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  4. A capacitor of capacitance 10 μ F is connected to an AC source and an AC Ammeter. If the source voltage varies as V = 50 √2 sin 100 t, the reading of the ammeter is

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Q. A capacitor of capacitance $10\, \mu F$ is connected to an $AC$ source and an $AC$ Ammeter. If the source voltage varies as $V = 50 \sqrt{2} \, \sin \, 100 t$, the reading of the ammeter is

KCETKCET 2016Alternating Current

Solution:

$C=10\, \mu F =10 \times 10^{-6} F $
$V=50 \sqrt{2} \sin\, 100 t $
Current, $ I =\frac{V}{R}, \omega=100 $
Current $=\frac{V_{ rms }}{X_{C}}$
$X_{C} =$ Impedance of capacitor
$X_{C} =\frac{1}{\omega C} $
$I =\frac{V_{ rms }}{1 / \omega C} $
$V_{ rma } =\frac{V}{\sqrt{2}}=\frac{50 \sqrt{2}}{\sqrt{2}}=50\, V$
Reading of ammeter
$=50 \times 100 \times 10 \times 10^{-6} A $
$=50 \times 10^{-3} A \Rightarrow 50\, mA$