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A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW ≤ P ≤ 2kW and 1 kVAR ≤ Q ≤ kVAR . A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is
  • a)
    0.447 lag
  • b)
    0.707 lag
  • c)
    0.894 lag
  • d)
    1
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A load is supplied by a 230 V, 50 Hz source. The active power P and th...
Under worst case,
Pmax = 2 kW
Qmax = 2 kVAR
cos 45 = 0.707lag
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Most Upvoted Answer
A load is supplied by a 230 V, 50 Hz source. The active power P and th...
Given Information:
- Supply voltage (V) = 230 V
- Frequency (f) = 50 Hz
- Active power (P) = 1 kW to 2 kW
- Reactive power (Q) = 1 kVAR
- Capacitor generates 1 kVAR reactive power for power factor correction.

Calculating Power Factor (PF) without Correction:
The power factor (PF) is the ratio of active power to apparent power. Apparent power (S) can be calculated using the following formula:

S = √(P^2 + Q^2)

Given that P = 1 kW and Q = 1 kVAR, we can calculate S:

S = √(1^2 + 1^2) = √2 kVA

The power factor (PF) can be calculated as:

PF = P / S = 1 / √2 ≈ 0.707 lagging

Calculating Power Factor (PF) with Correction:
To correct the power factor, a capacitor is connected across the load. The capacitor generates 1 kVAR reactive power.

The new reactive power (Q') after power factor correction can be calculated as:

Q' = Q - Capacitor Reactive Power

Q' = 1 kVAR - 1 kVAR = 0 kVAR

The new apparent power (S') can be calculated using the formula:

S' = √(P^2 + Q'^2) = √(1^2 + 0^2) = 1 kVA

The new power factor (PF') after power factor correction can be calculated as:

PF' = P / S' = 1 / 1 = 1 (unity power factor)

Worst Case Power Factor:
The worst case power factor after power factor correction occurs when the load is purely resistive (Q' = 0 kVAR). In this case, the power factor is 1 (unity power factor).

Therefore, the correct option is d) 1.
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A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW ≤P ≤2kW and 1 kVAR ≤Q ≤kVAR . A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction isa)0.447 lagb)0.707 lagc)0.894 lagd)1Correct answer is option 'B'. Can you explain this answer?
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