sinusoidal steady state circuit analysis topic 3

8/3/2019 Sinusoidal Steady State Circuit Analysis Topic 3

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TOPIC 3 : SINUSOIDAL STEADY STATE CIRCUIT ANALYSIS

A) i. Purely Resistive Circuit with AC Source

As shown in Figure 3(a) above the resistor voltage (VR), and the current (IR) are allsine waves with the same frequency from the voltage source VT.

The resistor voltage (VR), and current (I) are in phase with each other.. The phaseangle is always 0.

The Resistance R does not change with the frequency. R remains constant even if the frequency is increase or decrease. Ohm's law can be applied to resistors in AC circuits:

VR= IRx R

ii) Pure Inductive Circuit with Ac Source

Figure 3(b)

Phasor diagram: VR is in phase withIR(Phase angle = 0)

VR= IVRI< 0IR = IIRI < 0

Voltage VR

Current IR Sinusoldal waveform

showing VR is in phasewith IR:VR= Vm sin t

IR= Im sin t

Figure 3(a)

Sinusoldal waveform showing

VLlead IL by 90 or /2:VL= Vm sin (t + 90 )IL = Im sin t

Phasor diagram: Voltage VL leadcurrent IL by a phase angle of 90(Phase angle difference = 90)VR= IVRI< 90IR = IIRI < 0

IL

VL

VL

IL90


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Inductive Reactance

Inductive reactance is the opposition to AC current flow in an inductive circuit,

The symbol for inductive reactance is XL.Its unit is the ohm. Inductive reactance XL increase as the frequency across it increases, therefor inductive reactance is directly proportional to frequency,

XL is given by the equation:XL= 2fL

Since 2fis equal to (the angular frequency),XL= L

An inductor's current and voltage are related by an equation similar to Ohm's law:

VL= IL x XL

iii) Pure Capacitive Circuit with Ac Source

Figure 3(c)

Capacitive Reactance

Capacitive Reactance in a purely capacitive circuit is the opposition to current flow

in AC circuits only. The symbol for capacitive reactance is XC.Its unit is the ohm. The capacitive reactance XC of the capacitor decreases as the frequency across it

increases. therefore capacitive reactance is inversely proportional to frequency

XC is given by the equation:XC= 1/2fC

Since 2fis equal to (the angular frequency), XC= 1/C

A Capasitor current and voltage are related by an equation similar to Ohm's law:

Vc= ICx XC

iv) Graph showing relationship between R,XL and XC with respect to frequency

IC

VC90

Phasor diagram: IC lead VC by 90 or /2:(Phase angle = 90)VC = IVCI< 0

IC = IICI


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B) Series Circuit with inductive and capacitive load In a series circuit, current is taken as reference because current is the same in

series circuit.

i) R-L series

When a sinusoidal voltage is applied to an RL circuit, the current and all the voltagedrops are also sine waves.

Total current IT in an RL circuit always lags the source voltage VT

The resistor voltage VR is always in phase with the current IT. In an ideal inductor, the voltage VL always leads the current IT by 90

.

The impedance (Z) of an RL circuit is the total opposition to AC current flow caused bythe resistor (R) and the reactance of the inductor (XL).

The equation for the total impedance (ZT) of an RL circuit is:

or in complexs form : Z = R + j XL

The total voltage in a series RL circuit is given by this equation:

Or in complex form : VT = VR + jV L

The total phase angle can be determined by the equation:

= tan-1(XL/ R) or = tan-1 (VL/VR)

Complete Analysis of a Series RL Circuit:

R

frequency

R

VR

IR VR IL

VL VL

XL

IT

VAC , f

VR IT

+jVL

Where: ZT = the total impedance in ohmsX

L= the inductive reactance in ohms

R = the resistance in ohms R

ZT+jXL

Where:VT = total voltageVR = voltage across resistor RV = volta e across inductor L


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= -tan-1(XC/ R) or = -tan-1 (VC/VR)

Complete Analysis of a Series RC Circuit

Here is the procedure for doing a complete analysis of a series RC circuit, given the values ofR, C, f, and VT.

Step 1. Calculate the value of XC:

XC = 1 / (2fC)Step 2. Calculate the total impedance: or in complexs form Z= R jXC

Step 3. Use Ohm's Law to calculate the total current:IT = VT / ZT

Step 4. Calculate the currents through R and C. Since this is a series circuit:IT= IC = IR

Step 5. Calculate the voltages across R and C. By Ohm's Law:VR= IRxR and VC = IRxXC

Step 6. Determine the phase angles for R and C. Phase angles for these components arealways:

R= 0C = -90

Step 7. Calculate the total phase angle for the RC circuit:T =- tan

-1(XC/ R)

iii) Series R- LC

We have seen that the three basic passive components, R, L and C have very differentphase relationships to each other when connected to a sinusoidal AC supply. In a pure

ohmic resistor the voltage is "in-phase" with the current, in a pure inductancethe voltage "leads" the current by 90o, ( ELI ) and with a pure capacitance thevoltage "lags" the current by 90o, ( ICE ).

ThisPhase Difference, depends upon the reactive value of the components beingused and we know that reactance, ( X ) is zero if the element is resistive, positive ifthe element is inductive and negative if the element is capacitive giving the resultingimpedance values as:

o ZR = R = R


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Step 1. Calculate the value of XC and XL :XC = 1 / (2fC) and XL = 2fL

Step 2. Calculate the total impedance:

ZT = R + jXL -jXCStep 3. Use Ohm's Law to calculate the total current:

IT = VT / ZTStep 4. Calculate the currents through R, L and C. Since this is a series circuit:

IT =IC = IL =IRStep 5. Calculate the voltages across R, L and C. By Ohm's Law:

VR= IT xR, VC = IT xXC and VL = IT x XLStep 6 Prove that supply voltage in a series circuit is equivalent to the total vector of

voltages across each component of R, L and C. (Kirchoffs Law)

VT = VR+ jVL -jVCStep 7. Calculate the total phase angle for the R-L-C series circuit:

T = - tan-1(XC - XL/ R) if XC > XL

T = tan-1(XL- XC/ R) if XL > XC

Phasor diagram for R-L-C series

XL > XC (the circuit is inductive) XC > XL (the circuit is capacitive)

C) Parallel Circuit with inductive and capacitive load In a parallel circuit, voltage is taken as reference because voltage is the same in

parallel circuit

i) R-L parallel

Complete Analysis of a Parallel RL Circuit

Here is the procedure for doing a complete analysis of a parallel RL circuit, given the values ofR, L, f, and VT.

+jXL

-j(XC-XL)

-jXC

R

ZT

-

+jXL

+j(XL-XC)

-jXC

R

ZT

IT IR IL

VR VL

R XL

VT, f

-jIL

IR VT

IT

-


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Step 1. Calculate the value of XL and

XL = 2fL or j LStep 2. Determine the voltages for R and L. Since this is a parallal circuit:

VT = VL = VRStep 3. Use Ohm's Law to calculate the currents for R and L:

IR= VT / R and IL = VT / XL

Step 4. Calculate the total current:

or in complexs form ; IT = IR- jILStep 5. Use Ohm's Law to calculate the total impedance:

ZT = VT / ITStep 6. Determine the phase angles for R and L. Phase angles for these components in aparallel circuit are always:

R= 0 and L = -90Step 7. Calculate the total phase angle for the circuit:

T = -tan-1(IL/ IR)

Total Impedance(ZT) for Parallel RL Circuits

Or Total Impedance 1/ZT = (1/R)2 +(1/XL)2

ii) R-C parallel

Complete Analysis of a Parallel RC Circuit

Here is the procedure for doing a complete analysis of a parallel RC circuit, given the values of

R, C, f, and VT.

Step 1. Calculate the value of XC: XC = 1/(2fL)

Step2. Determine the voltages for R and L. Since this is a parallal circuit:

VT = VL = VRStep 3. Use Ohm's Law to calculate the currents for R and L:

IR= VT / R and IC = VT / XC

A simple parallel RL circuit is one that look like theone shown here --

Total Impedance ZT = R x (jXL)R + jXL

ITIT

R

VR

IC

VC

C

+jIC

IR VT

IT


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Step 4. Calculate the total current:

or in complexs form ; IT = IR+ jICStep 5. Use Ohm's Law to calculate the total impedance:

ZT = VT / ITStep 6. Determine the phase angles for R and L. Phase angles for these components in aparallel circuit are always:

R= 0 and C = 90Step 7. Calculate the total phase angle for the circuit:T = tan

-1(IC / IR)

Total Impedance 1/ZT= (1/R)2 +(1/XC)

2

or Total Impedance ZT = R x (- jXC)R +(jXC)

iii) Parallel R-L-C

Complete Analysis of a Parallel R-L-C Circuit

Here is the procedure for doing a complete analysis of a parallel RLC circuit, given the valuesof R, L, C, f, and VT.Step 1. Calculate the value of XL and XC:

XL = 2fL XC = 1/(2fL) Step2. Determine the voltages for R , L and C. Since this is a parallal circuit:

VT = VL = VC = VRStep 3. Use Ohm's Law to calculate the currents for R and L:

IR= VT / R , IL = VT / XL and IC = VT / XCStep 4. Calculate the total current:

or in complexs form ; IT = IR+ jIC + (-jIL)

Step 5. Use Ohm's Law to calculate the total impedance:

ZT = VT / IT

Step 6. Determine the phase angles for R and L. Phase angles for these components in aparallel circuit are always:

R= 0 , C = 90 and L = -90

Step 7. Calculate the total phase angle for the circuit:

T = tan-1 (IC IL)/ IR

IS IR

VR

R

IL

VL

L

IC

VC

C

jIC

jIL

IR VT


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Total Impedance :

1/ZT= (1/R)2 +(1/XL 1/XC)

2

Phasor diagram for R-L-C parallel

IC> IL IL > IC

iv) Impedance Triangle

The opposition to current flow in AC series circuits is impedance, Z which has two

components, resistance R and reactance, X and from this we can construct an impedancetriangle.,

D) Combination of series- parallel R-L-C circuits.

XC

jIC

j(IC-IL)

-jIL

IR V

IT

jIC

- j(IL-IC)

-jIL

IR V

-

Z1IT

V1

Z2

V2V3

Z3

VT, f

I2

I3


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Components connected in series must have the same current.

Components connected in parallel must have the same voltage These two rules apply not only to individual components but also to portions of

circuits.

i) Total impedance in a series-parallel circuit.

ii) Total current: IT = VT/ZT and I2 =V2/Z2

I3 = V3/Z3Supply current, IT will be the vector sum of the branch currents I2 and I3 (KCL)

iii) The voltage drop across each resistor in a series-parallel circuit.V2 = V3 (Z2 parallel to Z3 so the voltage drop is the same)

Supply Voltage VT will be the vector sum of V1 and V2or V1 and V3 (KVL)

V1 = IT x Z1V2 = V3 is the vector of VT minus V1

or V2=V3= IT x Z2xZ3Z2 +Z3

E) i) Work and Energy In physics, work is the expenditure of energy to overcome a restraint or to achieve a

change in the physical state of a body. Energy is the ability to do work. Energy comes in many forms. A few of these are:

o heat energy

o light energyo sound energy

o mechanical energy

o chemical energyo electrical energy.

Many everyday devices convert energy from one form to another. For example:

o A loudspeaker converts electrical energy to sound energy.

o A toaster converts electrical energy to heat energy.

o A light bulb or LED converts electrical energy to light energy. Energy(E) is measured in units calledjoules, symbol J to stand for the joule.

ii) Power Power, abbreviated P, is the rate at which energy is spent. In other words, power is the amount of energy that's used in a given amount of time,

or energy per time. In equation form, that's

P = E / t joules per second or WATT (W)

ZT = Z1 + Z2 // Z3

ZT =Z1 + Z2 X Z3Z2 + Z3


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Power Formulas

When current flows through resistance, electric energy is converted to heat energy. The rate of this energy conversion is the power. P= I2 R

P = I V

P= V2 / R In each of these equations, R is the size (in ohms) of the resistance, Vis the voltage

(in volts) across the resistance, and Iis the current (in amps) through the resistance.

F) Power consumption in AC circuits

i) Power in Pure Resistive

True orActual power,. symbol Ptrue unit Watt (W) is a measure of the rate at which a

component or circuit loses energy. This energy loss is usually due to dissipation of heat(as in a resistor) or conversion to some other form of energy (as in a motor that

converts electrical energy to motion). In a resistive circuit : Current and Voltage waveforms are in phase,

Average power: P = IRMS x ERMS

In AC circuits, the resistor's voltage and current must be in rms values(effective values):

Ptrue = Irms2 x R

Ptrue = Vrms2 / R

Ptrue = Vrms x Irms

Power dissipation in a resistor means that the resistor is converting some of the

circuit's electrical energy to heat energy. This energy is lost from the circuit as the air

around the resistor is heated. This is True Power

ii) Pure capacitance and inductancea) Power in an Inductor

An inductor stores energy in its magnectic field when there is current through it. An ideal inductor (assuming no winding resistance) does not dissipate energy, it only

stores it.

When an AC voltage is applied to an ideal inductor, energy is stored by the inductorduring a portion of a cycle, then the stored energy is returned to the source duringanother portion of the cycle.

Ideally all of the energy stored by an inductor during the positive portion of the power

cycle is returned to the source during the negative portion No net energy is lost in an ideal inductor due to conversion of heat. So the True power (Ptrue), unit watt is zero. The rate at which an inductor stores or returns energy is called its Reactive

power(Q), unit VAR (volt ampere reactive). These formulas apply:

POWER P


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Q = Vrms x IrmsQ = V2rms/XLQ = I2rms x XL

P = I V = 0 watt (because +cycle =-cycle so

average power =0)

0 - 90 90180 180-270 270 - 360

VL= +. VL= - VL= - VL= +IL= + IL= + IL= - IL= -

b) Power in a Capaciotr

In capacitor, true power is zero, capacitors don't lose any energy as heat. Reactive power is a measure of the rate at which a component is storing energy or

returning energy to the circuit. symbol Qfor reactive power, unit called the VAR. (volt-ampere reactive) These formulas apply:

Q = VrmsxIrms

Q = Vrms2/XC

Q =Irms2xXC

Question: Suppose a 1 F capacitor has a voltage drop of 3 V rms, and the voltage's

frequency is 2 kHz. Calculate the capacitor's reactive power.

For sinusoidal waves, the current through a capacitor leads the voltage across it by 90,

as shown in this example below.

Reactive Power Reactive power(Q) is the power

consumed in an AC circuit because of theexpansion and collapse of magnetic

(inductive) and electrostatic (capacitive)fields.

Reactive power is expressed in volt-

amperes-reactive (VAR). Unlike true power, reactive power is not useful power because it is stored in the circuit

itself.

90

vL = Vmax sin (t+ 90)iL = Imaxsin t

VL lead IL by 90

vc = Vmax sin tiC = Imax sin (t + 90)Ic lead VC by 90

P = I V = 0 watt (because +cycle =-cycle so

average power (True Power) =0)

0 - 90 90

180

180

-270

270 - 360Vc= +. VC= + VC= - IC = +

IC= + IC= - Ic = - VC= -

Power P

Power P


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Total Power (apparent power)

The total powerdelivered by the source is the apparent power.

Apparent power is expressed in volt-amperes (VA). Part of this apparent power,called true power, is dissipated by the circuit resistance in

the form of heat.

The rest of the apparent power is returned to the source by the circuit inductance andcapacitance

The phasor combination of resistive power (true power) and reactive power is the

apparent power.

Power Triangle

In an RL and RC circuit, part of the power is resistive and part reactive.

Power Factor Power factor(pf) is the ratio between true power and apparent power. True power is

the power consumed by an AC circuit, and reactive power is the power that is stored inan AC circuit.

Cos is called the power factor (pf) of an AC circuit. where is the phase anglebetween the applied voltage and current sine waves and also between P and S on apower triangle (Refer power triangle)

The power factor (pf=cos ) indicates how much of the apparent power is true power. A power factor of 1 indicates a purely resistive circuit, and a power factor of 0 indicates

a purely reactive circuit. Power Factor (pf) = Cos = P/S = Watt/VA.

wherecos = power factor (pf)P = true power (watts)

S = apparent power (VA) In an inductive circuit, the current lags the voltage and is said to have a lagging power

factor

In a capacitive circuit, the current leads the voltage and is said to have a leadingpower factor

Power Factor Correction

In an electric power system, a load with a low power factor draws more currentthan a load with a high power factor for the same amount of useful powertransferred.

The higher currents increase the energy lost in the distribution system, and requirelarger wires and other equipment.


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Because of the costs of larger equipment and wasted energy, electrical utilities willusually charge a higher cost to industrial or commercial customers where there is alow power factor.

Linear loads with low power factor (such as induction motors) can be corrected with a

passive network ofcapacitors or inductors. Power factor correction brings the power factor of an AC power circuit closer

to 1 by supplying reactive power of opposite sign, adding capacitors or inductorswhich act to cancel the inductive or capacitive effects of the load.

Questions

1. A series circuit with resistance 100 and capacitance 50F connected to a voltagesupply of 200V, 50 Hz. Calculate:

a) the circuit impedanceb) total circuit currentc) power factor of the circuitd) the phase nagle between the total voltage and the line currente) voltage across resistor

f) voltage across capacitorg) draw the related vector diagram for the circuit,

2. A pure inductor 0,08Henry is in series with a resistor 15. The voltage supply is 240 V,50 Hz, calculate:

a) the current in the circuitb) the power factor of the circuitc) Voltage drop across R and L

d) The actual power ,e) the reactive power andf) the apparent powerg) draw the power triangle

2. A single phase motor of 50HZ, takes 30 kWatt and 20 kVAR from the supply of 240 VCalculate the value of the capacitance needed to be connected parallel to the motor

for the correction of the power factor to be 0.95 lagging
http://en.wikipedia.org/wiki/Induction_motorhttp://en.wikipedia.org/wiki/Capacitorhttp://en.wikipedia.org/wiki/Inductorhttp://en.wikipedia.org/wiki/Inductorhttp://en.wikipedia.org/wiki/Capacitorhttp://en.wikipedia.org/wiki/Induction_motor

sinusoidal steady state circuit analysis topic 3

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