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Alternating Current
Capacitive Circuit (Peak Current)
1 question, with its answer.
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A 10 $\mu F$ capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly ($\pi = 3.14$):
$V_{rms} = 210$ V, so $V_0 = 210\sqrt{2}$ V $I_0 = \frac{V_0}{X_C} = V_0\omega C = V_0 \times 2\pi f \times C$ $= 210\sqrt{2} \times 2\pi \times 50 \times 10 \times 10^{-6}$ $\approx 0.93$ A
That is every question in Capacitive Circuit (Peak Current).
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