Electronics - Capacitors - Discussion

54. 

What is the reactive power?

[A]. 6.8 mW
[B]. 9.8 mW
[C]. 12.8 mW
[D]. 15.8 mW

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*E-6.
= 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.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.