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

Answer:

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

Explanation:

Given:

$C =\frac{10^2}{\pi}\times 10^{-6} \ F$

$V_{RMS}=220 \ V$

$f = 50 \ Hz$

Now,

The capacitive reactance will be:

⇒ $X_c=\frac{1}{\omega C} =\frac{1}{2 \pi f C}$

                   $=\frac{1}{2\times \pi\times 50\times \frac{10^{-4}}{\pi} }$

                   $=100 \ \Omega$

RMS value of current will be:

⇒ $I_{RMS}=\frac{V_{RMS}}{X_c}$

             $=\frac{220}{100}$

             $=2.2 \ A$

So that,

⇒ $V_m=220\times \sqrt{2}$

         $=311 \ V$

⇒ $I_m=2.2\times \sqrt{2}$

         $=3.1 \ A$

hence,

The equation will be:

⇒ $\nu=311 sin31 \ 4t$

and,

⇒ $i=3.1 sin(314t+\frac{\pi}{2} )$