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The current taken from a 230 V, 50 Hz supply is measured as 10 A with a lagging p.f. of 0.7. A capacitor is connected in parallel with the load. The true power
- a)increases
- b)decreases
- c)remains unchanged
- d)cannot be predicted
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The current taken from a 230 V, 50 Hz supply is measured as 10 A with ...
The power triangle is shown below.
P = Active power (or) Real power in W = Vrms Irms cos ϕ
Q = Reactive power in VAR = Vrms Irms sin ϕ
S = Apparent power in VA = Vrms Irms
S = P + jQ
ϕ is the phase difference between the voltage and current.
Calculation:
Given Vrms = 230 V, Irms = 10 A, cos ϕ = 0.7
Active power PL = 230 × 10 × 0.7 = 1610 W
Reactive power QL = 230 × 10 × 0.714 = 1642.52 VAR
When the capacitor is added in parallel to the load it supplies the reactive power (QC) to the load.
So the total reactive power supplied by the source (Q) to load will decreases, it can be shown as
Q = QL - QC
But the total active power supplied by the source to the load remains same.
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The current taken from a 230 V, 50 Hz supply is measured as 10 A with a lagging p.f. of 0.7. A capacitor is connected in parallel with the load. The true powera)increasesb)decreasesc)remains unchangedd)cannot be predictedCorrect answer is option 'C'. Can you explain this answer?
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