Question 379: AC fundamentals

Question

A capacitor of capacitance $50 \mu \mathrm{F}$ is connected to a.c. source $\mathrm{e}=220 \sin 50 \mathrm{t}$ ($\mathrm{e}$ in volt, $\mathrm{t}$ in second). The value of peak current is

Accepted Answer

Given $\mathrm{e}=220 \sin (50 \mathrm{t})$
Comparing with $\mathrm{e}=\mathrm{e}_0 \sin \omega \mathrm{t}$, we get
$\begin{aligned}
& \mathrm{e}_0=220 \mathrm{~V}, 	ext { peak voltage } \
& \mathrm{C}=50 \mu \mathrm{F}=50 	imes 10^{-6} \mathrm{~F} \
& 	herefore \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{50 	imes 50} 	imes 10^6=400 \Omega
\end{aligned}$
$	herefore \mathrm{I}_0=\frac{\mathrm{e}_0}{\mathrm{X}_{\mathrm{C}}}=\frac{220}{400}=0.55 \mathrm{~A}$
Peak current $=0.55 \mathrm{~A}$

Answer

$\frac{0.55}{\sqrt{2}} \mathrm{~A}$

Answer

$\frac{\sqrt{2}}{0.55} \mathrm{~A}$

Answer

$0.55 \mathrm{~A}$

Answer

$\sqrt{2} \mathrm{~A}$

AC fundamentals

A capacitor of capacitance 50 \mu \mathrm{F} is connected to a.c. source \mathrm{e}=220 \sin 50 \mathrm{t} (\mathrm{e} in volt, \mathrm{t} in second). The value of peak current is

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