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]
