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A capacitor of capacitance 102/π µF is connected across a 220 V, 50 Hz A.C. mains. Calculate the capacitive reactance, RMS value of current and write down the equations of voltage and current.

User Spacey
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1 Answer

14 votes
14 votes

Answer:

The capacitive reactance will be "100Ω" and RMS value of current is "2.2 A". A further explanation is provided below.

Step-by-step explanation:

Given:


C =(10^2)/(\pi)* 10^(-6) \ F


V_(RMS)=220 \ V


f = 50 \ Hz

Now,

The capacitive reactance will be:


X_c=(1)/(\omega C) =(1)/(2 \pi f C)


=(1)/(2* \pi* 50* (10^(-4))/(\pi) )


=100 \ \Omega

RMS value of current will be:


I_(RMS)=(V_(RMS))/(X_c)


=(220)/(100)


=2.2 \ A

So that,


V_m=220* √(2)


=311 \ V


I_m=2.2* √(2)


=3.1 \ A

hence,

The equation will be:


\\u=311 sin31 \ 4t

and,


i=3.1 sin(314t+(\pi)/(2) )

User Sharel
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