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:

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

Explanation:

Given:

[tex]C =\frac{10^2}{\pi}\times 10^{-6} \ F[/tex]

[tex]V_{RMS}=220 \ V[/tex]

[tex]f = 50 \ Hz[/tex]

Now,

The capacitive reactance will be:

⇒ [tex]X_c=\frac{1}{\omega C} =\frac{1}{2 \pi f C}[/tex]

                   [tex]=\frac{1}{2\times \pi\times 50\times \frac{10^{-4}}{\pi} }[/tex]

                   [tex]=100 \ \Omega[/tex]

RMS value of current will be:

⇒ [tex]I_{RMS}=\frac{V_{RMS}}{X_c}[/tex]

             [tex]=\frac{220}{100}[/tex]

             [tex]=2.2 \ A[/tex]

So that,

⇒ [tex]V_m=220\times \sqrt{2}[/tex]

         [tex]=311 \ V[/tex]

⇒ [tex]I_m=2.2\times \sqrt{2}[/tex]

         [tex]=3.1 \ A[/tex]

hence,

The equation will be:

⇒ [tex]\nu=311 sin31 \ 4t[/tex]

and,

⇒ [tex]i=3.1 sin(314t+\frac{\pi}{2} )[/tex]

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