# If a current of 40 mA flows through a series circuit consisting of a 0.4 µF capacitor and a resistor in series with a 4 kHz, 40 Vac source, what is the total circuit impedance?

Oct 1, 2015

${Z}_{T} = 1000 \angle - 5 , {71}^{\circ} \Omega$

#### Explanation:

Impedance of capacitor is Z_c=1/(2pifC)/_-90^@ =99,472/_-90^@Omega

From Ohm's Law ${Z}_{T} = \frac{{v}_{T}}{{i}_{T}} = 1000 \angle {\theta}_{v} \Omega$

Hence from Pythagoras and the impedance diagram, $| {Z}_{R} | = \sqrt{{1000}^{2} - 99 , {47}^{2}} = 995 \Omega$

From this same impedance diagram we get that the angle of the total impedance is ${\sin}^{- 1} \left(\frac{99 , 47}{1000}\right) = - 5 , {71}^{\circ}$, hence a leading power factor since the circuit is capacitive overall.