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.
Krishna said:
1 decade ago
Reason behind sending back is because there is no dissipative path (load). In ideal case capacitor cannot dissipate.
Apparent power is pure resistive power whereas reactive power comes into picture because of capacitive reactance in the circuit.
P = v^2/R but now it is v^2/Z where z = R+Xc (Xc=1/2*pi*f*c)here R = 0 ideal case.
Xc=1/2*pi*f*c.
= 1/6.28*5000*0.022*10^-6.
= 1/31400*0.022*10^-6.
= 1447.59.
P = R^2/Xc = 64/1447.59 = 0.0442W = 44.21mW.
Apparent power is pure resistive power whereas reactive power comes into picture because of capacitive reactance in the circuit.
P = v^2/R but now it is v^2/Z where z = R+Xc (Xc=1/2*pi*f*c)here R = 0 ideal case.
Xc=1/2*pi*f*c.
= 1/6.28*5000*0.022*10^-6.
= 1/31400*0.022*10^-6.
= 1447.59.
P = R^2/Xc = 64/1447.59 = 0.0442W = 44.21mW.
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)
Shashi said:
1 decade ago
Power reactive capacitance's formula is "P = V^2/Xc".
but now we don't know What is the Xc(Capacitance reactance)so first we should find Xc.
Xc = 1/2*pi*F*C.
Xc = 1446.86 ohm.
P = V^2/Xc.
= 8^2/1446.86.
= 64/1446.86.
= 0.4423 VAR.
= 44.23 mVAR.
but now we don't know What is the Xc(Capacitance reactance)so first we should find Xc.
Xc = 1/2*pi*F*C.
Xc = 1446.86 ohm.
P = V^2/Xc.
= 8^2/1446.86.
= 64/1446.86.
= 0.4423 VAR.
= 44.23 mVAR.
San said:
1 decade ago
R IS THE REACTANCE OF THE CAPACITANCE Xc.
XC = 1/2*3.14*F*C.
= 1/6.28*5000*0.022*10^-6.
= 1/31400*0.022*10^-6.
= 1447.59.
P = R^2/Xc = 64/1447.59 = 0.0442W = 44.21mW.
XC = 1/2*3.14*F*C.
= 1/6.28*5000*0.022*10^-6.
= 1/31400*0.022*10^-6.
= 1447.59.
P = R^2/Xc = 64/1447.59 = 0.0442W = 44.21mW.
Brajesh singh said:
1 decade ago
Reactive power = V(rms)*I(rms)*sin(angle b/w current and voltage).
Angle = 90 (bcoz capacitor here).
So P(reactive )= 8*(8*2*3.14*50k*.022micro).
P = 44.21 mW.
Angle = 90 (bcoz capacitor here).
So P(reactive )= 8*(8*2*3.14*50k*.022micro).
P = 44.21 mW.
Rajasree said:
1 decade ago
We know that,
p=v^2/R
Here R is the reactance of capacitor which denoted by Xc
Xc=1/(2*pi*f*c)
Xc=1446.8631 ohm
P=V^2/Xc
P=64/1446.8631
P=44.233 mw
p=v^2/R
Here R is the reactance of capacitor which denoted by Xc
Xc=1/(2*pi*f*c)
Xc=1446.8631 ohm
P=V^2/Xc
P=64/1446.8631
P=44.233 mw
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.
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.
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?
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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