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.

R

VR

C

VR leads VS VC lags VS

I leads VS

I

VS

VC

Sinusoidal Response of RC Circuits

• In a series RC circuit, the total impedance is the phasor sum of R and XC.• R is plotted along the positive x-axis.

R

• XC is plotted along the negative y-axis.

XC

1tan CX

R

Z

• It is convenient to reposition the phasors into the impedance triangle.

R

XC

Z

• Z is the diagonal

Impedance

Impediance

R = 1.2 k

XC = 960

Sketch the impedance triangle and show the values for R = 1.2 k and XC = 960 .

2 21.2 k + 0.96 k

1.33 k

Z

1 0.96 ktan

1.2 k39

Z = 1.33 k

39o

Impediance

Examples

• Ohm’s law is applied to series RC circuits using Z, V, and I.

V V

V IZ I ZZ 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.

Analysis of Series RC

Examples

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.

VR = 12 V

VC = 9.6 V

The voltage phasor diagram can be found from Ohm’s law. Multiply each impedance phasor by 10 mA.

x 10 mA =

R = 1.2 k

XC = 960 Z = 1.33 k

39o

VS = 13.3 V

39o

Analysis of Series RC

Phase Relationships

• Phasor diagrams that have reactance phasors can only be drawn for a single frequency because X is a function of frequency.

• As frequency changes, the impedance triangle for an RC circuit changes as illustrated here because |XC| decreases with increasing f. This determines the frequency response of RC circuits.

Z3

XC1

XC2

XC3

Z2

Z1

12

3

1

2

f

f

f

3

Increasing f

R

Variation of Phase Angle

Examples

(phase lag)

f

f

(phase lag)

V

RVR

Vout

C VoutVin

Vin

Vout

Vin

• 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.

Applications

Examples

(phase lead)

(phase lead)

V

R

VC

VoutC

VoutVin

Vin

Vout

Vin

• 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.

Applications

Examples

• Explain why the capacitance of parallel capacitors is the sum of capacitance of each capacitor.

1C 2C 3C

321

11111

CCC

C

jXjXjXjX

Parallel Capacitor Impedance

• For parallel circuits, it is useful to introduce two new quantities (susceptance and admittance) and to review conductance.

• Conductance is the reciprocal of resistance. 1

GR

• Admittance is the reciprocal of impedance.

• Capacitive susceptance is the reciprocal of capacitive reactance.

1C

C

BX

1Y

Z

Sinusoidal Response of Parallel RC

Admittance

• In a parallel RC circuit, the admittance phasor is the sum of the conductance and capacitive susceptance phasors. The magnitude can be expressed as 2 2 + CY G B

VS G BC

Y

G

BC

• From the diagram, the phase angle is 1tan CB

G


VS G BC

Y

G

BC

• G is plotted along the positive x-axis.

• BC is plotted along the positive y-axis. 1tan CB

G

• Y is the

diagonal

• Some important points to notice are:


VS C

0.01 F Y = 1.18 mSG = 1.0 mS

BC = 0.628 mS

Draw the admittance phasor diagram for the circuit.

f = 10 kHz

R

1.0 k

1 11.0 mS

1.0 kG

R

2 10 kHz 0.01 F 0.628 mSCB

The magnitude of the conductance and susceptance are:

2 22 2 + 1.0 mS + 0.628 mS 1.18 mSCY G B


• Ohm’s law is applied to parallel RC circuits using Y, V, and I.

I I

V I VY YY V

• 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.

Analysis of Parallel RC

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.

x 10 V =Y =

1.18 mS

G = 1.0 mS

BC = 0.628 mS

IR = 10 mA

IC = 6.28 mA

IS = 11.8 mA

Analysis of Parallel RC

• Notice that the formula for capacitive susceptance is the reciprocal of capacitive reactance. Thus BC and IC are directly proportional to f: 2CB fC

IR

ICIS

• As frequency increases, BC and IC must also increase, so the angle between IR and IS must increase.

Phase Angle of 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:

Z

Req = Z cos

XC(eq) = Z sin

Equivalent Series and Parallel RC

Examples

Phase Relationships

Examples

• 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:

2 22 2 2

2 2

C

C

R XZ

R X

R1 C1

R2 C2

Z1Z2

• The total impedance can be found by converting the parallel components to an equivalent series combination, then adding the result to R1 and XC1 to get the total reactance.

2 21 1 1CZ R X

Series-Parallel RC Circuits

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

VC = 9.6 V

x 10 mA =

R = 1.2 k

XC = 960 Z = 1.33 k

39o

VS = 13.3 V

39o

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.

The rms current in the earlier example was 10 mA. Show the power triangle.

Ptrue = 120 mW

Pr = 96 mVAR

x 10 mA = 39o

Pa = 133 mVA

VR = 12 V

VC = 9.6 VVS = 13.3 V

39o

The Power Triangle

Examples

• The power factor is the relationship between the apparent power in volt-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.

Power Factor

• 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)

Pr (VAR)

• Some components such as transformers, motors, and generators are rated in VA rather than watts.

Pa (VA)

Apparent

Impedance

Phase angle

Capacitive suceptance (BC)

Admittance (Y)

The total opposition to sinusoidal current expressed in ohms.

The ability of a capacitor to permit current; the reciprocal of capacitive reactance. The unit is the siemens (S).

The angle between the source voltage and the total current in a reactive circuit.

A measure of the ability of a reactive circuit to permit current; the reciprocal of impedance. The unit is the siemens (S).

Quiz

Power factor

Frequency response

Cutoff frequency

The frequency at which the output voltage of a filter is 70.7% of the maximum output voltage.

In electric circuits, the variation of the output voltage (or current) over a specified range of frequencies.

The relationship between volt-amperes and true power or watts. Volt-amperes multiplied by the power factor equals true power.

Selected Key Terms

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

Quiz

2. A series RC circuit is driven with a sine wave. If the output voltage is taken across the resistor, the output will

a. be in phase with the input.

b. lead the input voltage.

c. lag the input voltage.

d. none of the above

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

Quiz

4. If you increase the frequency in a series RC circuit,

a. the total impedance will increase

b. the reactance will not change

c. the phase angle will decrease


Quiz

5. Admittance is the reciprocal of

a. reactance

b. resistance

c. conductance

d. impedance

Quiz

6. Given the admittance phasor diagram of a parallel RC circuit, you could obtain the current phasor diagram by

a. multiplying each phasor by the voltage

b. multiplying each phasor by the total current

c. dividing each phasor by the voltage

d. dividing each phasor by the total current

Quiz

7. If you increase the frequency in a parallel RC circuit,

a. the total admittance will decrease

b. the total current will not change

c. the phase angle between IR and IS will decrease


Quiz

8. The magnitude of the admittance in a parallel RC circuit will be larger if

a. the resistance is larger

b. the capacitance is larger

c. both a and b


Quiz

9. The maximum power factor occurs when the phase angle is

a. 0o

b. 30o

c. 45o

d. 90o

Quiz

10. When power is calculated from voltage and current for an ac circuit, the voltage and current should be expressed as

a. average values

b. rms values

c. peak values

d. peak-to-peak values

Quiz

Answers:

1. a

2. b

3. c

4. c

5. d

6. a

7. d

8. d

9. a

10. b

Quiz

sinusoidal response of rc circuits

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