Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 54)
54.
What is the reactive power?
Discussion:
12 comments Page 1 of 2.
Nohj said:
5 years ago
@Lex,
CURRENT should be in mA to come up with the answer of 9.8mA.
CURRENT should be in mA to come up with the answer of 9.8mA.
Mayuresh said:
5 years ago
Please let me explain the formula.
Kal said:
6 years ago
It's 0.785 mA. So mW, not kW.
Hani said:
6 years ago
P = iv cos (θ).
θ= tan(inverse)xc\R.
θ= tan(inverse)xc\R.
Chuchu said:
7 years ago
If you're going to use V^2/Xc, you must first find the voltage in the capacitor since the circuit is in series.
So it is more reliable to use I^2 * Xc in the given.
So it is more reliable to use I^2 * Xc in the given.
Lex said:
7 years ago
9812.4 Watts is not 9.8mW. It should be kW.
Ghufran said:
1 decade ago
VAR or Watts?
Pritee said:
1 decade ago
The same is calculated by formula, Pr=v*v/Xc.
San said:
1 decade ago
To find reactive power Pr = I^2*Xc.
Where Xc = 1/wc, w = 2pif.
XC = 1/2*PI*fc.
= 1/6.28*10000*0.001*E-6.
= 15923.5 ohms.
Pr = I^2*Xc.
= 0.785^2*15923.5.
= 9812.4W.
= 9.8mW.
Where Xc = 1/wc, w = 2pif.
XC = 1/2*PI*fc.
= 1/6.28*10000*0.001*E-6.
= 15923.5 ohms.
Pr = I^2*Xc.
= 0.785^2*15923.5.
= 9812.4W.
= 9.8mW.
Vasu said:
1 decade ago
The formula being used here is Power =I^2 * Xc
= (0.785mA)^2 * (1/WC)
= (0.785 mA)^2 * (1/ (2*3.14 *Freq * Capacitance)
=9.8 mWatts
= (0.785mA)^2 * (1/WC)
= (0.785 mA)^2 * (1/ (2*3.14 *Freq * Capacitance)
=9.8 mWatts
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers