Example 2 is a series circuit illustrated in which a capacitor  of 200 μf is connected in series with a 10 ohm resistor.  [Figure 9-26]  What is the value of the impedance, the current  flow, and the voltage drop across the resistor?  First, the capacitance is changed from microfarads to farads.  Since 1 million microfarads equal 1 farad, then 200 μf =  0.000200 farads. Next solve for capacitive reactance: To find the voltage drop across the resistor (E$_{R}$): To find the voltage drop over the capacitor (E$_{C}$): The sum of these two voltages does not equal the applied  voltage, since the current leads the voltage. Use the following  formula to find the applied voltage: Since this circuit is resistive and capacitive, there is a phase  shift where current leads voltage: To find the impedance, When the circuit contains resistance, inductance, and  capacitance, the following equation is used to find the  impedance. To find the current: Figure 9-26.  A circuit containing resistance and capacitance. Solution: X$_{C}$ = 1 2πfC X$_{C}$ = 1 2π(60)(.00020) X$_{C}$ = 1 0.07536 X$_{C}$  = 13Ω I$_{T}$ =  E Z Z =  √  R$^{2}$ + X 2 $_{C}$  Z =  √  10$^{2}$ + 13 2 Z = 16.4Ω E =  √  (E$_{R}$)$^{2}$ + (E$_{C}$)2 E =  √  67$^{2}$ + 86.1 2 E =  √  4,489 + 7,413 E =  √  11,902 E = 110 volts E$_{R}$ = I × R E$_{R}$  = 6.7A × 10Ω E$_{R}$ = 67 volts E$_{C}$ = I × X$_{C}$ E$_{C}$  = 6.7A × 13Ω E$_{C}$ = 86.1 volts I$_{T}$ = 110V 6.4Ω I$_{T}$ = 6.7 amps Z =  √  R$^{2}$ + (X$_{L}$ – X$_{C}$) 2 R = 10Ω A C = 200  µ F 110V AC 60 cycles  9-17