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 2 of 4.
Savi said:
1 decade ago
I didn't understand please tell me easy way.
Siva said:
1 decade ago
In our home we see 230 v whether it is RMRS value or anything.
Kiran said:
1 decade ago
VAR - Volt Ampere Reactance (Units of reactive power).
Rachna said:
1 decade ago
How the problem can be solve?
Gomathi said:
1 decade ago
Why they use the unit VAR (IN ANS). If it's one of the unit of mean where we use?
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.
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.
Makkk said:
1 decade ago
What is the reason behind sending the power to source back?
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.
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