3.1 Study of AC circuits consisting of pure resistance pure inductance pure capacitance series RL RC and RLC circuits | unit 3 single phase ac circuits 21

BEE

Unit 3

Single phase AC circuit

3.1 Study of AC circuits consisting of pure resistance, pure inductance, pure capacitance, series R-L, R-C and R-L-C circuits

3 Basic element of AC circuit.

1] Resistance

2] Inductance

3] Capacitance

Each element produces opposition to the flow of AC supply in forward manner.

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

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 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

Phases diagram

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 = Vrms Irms

Power ware form [Resultant]

Ac circuit containing pure Inductors

Consider pure Inductor (L) is connected across alternating voltage. Source

V = Vm Sin ωt

When an alternating current flow through inductance it set ups alternating magnetic flux around the inductor.

This changing the flux links the coil and self-induced emf is produced

	According to faradays Law of E M I e =
at all instant applied voltage V is equal and opposite to self-induced emf [ lenz’s law] V = -e = But V = Vm Sin ωt dt Taking integrating on both sides dt dt
		(-cos ) but sin (– ) = sin (+ ) sin ( - /2) And Im= /2) /2 = -ve

= lagging

= I lag v by 900

Waveform:

Phasor:

Power P = Ѵ. I

= Vm sin wt Im sin (wt /2)

= Vm Im Sin wt Sin (wt – /s)

①

And

Sin (wt - /s) = - cos wt ②

Sin (wt – ) = - cos

sin 2 wt from ① and ②

The average value of sin curve over a complete cycle is always zero

Pavg = 0

Ac circuit containing pure capacitors:

$C:\Users\ManishM\Downloads\be3_copy$

Consider pure capacitor C is connected across alternating voltage source

Ѵ = Ѵm Sin wt

Current is passing through capacitor the instantaneous charge ɡ produced on the plate of capacitor

ɡ = C Ѵ

ɡ = c Vm sin wt

the current is rate of flow of charge

i= (cvm sin wt)

i = c Vm w cos wt

then rearranging the above eqth.

i = cos wt

= sin (wt + X/2)

i = sin (wt + X/2)

but

X/2)

= leading

= I leads V by 900

Waveform:

Phase

$C:\Users\ManishM\Downloads\be3_2$

Power P= Ѵ. i

= [Vm sinwt] [ Im sin (wt + X/2)]

= Vm Im Sin wt Sin (wt + X/2)]

(cos wt)

to charging power waveform [resultant].

Series R-L Circuit

$C:\Users\ManishM\Downloads\BE3_4$

Consider a series R-L circuit connected across voltage source V= Vm sin wt

As some I is the current flowing through the resistor and inductor due do this current voltage drops arcos R and L R VR = IR and L VL = I X L

Total V = VR + VL

V = IR + I X L V = I [R + X L]

Take current as the reference phasor: 1) for resistor current is in phase with voltage 2) for Inductor voltage leads current or current lags voltage by 90 0.

$C:\Users\ManishM\Downloads\BE3_5$

For voltage triangle

Ø is power factor angle between current and resultant voltage V and

V =

where Z = Impedance of circuit and its value is =

Impedance Triangle

Divide voltage triangle by I

$C:\Users\ManishM\Desktop\TRI.PNG$

Rectangular form of Z = R+ixL

and polar from of Z = L +

(+ j X L + because it is in first quadrant )

Where =

+ Tan -1

Current Equation:

From the voltage triangle we can sec. that voltage is leading current by or current is legging resultant voltage by

Or i = = [ current angles - Ø )

Resultant Phasor Diagram from Voltage and current eqth.

$C:\Users\ManishM\Downloads\BE_3_7$

Wave form

Power equation

P = V.I.

P = Vm Sin wt Im Sin wt – Ø

P = Vm Im (Sin wt) Sin (wt – Ø)

P = (Cos Ø) - Cos (2wt – Ø)

Since 2 sin A Sin B = Cos (A-B) – Cos (A+B)

P = Cos Ø - Cos (2wt – Ø)

①②

Average Power

pang = Cos Ø

Since ② term become zero because Integration of cosine come from 0 to 2ƛ

pang = Vrms Irms cos Ø watts.

Power Triangle:

$C:\Users\ManishM\Downloads\be3_8$

From

VI = VRI + VLI B

Now cos Ø in A =

①

Similarly Sin =

Apparent Power Average or true Reactive or useless power

Or real or active

-Unit (VI) Unit (Watts) C/W (VAR) denoted by (Ø)

Denoted by [S] denoted by [P]

Power for R L ekt.

$C:\Users\ManishM\Downloads\BE3_9$

Series R-C circuit

$C:\Users\Vidya.Tamhane\Downloads\BD3_12.jpg$

V = Vm sin wt

$C:\Users\Vidya.Tamhane\Downloads\bc3_13.jpg$

Consider a series R – C circuit in which resistor R is connected in series with capacitor C across a ac voltage so use V = VM Sin wt (voltage equation).
Assume Current I is flowing through

R and C voltage drops across.

R and C R VR = IR

And C Vc = Ic

V = lZl

Voltage triangle: take current as the reference phasor 1) for resistor current is in phase with voltage 2) for capacitor current leads voltage or voltage lags behind current by 900

$C:\Users\Vidya.Tamhane\Downloads\BE3_14.jpg$

Where Ø is power factor angle between current and voltage (resultant) V

And from voltage

V =

V = lZl

Where Z = impedance of circuit and its value is lZl =

Impendence triangle:

Divide voltage by as shown

$C:\Users\Vidya.Tamhane\Downloads\BE3_14 (1).jpg$

Rectangular from of Z = R - jXc

Polar from of Z = lZl L - Ø

( - Ø and –jXc because it is in fourth quadrant ) where

lZl =

and Ø = tan -1

Current equation:

from voltage triangle we can see that voltage is lagging current by Ø or current is leading voltage by Ø

i = IM Sin (wt + Ø) since Ø is +ve

Or i = for RC

LØ [ resultant current angle is + Ø]

Resultant phasor diagram from voltage and current equation

$C:\Users\Vidya.Tamhane\Downloads\bc3_115.jpg$

Resultant wave form :

Power Equation:

P = V. I

P = Vm sin wt. Im Sin (wt + Ø)

= Vm Im sin wt sin (wt + Ø)

2 Sin A Sin B = Cos (A-B) – Cos (A+B)

Average power

pang = Cos Ø

since 2 terms integration of cosine wave from 0 to 2ƛ become zero

2 terms become zero

pang = Vrms Irms Cos Ø

Power triangle RC Circuit:

$C:\Users\Vidya.Tamhane\Downloads\BE3_16.jpg$

R-L-C series circuit

Consider ac voltage source V = Vm sin wt connected across combination of R L and C. when I flowing in the circuit voltage drops across each component as shown below.

VR = IR, VL = I L, VC = I C

According to the values of Inductive and Capacitive Reactance I e XL and XC decides the behaviour of R-L-C series circuit according to following conditions

① XL> XC, ② XC> XL, ③ XL = XC

① XL > XC: Since we have assumed XL> XC

Voltage drop across XL> than XC

VL> VC A

Voltage triangle considering condition A

$C:\Users\Vidya.Tamhane\Downloads\BE3_17 (1).jpg$

VL and VC are 180 0 out of phase.

Therefore, cancel out each other

Resultant voltage triangle

$C:\Users\Vidya.Tamhane\Downloads\BE3_18.jpg$

Now V = VR + VL + VC c phasor sum and VL and VC are directly in phase opposition and VLVC their resultant is (VL - VC).

From voltage triangle

V =

V = I

Impendence : divide voltage

$C:\Users\Vidya.Tamhane\Downloads\BE3_19.jpg$

Rectangular form Z = R + j (XL – XC)

Polor form Z = l + Ø B

Where =

And Ø = tan-1

Voltage equation: V = Vm Sin wt
Current equation

i = from B

i = L-Ø C

as VLVC the circuit is mostly inductive and I lags behind V by angle Ø

Since i = L-Ø

i = Im Sin (wt – Ø) from c

$C:\Users\Vidya.Tamhane\Downloads\BE3_20.jpg$

XC XL :Since we have assured XC XL

the voltage drops across XC than XL

XC XL (A)

voltage triangle considering condition (A)

$C:\Users\Vidya.Tamhane\Downloads\BE3_21.jpg$

Resultant Voltage

$C:\Users\Vidya.Tamhane\Downloads\BC3_21.jpg$

Now V = VR + VL + VC phases sum and VL and VC are directly in phase opposition and VC VL their resultant is (VC – VL)

From voltage

V =

Impedance : Divide voltage

$C:\Users\Vidya.Tamhane\Downloads\BE3_23.jpg$

Rectangular form : Z + R – j (XC – XL) – 4th qurd

Polar form : Z = L -

Where

And Ø = tan-1 –

Voltage equation : V = Vm Sin wt
Current equation : i = from B
i = L+Ø C

as VC the circuit is mostly capacitive and leads voltage by angle Ø

since i = L + Ø

Sin (wt – Ø) from C

Power :

$C:\Users\Vidya.Tamhane\Downloads\BE3_22.jpg$

XL= XC (resonance condition):

ɡȴ XL= XC then VL= VC and they are 1800 out of phase with each other they will cancel out each other and their resultant will have zero value.

Hence resultant V = VR and it will be in phase with I as shown in below phasor diagram.

$C:\Users\Vidya.Tamhane\Downloads\be3_24.jpg$

From above resultant phasor diagram

V =VR + IR

Or V = I lZl

Because lZl + R

Thus Impedance Z is purely resistive for XL = XC and circuit current will be in phase with source voltage.

Since VR=V Øis zero when XL = XC power is unity

ie pang = Vrms I rms cos Ø = 1 cos o = 1

maximum power will be transferred by condition. XL = XC

3.2 Resonance in series RLC circuits

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 equal 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.

$C:\Users\Vidya.Tamhane\Downloads\be3_25.jpg$

$C:\Users\Vidya.Tamhane\Downloads\be3_26.jpg$

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

3.3 Concept of impedance

Two impedances in parallel

$C:\Users\Vidya.Tamhane\Downloads\BE3_28.jpg$

I1 an I2 can be founded using current division rules

It states that the current in one branch is the products, ratio of total current and opposite branch (impedance / reactance / resistance) to the total (impedance / reactance / resistance)

above ckt. Can be found using following steps

Find total impedance Z1 + Z2 = Z using conversion from polar to rectangular (if given Z1/Z2 is in polar form) and then finding Z = Z1 + Z2
Find total current I using formulas
Then find I1 using current division rule or Z
Find I2 using current division Rule

or z

3.4 Concept of active, reactive, apparent, complex power and power factor

Apparent power (S): it is defined as product of rms value of voltage (V) and current (I) or it is the total power / max power

S = V

Unit - volt – Ampere (VA)

2. In kilo – K. V. A.

3. Real power / true power / active power / useful power: [P] it is defined as the product of rms value active component or it is the average power or actual power consumed by the resistive part (R) in the given combinational circuit

It is measured in watts

P = VI Cos Ø watts /km where Ø is the power factor angle

4. Reactive power / Imaginary / useless power [Q]

It is defined as the product of voltage current and sine of angle between V and

Volt Amp Relative

Unit – V A R

Power triangle

$C:\Users\ManishM\Downloads\BE3_10$

As we know power factor is cosine of angle between voltage and current

i e P.F. = Cos

$C:\Users\ManishM\Downloads\BE3_11$

in other words also we can derive it from impedance triangle

now consider Impedance triangle in R – L- ckt.

$C:\Users\ManishM\Downloads\BE3_11(1)$

From now Cos = power factor =

power factor = Cos or

3.5 Parallel AC circuits

AC parallel circuit:

$C:\Users\Vidya.Tamhane\Downloads\BE3_27.jpg$

Total I = I1 + I2 + I3

As parallel circuit applied total voltahe V is same at each branch

I = I1 + I2 + I3

= = + +

= + + +……

3.6 Concept of admittance

Admittance is defined as reciprocal of the impedance (Z)

It is denoted by Y and its unit Siemens (S)

Admittance Y =

$C:\Users\Vidya.Tamhane\Downloads\BE3_29.jpg$

I = I1 + I2 + I3

= + +

Y = Y1 + Y2 + Y3

Rationalize the expression of Y as follows

Y =

But Z2 = R2 + X2

Y =

Let + G + conductance

= B = susceptance

Y = G jB

$C:\Users\Vidya.Tamhane\Downloads\be3_30.jpg$

Conductance (G) : it is defined as the ratio of resistance R and the squared impedance (Z2) and it is measured in Siemens or mho

G =

Theoretically G = reciprocal od resistance.

Susceptance (B) :it is defined as the ratio of reactance X and the squared impedance (Z2) or mho

B =

Theoretically B = reciprocal od resistance.

Reference Books:

H Cotton, Electrical technology, CBS Publications
L. S. Bobrow, ―Fundamentals of Electrical Engineering‖, Oxford University Press, 2011.
E. Hughes, ―Electrical and Electronics Technology‖, Pearson, 2010.
D. C. Kulshreshtha, ―Basic Electrical Engineering‖, McGraw Hill, 2009.

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