Example 3: What is the impedance of a series circuit  consisting of a capacitor with a capacitive reactance of 7  ohms, an inductor with an inductive reactance of 10 ohms,  and a resistor with a resistance of 4 ohms?  [Figure 9-27] Solution: To find total current: Remember that inductive and capacitive reactances can cause  a phase shift between voltage and current. In this example,  inductive reactance is larger than capacitive reactance, so the  voltage leads current.  It should be noted that since inductive reactance, capacitive  reactance, and resistance affect each other at right angles,  the voltage drops of any series AC circuit should be added  using vector addition.  Figure 9-28  shows the voltage drops  over the series AC circuit described in example 3 above.  Figure 9-27.  A circuit containing resistance, inductance, and  capacitance. 9-18 4Ω 10 Ω 7 Ω 110V AC 60 cycles  To determine the total impedance of the parallel circuit shown  in  Figure 9-29 , one would first determine the capacitive and  inductive reactances. (Remember to convert microfarads  to farads.) When solving parallel AC circuits, one must also use a  derivative of the Pythagorean Theorem. The equation for  finding impedance in an AC circuit is as follows: Z = 1  $^{2}$ +   1 – 1 2 R X$_{L}$ X$_{C}$ E$_{T}$ =  √  E$_{R}$$^{2}$+ (E$_{L}$ – E$_{C}$) 2 E$_{T}$  =  √ 88$^{2}$+ (220 – 154) 2 E$_{T}$ =  √  88$^{2}$+ 66 2 E$_{T}$ =  √ 12,100 E$_{T}$ = 110 volts I$_{T}$ = E  T   Z I$_{T}$ = 110V$_{ }$ 5Ω I$_{T}$ = 22 amps Z =  √  R$^{2}$ + (X$_{L}$ – X$_{C}$) 2 Z  =  √ 4$^{2}$ + (10 – 7) 2 Z =  √  25 Z = 5Ω E$_{R}$ = I × R E$_{X}$$_{L}$ = I × X$_{L}$ E$_{X}$$_{C}$ = I × X$_{C}$ To determine the total applied voltage for the circuit, each  individual voltage drop must be added using vector addition.  Parallel AC Circuits To calculate the individual voltage drops, simply use the  equations: Figure 9-28.  Voltage drops.  110V AC 60 cycles  E$_{R}$ = 88V E$_{L}$ = 220V E$_{C}$ = 154V