content_image/512523.jpg content_image/512525.jpg content_image/512524.jpg content_image/512531.jpg content_image/512529.jpg content_image/512527.jpg content_image/512528.jpg content_image/512526.jpg content_image/512530.jpg
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? 

Solution: 

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: 

<img src='content_image/512523.jpg'>

To find the impedance, 

<img src='content_image/512525.jpg'>

Since this circuit is resistive and capacitive, there is a phase shift where current leads voltage: 

To find the current: 

<img src='content_image/512524.jpg'>

<img src='content_image/512531.jpg'>

<img src='content_image/512529.jpg'>

To find the voltage drop across the resistor (E$_{R}$): 

<img src='content_image/512527.jpg'>

To find the voltage drop over the capacitor (E$_{C}$): 

<img src='content_image/512528.jpg'>

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: 

<img src='content_image/512526.jpg'>

When the circuit contains resistance, inductance, and capacitance, the following equation is used to find the impedance. 

<img src='content_image/512530.jpg'>

