Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 74)
74.
If the reactive power of a circuit is 50 mW while the apparent power is 64 mW, then what is the true power of the circuit?
Discussion:
6 comments Page 1 of 1.
Mark j said:
3 years ago
You are correct @Gomathi.
You have to use the power triangle. Solve it for the true power.
You have to use the power triangle. Solve it for the true power.
Gomathi said:
10 years ago
True power = sqrt(apparent power^2 - reactive power^2).
= sqrt(4096 - 2500).
= sqrt(1596) = 39.94 mW.
= sqrt(4096 - 2500).
= sqrt(1596) = 39.94 mW.
(2)
Trushal said:
1 decade ago
How is this possible:
Reactive power square = 4096.
True power square = 2500.
And answer is sort of 1.638 = 1.279.
Reactive power square = 4096.
True power square = 2500.
And answer is sort of 1.638 = 1.279.
Zaheer said:
1 decade ago
Apparent power=srt of (real power) ^2+ (reactive power) ^2.
Keerthi said:
1 decade ago
If any one please tell me the relation between apparent, true and reactive powers.
Shilpa said:
1 decade ago
(apparent power)^2=(real power)62+(reactive power)^2
64^2=x^2+50^2
x^2=1596
x=39.999~=40mW
64^2=x^2+50^2
x^2=1596
x=39.999~=40mW
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