Question

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

• A. 50 mA
• B. 70.7 mA
• C. 5.0 mA
• D. 7.07 mA

Answer 1

Answer: (A)

50 mA

Answer 2

$C=10\, \mu F =10 \times 10^{-6} F$
$V=50 \sqrt{2} \sin\, 100 t$
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
$=50 \times 100 \times 10 \times 10^{-6} A$
$=50 \times 10^{-3} A \Rightarrow 50\, mA$