A $100 μ F$ capacitor in series with a $40 Ω$ resistor is connected to a $110 V, 60 H z$ supply. The maximum current in the circuit is $x \times 10^{- 2} A$ . Find $x$ .

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Solution

$V = 110 \sqrt{2} s i n (2 π 60 t) V ω = 2 π 60 r a d / s e c i n p h a s o r, I_{m a x} = \frac{110 \sqrt{2}}{\sqrt{R^{2} + \frac{1}{C^{2} ω^{2}}}} = 3.24 A$
$x$ =324

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Similar questions

A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

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