If a 47 $\mu \vec { F }$ capacitor is connected to a 20V, 400Hz source, the current is__________
A) 0.851 A
B) 2.36 A
C) 8.51 A
D) 425 A

Question

If a 47 $\mu \vec { F }$ capacitor is connected to a  20V, 400Hz source, the current is__________
A)  0.851  A
B)  2.36  A
C)  8.51  A
D)  425  A

Quizplus · Accepted Answer

The impedance of a capacitor is given by the formula: 
$$ Z_c = \frac{1}{j \omega C} $$ 
where
$$ \omega = 2\pi f $$ 
and f is the frequency of the source.

Plugging in the numbers,
$$ \omega = 2\pi (400) = 800\pi  $$
$$ Z_c = \frac{1}{j(800\pi)(47 	imes 10^{-6})} = -j555.55\ \Omega $$

The negative sign indicates that the impedance is capacitive.

By Ohm's Law, the current is given by the formula:
$$ I = \frac{V}{Z} $$
where V is the voltage of the source and Z is the impedance of the capacitor.

Plugging in the numbers,
$$ I = \frac{20}{-j555.55} = 0.036 + j0.072\ 	ext{A} $$

The magnitude of the current is 
$$ |I| = \sqrt{(0.036)^2 + (0.072)^2} = 0.080\ 	ext{A} $$

Converting to amperes,
$$ |I| = 80\ 	ext{mA} $$

Therefore, the best choice is B) 2.36 A because this is the nearest answer in amperes to the calculated magnitude of the current.

If a 47 μF⃗\mu \vec { F }μF Capacitor Is Connected to a 20V, 400Hz Source, the Connected

If a 47 $\mu \vec { F }$ Capacitor Is Connected to a 20V, 400Hz Source, the Connected