Unit-2

AC Circuits

Average 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.

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.

Reactance

- Inductive Reactance (XL)

It is opposition to the flow of an AC current offered by inductor.

XL = ω L But ω = 2 ᴫ F

XL = 2 ᴫ F L

It is measured in ohm

XL∝FInductor blocks AC supply and passes dc supply zero

2. Capacitive Reactance (Xc)

It is opposition to the flow of ac current offered by capacitor

Xc =

Measured in ohm

Capacitor offers infinite opposition to dc supply

Impedance (Z)

The ac circuit is to always pure R pure L and pure C it well attains the combination of these elements. “The combination of R1 XL and XC is defined and called as impedance represented as

Z = R +i X

Ø = 0

only magnitude

R = Resistance, i = denoted complex variable, X =Reactance XL or Xc

Polar Form

Z = L I

Where =

Measured in ohm

Ac circuit containing pure resisting

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.

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

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)

E=I

Z=R+j(XL+XC)

Z=

Φ= tan-1

I=

CosΦ= R/Z

Active Power= EI CosΦ

Reactive Power = EI SinΦs

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.

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.

The resonant frequency for above circuit is fr = Hz

The current at resonance is I=

The value L/RC is known as dynamic impedance.

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.

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 25Hz,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

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.

Phase sequence

The change in voltage is in orderVRR’- 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)

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

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.