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

Q7.15    $100\mu F$ capacitor in series with a $40\Omega$ resistance is connected to a
$110\: V$, $60\: Hz$supply.
What is the maximum current in the circuit?

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Given,

The capacitance of the capacitor $C=100\mu F$

The resistance of the circuit $R=40\Omega$

Voltage Supply $V = 100V$

Frequency of voltage supply $f=60Hz$

The maximum current in the circuit

$I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}V}{\sqrt{R^2+\left ( \frac{1}{wC} \right )^2}}=\frac{\sqrt{2}*110}{\sqrt{40^2+\left ( \frac{1}{2\pi *60*100*10^{-6}} \right )^2}}=3.24A$

Hence Maximum Current in the circuit is 3.24A.

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