Question

What voltage is needed to create a current of 43 mA in a circuit containing only a 1.3-μF capacitor, when the frequency is 3.9 kHz? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.

Answer 1

Answer:

1.35 V

Explanation:

Given:

Capacitance of the capacitor (C) = 1.3 μF = $1.3\times 10^{-6}\ F\ \ [1\mu F=10^{-6}\ F]$

Frequency (f) = 3.9 kHz= 3.9 × 1000 Hz = 3900 Hz [1 kHz = 1000 Hz]

Current flowing in the circuit (I) = 43 mA = 0.043 A [1 mA = 0.001 A]

Now, as only capacitor is there in the circuit, the impedance of the circuit is equal to the reactance of the capacitor.

So, $Z=X_C$

The reactance of the capacitor is given as:

$X_C=\frac{1}{2\pi fC}$

Plug in the values given and solve for $X_C$. This gives,

$X_C=\frac{1}{2\pi\times 3900\times 1.3\times 10^{-6}}\\\\X_C=31.39\ \Omega$

Therefore, the impedance is, $Z=31.39\ \Omega$

Now, the voltage needed to create the given current is obtained using the formula from Ohm's law and is given as:

$V=IZ\\\\V=(0.043\ A)(31.39\ \Omega)\\\\V=1.35\ V$

Therefore, the voltage needed to create a current of 43 mA is 1.35 V.