Discussion :: Capacitors  General Questions (Q.No.54)
54.  What is the reactive power? 

Answer: Option B Explanation: No answer description available for this question.

Saranya said: (Sep 22, 2011)  
Can anyone explain this..? 
Saranya said: (Sep 22, 2011)  
Can anyone tell me what is the formula used for this ? 
Vasu said: (Sep 27, 2011)  
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 
San said: (Apr 10, 2013)  
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*E6. = 15923.5 ohms. Pr = I^2*Xc. = 0.785^2*15923.5. = 9812.4W. = 9.8mW. 
Pritee said: (Oct 4, 2013)  
The same is calculated by formula, Pr=v*v/Xc. 
Ghufran said: (Nov 4, 2014)  
VAR or Watts? 
Lex said: (Jun 18, 2018)  
9812.4 Watts is not 9.8mW. It should be kW. 
Chuchu said: (Sep 20, 2018)  
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. 
Hani said: (Mar 19, 2019)  
P = iv cos (θ). θ= tan(inverse)xc\R. 
Kal said: (Oct 13, 2019)  
It's 0.785 mA. So mW, not kW. 
Mayuresh said: (Mar 29, 2020)  
Please let me explain the formula. 
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