1. Overview of Impedance and Phases of RC Circuits for AC Sources….w/ Examples! PHY230 – Electronics Prof. Mitchell

2. What Are We Reviewing???  The relationship between current and voltage in an RC circuit  How to determine the impedance and phase angle in a series RC circuit  How to analyze a series RC circuit  How to determine the impedance and phase angle in a parallel RC circuit  How to analyze a parallel RC circuit REMEMBER: i = j sometimes for electrical engineers, which is where some of these figures are from!

3. AC Response of a Series RC Circuit  The current through a capacitor leads the voltage across it.  The current must be the same everywhere because it is a series circuit.  Thus, the capacitor voltage lags the source voltage.  Capacitance causes a phase shift between voltage and current that depends on the relative values of the resistance and the capacitive reactance

4. Adding Impedances  The phase angle is the phase difference between the total current and the source voltage, or the impedances of the resistor (R) and the capacitive reactance (X C )  The impedance of a series RC circuit is determined by adding the R and X C, the same way you would add resistors in a DC circuit.

5. Graphing It…The Impedance Triangle

6. Example: Determine the impedance and the phase angle Z = √[(47) 2 + (100) 2 ]= 110 Ω θ= tan -1 (100/47) = tan -1 (2.13) = 64.8

7. Example #2: If the current is 0.2 mA, determine the source voltage and the phase angle Just use Ohm’s Law: V=IZ X C = 1/[2 π(1×10 3 )(0.01×10 -6 ) ] = 15.9 k Ω Z = √[(10×10 3 ) 2 + (15.9×10 3 ) 2] = 18.8 k Ω V S = IZ = (0.2mA)(18.8k Ω) = 3.76 V θ= tan -1 (15.9k/10k) = 57.8

8. Current, Voltage and Impedance  In a series circuit, the current is the same through both the resistor and the capacitor.  The resistor voltage is in phase with the current, and the capacitor voltage lags the current by 90.  X C and R also follow the voltage so, X C lags R!

9. Kirchhoff's Voltage Law (KVL)  From the KVL, the sum of the voltage drops must equal the applied voltage (V S ).  Since V R and V C are 90 o out of phase with each other, they must be added as phasor quantities  V S = √[V R 2 + V C 2 ] & θ= tan -1 (V C /V R )

10. Example #3: Determine the source voltage and the phase angle V S = √[(10) 2 + (15) 2 ]= 18 V θ= tan -1 (15/10) = tan -1 (1.5) = 56.3

11. How Do Things Change With Frequency? For a series RC circuit, as frequency increases: –R remains constant –X C decreases –Z decreases –θ decreases

12. Example #4: Determine the impedance and phase angle for each of the following values of frequency: (a) 10 kHz (b) 30 kHz (a)X C = 1/2π(10×10 3 )(0.01×10 -6 ) = 1.59 kΩ Z = √[(1.0×10 3 ) 2 + (1.59×10 3 ) 2 ]= 1.88 kΩ θ= tan -1 (1.59k Ω /1.0k Ω ) = 57.8 (b)X C = 1/2π(30×10 3 )(0.01×10 -6 ) = 531 kΩ Z = √[(1.0×10 3 ) 2 + (531) 2 ]= 1.13 kΩ θ= tan -1 (531 Ω/ 1.0k Ω ) = 28.0

13. What About Parallel RC Circuits? Total impedance in parallel RC circuit: Z = (RX C ) / (√[R 2 +X C 2 ]) …add like you would resistors in parallel (TRY FOR YOURSELVES!) Phase angle between the applied V and the total I: θ= tan -1 (R/X C )

14. Current and Voltage  The applied voltage, V S, appears across both the resistive and the capacitive branches  Total current, I tot, divides at the junction into the two branch current, I R and I C

15. The Node Rule is a Bit Tricky! Total current (I S ) is the phasor sum of the two branch currents. Since I R and I C are 90 º out of phase with each other, they must be added as phasor quantities, just like we have been: Thus, I tot = I 1 +I 2 (DC) => I tot = √[I R 2 + I C 2 ] (AC) w/ θ= tan -1 (I C /I R )

16. Example #5: Determine the value of each current, and describe the phase relationship of each with the source voltage I R = 12/220 = 54.5 mA I C = 12/150 = 80 mA I tot = √(54.5) 2 + (80) 2 = 96.8 mA θ= tan -1 (80/54.5) = 55.7