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 4 of 4.
Prashant said:
1 decade ago
Xc=1/(2(pi)fc)
f=5k
c=0.022micro
Xc=1446.86
P=V^2/Xc=8^2/1446.86
P=0.04423VAR ~ 44.23m VAR
f=5k
c=0.022micro
Xc=1446.86
P=V^2/Xc=8^2/1446.86
P=0.04423VAR ~ 44.23m VAR
Jithin Kk said:
1 decade ago
So what about the rms value of voltage.
Engr. Shahbul said:
1 decade ago
Xc=1000/2*3.14*5*.022=1447.597ohm. P=v2/Xc=8*8/1447.597=0.0442112w, =44.21mW.
Salman said:
1 decade ago
Xc=1/wc.
w=2*pi*f.
Reactive power Q=Vrms/Xc.
w=2*pi*f.
Reactive power Q=Vrms/Xc.
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.
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.
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