Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 4)
4.
What is the reactive power in the given circuit?
0 VAR
691 
VAR
VAR44.23 mVAR
1.45 kVAR
Discussion:
36 comments Page 1 of 4.
KIRAN V said:
5 years ago
Given : F = 5 kHz; C = 0.22F
Where XC = CAPACITIVE LOAD.
XC = 1/ (2 * π * F * C).
XC = 1 / (6.28 * 5 * 103 * 0.022 * 10-6)
XC = 1 / 0.6908 * 10-3
XC = 1.4475 * 103 = 1447.5Ω.
P = V * I.
P = V2 / R or P = V2 / XC.
Here:
P = V2/XC.
P = 64 V / 1447.5,
P = 0.04421 Var.
P = 44.21 m Var.
Where XC = CAPACITIVE LOAD.
XC = 1/ (2 * π * F * C).
XC = 1 / (6.28 * 5 * 103 * 0.022 * 10-6)
XC = 1 / 0.6908 * 10-3
XC = 1.4475 * 103 = 1447.5Ω.
P = V * I.
P = V2 / R or P = V2 / XC.
Here:
P = V2/XC.
P = 64 V / 1447.5,
P = 0.04421 Var.
P = 44.21 m Var.
(3)
Rasika chougale said:
6 years ago
Good discussion, thanks everyone for explaining it.
(1)
Asha said:
8 years ago
Here RMS value is given and everyone is taking it directly. Why we are not converting it into max value?
Kiran V said:
9 years ago
Xc = 1/2 * π * f * c.
Xc = 144.75 ohm.
P = 64/144.75.
P = 0.4421 Var.
Xc = 144.75 ohm.
P = 64/144.75.
P = 0.4421 Var.
Md asif said:
9 years ago
P = vi.
P = v^2/R.
Where r = capacitive load .
Xc = 1/2 * pi * f * c.
Xc = 1446.86 ohm.
P = 64/1446.86.
P = 44.21mw.
P = v^2/R.
Where r = capacitive load .
Xc = 1/2 * pi * f * c.
Xc = 1446.86 ohm.
P = 64/1446.86.
P = 44.21mw.
Chandra said:
9 years ago
@Yashu.
22/7 = 3.142.
22/7 = 3.142.
Yashu said:
9 years ago
What is the value of Pi?
ADARSH said:
10 years ago
Yes as we know that:
P = v^2/R.
Where R is Reactant denoted by Xc.
Xc = 1/2*pi*f**c.
Where pi-constant, f-frequency & c-capacitance.
P = v^2/R.
Where R is Reactant denoted by Xc.
Xc = 1/2*pi*f**c.
Where pi-constant, f-frequency & c-capacitance.
Harika said:
10 years ago
Please can anyone explain the derivation of P=V^2/Xc?
Pavan Kumar said:
1 decade ago
Reactive Power is the power which is lost in the circuit and hence not delivered to the Load.
(1)
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