could be done by installing a new capacitor with a rating of 1.478 MVAR, or by adjusting the existing equipment to provide the necessary reactive power.
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At an Industrial substation with a 4 MW load a capacitor of 2.5 MVAR is installed to maintain the load power factor at 0.97 lag. If the capacitor goes out of service, the load power factor becomes –a)1.00 lagb)1.00 leadc)0.75 lagd)0.65 lagCorrect answer is option 'C'. Can you explain this answer?
Question Description
At an Industrial substation with a 4 MW load a capacitor of 2.5 MVAR is installed to maintain the load power factor at 0.97 lag. If the capacitor goes out of service, the load power factor becomes –a)1.00 lagb)1.00 leadc)0.75 lagd)0.65 lagCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about At an Industrial substation with a 4 MW load a capacitor of 2.5 MVAR is installed to maintain the load power factor at 0.97 lag. If the capacitor goes out of service, the load power factor becomes –a)1.00 lagb)1.00 leadc)0.75 lagd)0.65 lagCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At an Industrial substation with a 4 MW load a capacitor of 2.5 MVAR is installed to maintain the load power factor at 0.97 lag. If the capacitor goes out of service, the load power factor becomes –a)1.00 lagb)1.00 leadc)0.75 lagd)0.65 lagCorrect answer is option 'C'. Can you explain this answer?.
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