Unit - 2

AC Circuits

Representation of sinusoidal waveforms

Peak Value:

The arithmetic mean of all the value over complete one cycle is called as average value

=

For the derivation we are considering only hall cycle.

Thus varies from 0 to ᴫ

i = Im Sin

Solving

We get

Similarly, Vavg=

The average value of sinusoid ally varying alternating current is 0.636 times maximum value of alternating current.

RMS value: Root mean square value

The RMS value of AC current is equal to the steady state DC current that required to produce the same amount of heat produced by ac current provided that resistance and time for which these currents flows are identical.

I rms =

Direction for RMS value:

Instantaneous current equation is given by

i = Im Sin

But

I rms =

=

=

=

Solving

=

=

Similar we can derive

V rms= or 0.707 Vm

the RMS value of sinusoidally alternating current is 0.707 times the maximum value of alternating current.

Phasor for Sinusoidal wave

A sinusoidal quantity, i.e. current, i(t) = Imsin(ωt), is taken up as an example. In Fig.a, the length, OP, along the x-axis, represents the maximum value of the current Im, on a certain scale. It is being rotated in the anti-clockwise direction at an angular speed, ω, and takes up a position, OA after a time t (or angle, θ = ωt, with the x-axis). The vertical projection of OA is plotted in the right hand side of the above figure with respect to the angle θ. It will generate a sine wave (Fig.b), as OA is at an angle, θ with the x-axis, as stated earlier. The vertical projection of OA along y-axis is OC = AB = i(θ)=Im sin θ, which is the instantaneous value of the current at any time t or angle θ. The angle θ is in rad., i.e. θ = ω t.

The angular speed, ω is in rad/s, i.e. ω = 2π f, where f is the frequency in Hz or cycles/sec. Thus, i=Im sinθ = Im sinωt = Im 2sinπft.

So, OP represents the phasor with respect to the above current, i. The line, OP can be taken as the rms value I=Im/√2, instead of maximum value, Im. Then the vertical projection of OA, in magnitude equal to OP, does not represent exactly the instantaneous value of I, but represents it with the scale factor of 1/√2=0.707. The reason for this choice of phasor as given above.

Fig 1(a) Phasor for current (b) Waveform

The current can be of the form, i(t)= Im (sinωt −α) as shown in Fig d. The phasor representation of this current is the line, OQ, at an angle, α (may be taken as negative), with the line, OP along x-axis (Fig c). One has to move in clockwise direction to go to OQ from OP (reference line), though the phasor, OQ is assumed to move in anti-clockwise direction as given earlier. After a time t, OD will be at an angle θ with OQ, which is at an angle (θ −α = ωt −α), with the line, OP along x-axis. The vertical projection of OD along y-axis gives the instantaneous value of the current

i=√2 I sin (ωt −α) = Im sin (ωt −α).

Fig 2(c) Phasor of phase shifted sinusoidal current (d) waveform

Key takeaway

V rms= or 0.707 Vm

the RMS value of sinusoidally alternating current is 0.707 times the maximum value of alternating current.

Real Power: [P]

It is nothing but the actual power being used in a circuit.

P= = I2R Watts

Reactive Power: [Q]

It is the function of reactance in the circuit X. Mainly reactive loads are inductor and capacitors. These elements dissipate zero power. These element shows that they dissipate power. This is called as reactive power.

Q= = I2X VAR (volt-Ampere-Reactive)

Apparent Power: [S]

It is the product of a circuit voltage and current without reference to phase angle. It is the combination of both reactive and real power.

S= = I2Z VA (volt-Ampere)

Power factor (P.F.)

It is the cosine of angle between voltage and current

If Ɵis –ve or lagging (I lags V) then lagging P.F.

If Ɵ is +ve or leading (I leads V) then leading P.F.

If Ɵ is 0 or in phase (I and V in phase) then unity P.F.

Ac circuit containing pure resistance (R)

Fig 3 Circuit containing R

Consider Circuit Consisting pure resistance connected across ac voltage source

V = Vm Sin ωt ①

According to ohm’s law i = =

But Im =

②

From ① and ②phase or represents RMD value.

Fig 4 Waveform with element R

Power P = V. i

Equation P = Vm sin ω t Im sin ω t

P = Vm Im Sin2 ω t

P = -

Constant fluctuating power if we integrate it becomes zero

Average power

Pavg =

Pavg =

Pavg = Vrms Irms

AC circuit with a pure inductance

Fig 5 Circuit with pure inductance

Consider an AC circuit with a pure inductance L as shown in the figure. The alternating voltage v is given by

V = Vm Sin ωt

The current flowing in the circuit is i. The voltage across the inductor is given as VL which is the same as v. We can find the current through the inductor as follows

I=L

Vm Sin ωt= L

Di= Sin ωt

i =

i = sin (t-)

i= im sin (t-)

im =

We observe that in a pure inductive circuit, the current lags behind the voltage by 90⁰. Hence the voltage and current waveforms and phasors can be drawn as below

Fig 6 Phasor for Pure L

Inductive reactance

The inductive reactance XL is given as

XL = 2πfL

im =

AC circuit with a pure capacitance

Fig 7 Circuit with pure Capacitance

Consider an AC circuit with a pure capacitance C as shown in the figure. The alternating voltage v is given by

V = Vm Sin ωt

= Vm 00

We can find the current through the capacitor as follows

q=CV

q= Cvm Sin ωt

= ωCvm cos(ωt)

i= ωCvm cos(ωt)

i=im sin (t+)

im= ωCvm

Xc = vm/im = 1/ωC

We observe that in a pure capacitive circuit, the current leads the voltage by 90⁰. Hence the voltage and current waveforms and phasors can be drawn as below.

Fig 8 Phasor for C

Capacitive reactance

The capacitive reactance XC is given as

Xc= 1/2πfC

im = vm/Xc

R-L Series Circuit:

Fig 9 Series RL Circuit

Consider an AC circuit with a resistance R and an inductance L connected in series as shown in the figure. The alternating voltage v is given by

V = Vm Sin ωt

The current flowing in the circuit is i. The voltage across the resistor is VR and that across the inductor is VL

VR=IR is in phase with I

VL=IXL leads current by 90 degrees

With the above information, the phasor diagram can be drawn as shown.

Fig 10 Phasor for RL Circuit

The current I is taken as the reference phasor. The voltage VR is in phase with I and the voltage VL leads the current by 90⁰. The resultant voltage V can be drawn as shown in the figure. From the phasor diagram we observe that the voltage leads the current by an angle Φ or in other words the current lags behind the voltage by an angle Φ. The waveform and equations for an RL series circuit can be drawn as below

Fig 11 Waveform for R-L circuit

V= Vm Sin ωt

I= Im Sin (ωt-ɸ)

From the phasor diagram, the expressions for the resultant voltage V and the angle Φ can be derived as follows.

V=

VR =IR

VL= IXL

V=I

V=IZ

The impedance in an AC circuit is similar to a resistance in a DC circuit. The unit for impedance is ohms(Ω)

Phase angle:

Φ= tan-1

Φ= tan-1

Power Factor:

The power factor in an AC circuit is defined as the cosine of the angle between voltage and current ie., cosΦ

Impedance Triangle: We can derive a triangle called the impedance triangle from the phasor diagram of an RL series circuit as shown

The impedance triangle is right angled triangle with R and XL as two sides and impedance as the hypotenuse. The angle between the base and hypotenuse is Φ. The impedance triangle enables us to calculate the following things.

- Impedance Z=
- Power Factor cosΦ =R/Z
- Phase Angle Φ= tan-1
- Whether current is leading or lagging

RC Series Ciruit

Fig 12 RC Series Circuit

Consider an AC circuit with a resistance R and a capacitance C connected in series as shown in the figure. The alternating voltage v is given by

V= Vm Sin ωt

The current flowing in the circuit is i. The voltage across the resistor is VR and that across the capacitor is Vc

VR=IR is in phase wih I

Vc=IXc lags current by 90 degrees

With the above information, the phasor diagram can be drawn as shown.

Fig 13 Phasor for RC Circuit

The current I is taken as the reference phasor. The voltage VR is in phase with I and the voltage VC lags behind the current by 90⁰. The resultant voltage V can be drawn as shown in the figure. From the phasor diagram we observe that the voltage lags behind the current by an angle Φ or in other words the current leads the voltage by an angle Φ. The waveform and equations for an RC series circuit can be drawn as below.

Fig 14 Waveform for RC Circuit

V= Vm Sin ωt

I= Im Sin (ωt+ɸ)

From the phasor diagram, the expressions for the resultant voltage V and the angle Φ can be derived as follows.

V=

VR =IR

Vc= IXc

V=I

V=IZ

Z=

Phase angle

Φ= tan-1

Φ= tan-1

Phasor algebra for RC series circuit.

Series RLC circuit

The voltage across inductor = jIXL

Voltage across capacitance = -jIXC

Net voltage across L and C = jI(XL-XC) = j(EL-EC)

Voltage drop across R= IR=ER

Applied voltage E= IR+jI(XL-XC)

Fig 15 Series RLC Circuit

E=I

Z=R+j(XL+XC)

Z=

Φ= tan-1

I=

CosΦ= R/Z

Active Power= EI CosΦ

Reactive Power = EI SinΦs

Examples

Q) An alternating current is given by i= 141.4 sin(314t)

Find i) The maximum value ii) Frequency iii) Time Period iv) The instantaneous value when t= 3ms i= 141.4 (314).

A)

Compare given equation with eq-1.

Maximum value

Q) The voltage is applied to 0.1 H inductor. Find the steady-state current through the inductor.

Solution: from equation

Q) Find equivalent impedance of below circuit.

Solution:

and are in parallel after simplifying

Therefore

Q) Find impedance of below circuit.

A)

The delta network connected to nodes a, b, and c can be converted to the Y network of fig. We obtain the Y impedances as follows.

The total impedance at the source terminals is

Q) Find form factor of figure factor of figure show below

A)

From figure:

Average value

Put T=t/2

Definition: it is defined as the phenomenon which takes place in the series or parallel R-L-C circuit which leads to unity power factor

Voltage and current in R – L - C ckt. Are in phase with each other

Resonance is used in many communicate circuit such as radio receiver.

Resonance in series RLC series resonance in parallel RLC anti resonance / parallel resonance.

- Condition for resonance XL = XC
- Resonant frequency (Fr): for given values of R-L-C the inductive reactance XL become exactly aqual to the capacitive reactance Xc only at one particular frequency. This frequency is called as resonant frequency and denoted by (fr)
- Expression for resonant frequency(fr): we know thet XL = 2ƛ FL - Inductive reactance

Xc = - capacitive reactance

At a particular frequency ȴ = fr, the Inductive and capacitive reactance are exactly equal

XL = XC ……at ȴ = fr

Ie L =

fr2 =

fr = H2

And = wr = rad/sec

Quality factor / Q factor

The quality of resonance circuit is measured in terms of efficiency of L and C to stare energy and the efficiency of L and C to store energy as measured in terms of a factor called quality factor or Q factor it is expressed as

Q = and Q =

The sharpness of tuning of R-L-C series circuit or its selectivity is measured by value of Q. As the value of Q increases, sharpness of the curve also increases and the selectivity increases.

Fig 17 Bandwidth of Resonant Circuit

Bandwidth (BW) = f2 = b1

and are the frequency at which the power delivered to the resistor is reduced to 50% of the power delivered to it at resonance these frequency are called as half power frequency

Bw = fr/Q

Resonance in Parallel circuit:

When a coil is in parallel with a capacitor, as shown below. The circuit is said to be in resonance.

Fig 18 Series RLC circuit

The resonant frequency for above circuit is fr = Hz

The current at resonance is I=

The value L/RC is known as dynamic impedance.

Fig 19 Bandwidth of RLC Circuit

The current at resonance is minimum. The circuits admittance must be at its minimum and one of the characteristics of a parallel resonance circuit is that admittance is very low limiting the circuits current. Unlike the series resonance circuit, the resistor in a parallel resonance circuit has a damping effect on the circuit bandwidth making the circuit less selective.

Also, since the circuit current is constant for any value of impedance, Z, the voltage across a parallel resonance circuit will have the same shape as the total impedance and for a parallel circuit the voltage waveform is generally taken from across the capacitor.

Bandwidth and selectivity:

and are the frequency at which the power delivered to the resistor is reduced to 50% of the power delivered to it at resonance thesefrequency are called as half power frequency

Bw = fr/Q

Q = = fCR = R

Resonant Frequency:

The resonant frequency for parallel resonant circuit is given as

fR=

Where L= inductance of the coil

C = is the capacitance

Rs = Resistive value of coil.

Key takeaway

Summary of characteristics of resonant RLC circuits

Characteristic | Series circuit | Parallel Circuit |

Resonant frequency, | ||

Quality factor, Q | ||

Bandwidth B | ||

Half Power frequencies | ||

For Q |

Examples

Que) A coil takes a current of 6A when connected to 24V dc supply. To obtain the same current with 50HZ ac, the voltage required was 30V. Calculate inductance and p.f of coil?

Sol: The coil will offer only resistance to dc voltage and impedance to ac voltage

R =24/6 = 4ohm

Z= 30/6 = 5ohm

XL =

= 3ohm

Cosφ = = 4/5 = 0.8 lagging

Que) The potential difference measured across a coil is 4.5V, when it carries a dc current of 8A. The same coil when carries ac current of 8A at 25 Hz, the potential difference is 24V. Find current and power when supplied by 50V,50Hz supply?

Sol: R=V/I= 4.5/8 = 0.56ohm

At 25Hz, Z= V/I=24/8 =3ohms

XL =

= 2.93ohm

XL = 2fL = 2x 25x L = 2.93

L=0.0187ohm

At 50Hz

XL = 2x3 =6ohm

Z = = 5.97ohm

I= 50/5.97 = 8.37A

Power = I2R = 39.28W

Que) A coil having inductance of 50mH an resistance 10ohmis connected in series with a 25F capacitor across a 200V ac supply. Calculate resonant frequency and current flowing at resonance?

Sol: f0= = 142.3Hz

I0 = V/R = 200/10 = 20A

Que) A 15mH inductor is in series with a parallel combination of 80ohm resistor and 20F capacitor. If the angular frequency of the applied voltage is 1000rad/s find admittance?

Sol: XL = 2fL = 1000x15x10-3 = 15ohm

XL = 1/C = 50ohm

Impedance of parallel combination Z = 80||-j50 = 22.5-j36

Total impedance = j15+22.5-j36 = 22.5-j21

Admittance Y= 1/Z = 0.023-j0.022 siemens

Que) A circuit connected to a 100V, 50 Hz supply takes 0.8A at a p.f of 0.3 lagging. Calculate the resistance and inductance of the circuit when connected in series and parallel?

Sol: For series Z =100/0.8 = 125ohm

Cosφ =

R = 0.3 x 125 = 37.5ohm

XL = = 119.2ohm

XL = 2fL = 2x 50x L

119.2 = 2x 50x L

L= 0.38H

For parallel:

Active component of current = 0.8 cosφ = 0.3x0.3 = 0.24A

R = 100/0.24 =416.7ohm

Quadrature component of current = 0.8 sinφ = 0.763

XL= 100/0.763 = 131.06ohm

L= 100/0.763x2x50 = 0.417H

Q) A series RLC circuit is connected across a 50Hz supply. R=100, L-159.16mH and C=63.7F. If the voltage across C is 150 . Find the supply voltage.

Solution

Q) An impedance coil in parallel with a 100µF capacitor is connected across a 200V, 50Hz supply. The coil takes a current of 4A and the power loss in the coil is 600W. Calculate (i) the resistance of the coil (ii) the inductance of the coil (iii) the power factor of the entire circuit.

A)

Q) A parallel circuit comprises of a resistor of 20Ω in series with an inductive reactance 15Ω in one branch and a resistor of 30Ω in series with a capacitive reactance of 20Ω in the other branch. Determine the current and power dissipated in each branch if the total current drawn by the parallel circuit is 10∠ −30⁰A?

A)

According to KCL

Q) A current of (120-j50)A flows through a circuit when the applied voltage is (8+j12)V. Determine (i) impedance (ii) power factor (iii) power consumed and reactive power.

A)

i)

Ii)

Iii)

But

Three phase voltage sources

In three phase the windings are separated by 1200 each. The voltage produced in those windings are 1200 apart from each other. Below shown is one coil RR’ and two more coils YY’ and BB’ each having phase shift of 1200.

The instantaneous value of voltages is given as

VRR’ = Vmsinωt

VYY’ = Vmsin(ωt-120)

VBB’ = Vmsin(ωt-240)

The three phase voltages are of same magnitude and frequency.

Fig 20 Phase sequence of 3-phase

Phase sequence

The change in voltage is in order VRR’- VYY’- VBB’. So, the three-phase are changed in that order and are called as phase change.

VRR’ = Vmsinωt

VYY’ = Vmsin(ωt-120)

VBB’ = Vmsin(ωt-240)

Phasor Diagram

Consider equation ①

Note: we are getting resultant line current IR by subtracting 2 phase currents IRY and IBR take phase currents at reference as shown

Cos 300 =

=

- Complete phases diagram for delta connected balanced Inductive load.

Phase current IYB lags behind VYB which is phase voltage as the load is inductive

- Power relation for delta load star power consumed per phase

PPh = VPh IPh Cos Ø

For 3 Ø total power is

PT= 3 VPh IPh Cos Ø …….①

For star

VL and IL = IPh (replace in ①)

PT = 3 IL Cos Ø

PT = 3 VL IL Cos Ø – watts

For delta

VL = VPh and IL = (replace in ①)

PT = 3VL Cos Ø

PT VL IL Cos Ø – watts

Total average power

P = VL IL Cos Ø – for ʎ and load

K (watts)

Total reactive power

Q = VL IL Sin Ø – for star delta load

K (VAR)

Total Apparent power

S = VL IL – for star delta load

K (VA)

Three phase balanced circuits

Balanced Delta Load

Fig 22 Delta Connection

Let us consider a balanced 3-phase delta connected load Determination of phase voltages:

VAB = V∠00, VBC = V∠-1200, VCA = V∠ − 2400 = V∠1200

Phase current = Phase voltage/ Load impedance

IAB= VAB/Z

IBC= VBC/Z

ICA= VCA/Z

Line currents are calculated by applying KCL at nodes A,B,C

IA = IAB – ICA

IB = IBC - IAB

IC = ICA - IBC

Note: Line currents are also balanced and equal to √3phase current.

Balanced star connected load:

Fig 23 Star Connection

Let us consider a balanced 3-phase star connected load. For star connection,

Phase voltage= Line voltage/(√3)

For ABC sequence, the phase voltage is polar form are taken as

VAN = Vph∠ − 900

VCN = Vph∠1500

VBN = Vph∠300

For star connection line currents and phase currents are equal

IA = VAN/Z

IB = VBN/Z

IC = VCN/Z

To determine the current in the neutral wire, apply KVL at star point IN + IA + IB + IC =0

IN = - (IA + IB + IC) (since they are balanced) In a balanced system the neutral current is zero. Hence if the load is balanced, the current and voltage will be same whether neutral wire is connected or not. Hence for a balanced 3-phase star connected load, whether the supply is 3- phase 3 wire or 3-phase 4 wire, it is immaterial. In case of unbalanced load, there will be neutral current.

Star Connection:

In this type similar ends are connected to common point called as neutral and having a star shape. These connections are used in case of unbalanced current flowing in the three-phase. To avoid any kind of damage we use this connection.

Line voltage VL = Vphase

Line current IL = Iphase

Delta Connection:

There are three wires with no neutral. They are used for short distance due to unbalanced current in circuit.

Line voltage VL =Vphase

Line current IL = Iphase

Examples

Q) Three similar resistors are connected in star across 400V 3-phase lines. Line current is 4A. Calculate the value of each resistor.

Sol: For star connection:

IL=Iph=4A

Vph=VL/ = 400/ = 231V

Rph= 231/4= 57.75ohm

For Delta Connection:

IL=4A

Iph= IL/

=4/ ==2.30A

Zph=400/2.30=173.9ohm

Rph= 173.9/3 = 57.97ohm

Q) Three identical impedances are connected in delta 3-phase supply of 400V. The line current is 30A and total power taken from the supply is 10kW. Calculate the resistance and reactance value of each impedance?

Sol: VL=Vph=400V

IL=30A

Iph=IL/= 30/ =17.32A

Zph=Vph/Iph= 400/17.32=23.09ohm

P=VLILCos Ø

Cos Ø = 10000/ 400x30 = 0.48

Sin Ø =0.88

Rph=ZphCos Ø= 23.09x0.48=11.08ohm

Xph=ZphSin Ø = 23.09x0.88=20.32ohm

Q) A star connected alternator supplies a delta connected load. The impedance of the load branch is 6+j5 ohm/phase. The line voltage is 230V. Determine the current in the load branch and power consumed by the load.

Sol: Zph= = 7.8ohm

VL=Vph=230V

Iph=Vph/Zph=230/7.8=29.49A

Iph=IL/

IL=Iph=x29.49=51.07A

P=VLIL Cos Ø = x 230x51.07x0.768=15.62kW

Q) The load connected to a 3-phase supply comprise three similar coils connected in star. The line currents are 25A and the kVA and kW inputs are 18 and 10 respectively. Find the line and phase voltage, the kVAR input resistance and reactance of each coil?

Sol: IL= 25A

P= 10000W

Cos Ø = 10/18 = 0.56

P=VLIL Cos Ø

10000= x VLx25x0.56

VL =412.39V

Vph=VL/ = 412.39/=238.09V

KVAR= = 14.96

Zph=238.09/25=9.52ohm

Rph=Zph Cos Ø= 9.52x0.56=5.33ohm

Xph=Zph Sin Ø = 9.52x0.83=7.88ohm

Q) A balanced delta connected load consisting of three coils draws 8 A at 0.5 p.f from 100V 3-phase ac supply. If the coils are reconnected in star across the same supply. Find the line current and total power consumed?

Sol: For Delta connection:

IL=8A

Iph= IL/= 8A

Vph=100V

Zph=100/8=12.5ohm

Rph=Zph Cos Ø=12.5x0.5 = 6.25ohm

Xph=Zph Sin Ø = 12.5x0.866=10.825ohm

P=VLIL Cos Ø

= x 100x 8x0.5=1200W

For Star Connection:

Vph= VL/= 100/V=57.73V

Zph=100/8=12.5ohm

Iph=57.73/12.5=4.62A

P=VLIL Cos Ø

= x 100x 4.62x0.5

P= 400W

References:

- D. P. Kothari and I. J. Nagrath, “Basic Electrical Engineering”, Tata McGraw Hill, 2010.
- D. C. Kulshreshtha, “Basic Electrical Engineering”, McGraw Hill, 2009.
- L. S. Bobrow, “Fundamentals of Electrical Engineering”, Oxford University Press, 2011.
- E. Hughes, “Electrical and Electronics Technology”, Pearson, 2010.
- V. D. Toro, “Electrical Engineering Fundamentals”, Prentice Hall India, 1989.