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?

1 Answer
Oct 1, 2015

Answer:

#Z_T=1000/_-5,71^@Omega#

Explanation:

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

From Ohm's Law #Z_T=(v_T)/(i_T)=1000/_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)((99,47)/1000) = -5,71^@#, hence a leading power factor since the circuit is capacitive overall.