Question 430: AC fundamentals

Question

A $10 \mu \mathrm{F}$ capacitor is connected to a $210 \mathrm{~V}, 50 \mathrm{~Hz}$ source as shown in figure. The peak current in the circuit is nearly $(\pi=3.14)$ :

Accepted Answer

Capacitive Reactance
$\begin{aligned} X_C =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 	imes 3.14 	imes 50 	imes 10 	imes 10^{-6}} \ =\frac{1000}{3.14}\end{aligned}$
$
\begin{aligned}
& V_{\mathrm{rms}}=210 \mathrm{~V} \
& i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C} \
& 	ext { Peak current }=\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} 	imes \frac{210}{1000} 	imes 3.14=0.932 \
& \qquad =0.93 \mathrm{~A}
\end{aligned}
$

Answer

0.58 A

Answer

0.93 A

Answer

1.20 A

Answer

0.35 A

AC fundamentals

A 10 \mu \mathrm{F} capacitor is connected to a 210 \mathrm{~V}, 50 \mathrm{~Hz} source as shown in figure. The peak current in the circuit is nearly (\pi=3.14) :

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