Question

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.

Answer 1

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) )`