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Sinusoidal Response of RC Circuits

Sinusoidal Response of RC Circuits. When both resistance and capacitance are in a series circuit, the phase angle between the applied voltage and total current is between 0  and 90 , depending on the values of resistance and reactance. Impedance.

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Sinusoidal Response of RC Circuits

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  1. Sinusoidal Response of RC Circuits • When both resistance and capacitance are in a series circuit, the phase angle between the applied voltage and total current is between 0 and 90, depending on the values of resistance and reactance.

  2. Impedance • In a series RC circuit, the total impedance is the phasor sum of R and XC. • Ris plotted along the positive x-axis. • XC is plotted along the negative y-axis. R R • Z is the diagonal XC XC Z Z • It is convenient to reposition the phasors into the impedance triangle.

  3. Impediance

  4. Impediance

  5. Impediance Example Sketch the impedance triangle and show the values for R = 1.2 kW and XC = 960 W. R = 1.2 kW 39o XC = 960 W Z = 1.33 kW

  6. Examples

  7. Examples

  8. Analysis of Series RC • Ohm’s law is applied to series RC circuits using Z, V, and I. • Because I is the same everywhere in a series circuit, you can obtain the voltages across different components by multiplying the impedance of that component by the current as shown in the following example.

  9. Examples

  10. Examples

  11. x 10 mA = Analysis of Series RC Example Assume the current in the previous example is 10 mArms. Sketch the voltage phasor diagram. The impedance triangle from the previous example is shown for reference. The voltage phasor diagram can be found from Ohm’s law. Multiply each impedance phasor by 10 mA. VR = 12 V R = 1.2 kW 39o 39o XC = 960 W VC = 9.6 V VS = 13.3 V Z = 1.33 kW

  12. Phase Relationships

  13. Variation of Phase Angle • Phasor diagrams that have reactance phasors can only be drawn for a single frequency because X is a function of frequency. • As frequencychanges, the impedance triangle for an RC circuit changes as illustrated here because |XC| decreases with increasing f. This determines the frequency response of RCcircuits.

  14. Examples

  15. Examples

  16. Examples

  17. Examples

  18. Applications • For a given frequency, a series RC circuit can be used to produce a phase lag by a specific amount between an input voltage and an output by taking the output across the capacitor. This circuit is also a basic low-pass filter, a circuit that passes low frequencies and rejects all others. R VR Vout Vin Vout C Vin Vout Vin

  19. Examples

  20. Applications • Reversing the components in the previous circuit produces a circuit that is a basic lead network. This circuit is also a basic high-pass filter, a circuit that passes high frequencies and rejects all others. This filter passes high frequencies down to a frequency called the cutoff frequency. C Vout Vin Vin Vout R Vout VC Vin

  21. Examples

  22. Parallel Capacitor Impedance • Explain why the capacitance of parallel capacitors is the sum of capacitance of each capacitor.

  23. Sinusoidal Response of Parallel RC • For parallel circuits, it is useful to introduce two new quantities (susceptance and admittance) and to review conductance. • Conductance is the reciprocal of resistance. • Capacitive susceptance is the reciprocal of capacitive reactance. • Admittance is the reciprocal of impedance.

  24. Admittance

  25. Admittance

  26. Sinusoidal Response of Parallel RC • In a parallel RCcircuit, the admittance phasor is the sum of the conductance and capacitive susceptance phasors. The magnitude can be expressed as • From the diagram, the phase angle is BC Y BC VS G G

  27. Sinusoidal Response of Parallel RC • Some important points to notice are: • G is plotted along the positive x-axis. • BCis plotted along the positive y-axis. • Y is the diagonal BC Y BC VS G G

  28. Sinusoidal Response of Parallel RC Example Draw the admittance phasor diagram for the circuit. The magnitude of the conductance and susceptance are: BC= 0.628 mS VS R 1.0 kW C 0.01 mF f = 10 kHz Y = 1.18 mS G = 1.0 mS

  29. Analysis of Parallel RC • Ohm’s law is applied to parallel RC circuits using Y, V, and I. • Because V is the same across all components in a parallel circuit, you can obtain the current in a given component by simply multiplying the admittance of the component by the voltage as illustrated in the following example.

  30. x 10 V = Analysis of Parallel RC Example If the voltage in the previous example is 10 V, sketch the current phasor diagram. The admittance diagram from the previous example is shown for reference. The current phasor diagram can be found from Ohm’s law. Multiply each admittance phasor by 10 V. BC= 0.628 mS IC= 6.28 mA Y = 1.18 mS IS = 11.8 mA IR = 10 mA G = 1.0 mS

  31. IC IS q IR Phase Angle of Parallel RC • Notice that the formula for capacitive susceptance is the reciprocal of capacitive reactance. Thus BC and ICare directly proportional to f: • As frequency increases, BC and IC must also increase, so the angle between IR and ISmust increase.

  32. Equivalent Series and Parallel RC • For every parallel RC circuit there is an equivalent series RC circuit at a given frequency. • The equivalent resistance and capacitive reactance are shown on the impedance triangle: Req= Z cos q XC(eq)= Z sin q Z

  33. Examples

  34. Examples

  35. Phase Relationships

  36. Examples

  37. Examples

  38. Z1 Z2 R1 C1 R2 C2 Series-Parallel RC Circuits • Series-parallel RC circuits are combinations of both series and parallel elements. These circuits can be solved by methods from series and parallel circuits. For example, the components in the green box are in series: The components in the yellow box are in parallel: • The total impedance can be found by converting the parallel components to an equivalent series combination, then adding the result to R1 and XC1to get the total reactance.

  39. The Power Triangle

  40. x 10 mA = The Power Triangle • Recall that in a series RC circuit, you could multiply the impedance phasors by the current to obtain the voltage phasors. The earlier example is shown for review: VR = 12 V R = 1.2 kW 39o 39o XC = 960 W VC = 9.6 V VS = 13.3 V Z = 1.33 kW

  41. VR = 12 V x 10 mA = 39o VC = 9.6 V VS = 13.3 V The Power Triangle • Multiplying the voltage phasors by Irms gives the power triangle (equivalent to multiplying the impedance phasors by I2). Apparent power is the product of the magnitude of the current and magnitude of the voltage and is plotted along the hypotenuse of the power triangle. Example The rms current in the earlier example was 10 mA. Show the power triangle. Ptrue = 120 mW 39o Pr = 96 mVAR Pa = 133 mVA

  42. Examples

  43. Examples

  44. Power Factor • Thepower factor is the relationship between the apparent power involt-amperes and true power in watts. Volt-amperes multiplied by the power factor equals true power. • Power factor is defined mathematically as PF = cos  • The power factor can vary from 0 for a purely reactive circuit to 1 for a purely resistive circuit.

  45. Apparent • Apparent power consists of two components; a true power component, that does the work, and a reactive power component, that is simply power shuttled back and forth between source and load. Ptrue (W) • Some components such as transformers, motors, and generators are rated in VA rather than watts. Pr (VAR) Pa (VA)

  46. Quiz The total opposition to sinusoidal current expressed in ohms. Impedance Phase angle Capacitive suceptance (BC) Admittance (Y) The angle between the source voltage and the total current in a reactive circuit. The ability of a capacitor to permit current; the reciprocal of capacitive reactance. The unit is the siemens (S). A measure of the ability of a reactive circuit to permit current; the reciprocal of impedance. The unit is the siemens (S).

  47. Selected Key Terms The relationship between volt-amperes and true power or watts. Volt-amperes multiplied by the power factor equals true power. Power factor Frequency response Cutoff frequency In electric circuits, the variation of the output voltage (or current) over a specified range of frequencies. The frequency at which the output voltage of a filter is 70.7% of the maximum output voltage.

  48. Quiz 1. If you know what the impedance phasor diagram looks like in a series RC circuit, you can find the voltage phasor diagram by a. multiplying each phasor by the current b. multiplying each phasor by the source voltage c. dividing each phasor by the source voltage d. dividing each phasor by the current

  49. Quiz 2. A series RC circuit is driven with a sine wave. If the output voltage is taken across the resistor, the output will • be in phase with the input. • lead the input voltage. • lag the input voltage. • none of the above

  50. Quiz 3. A series RC circuit is driven with a sine wave. If you measure 7.07 V across the capacitor and 7.07 V across the resistor, the voltage across both components is a. 0 V b. 5 V c. 10 V d. 14.1 V

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