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Unit – 2Steady- State Analysis of Single-Phase AC Circuits Q1) Write the step to solve numerical.A1)

1. Calculate VPh from the given value of VL by relation

For star VPh =

For delta VPh = VL

2. Calculate IPh using formula

IPh =

3. Calculate IL using relation

IL = IPh - for star

IL = IPh - for delta

4. Calculate P by formula (active power)

P = VL IL Cos Ø – watts

5. Calculate Q by formula (reactive power)

Q = VL IL Sin Ø – VAR

6. Calculate S by formula (Apparent power)

S = VL IL– VA

Q2) Explain Three phase balanced circuits in detail.A2)

3Φ system in which three voltages are of identical magnitudes and frequency and are displaced by 120° from each other called as symmetrical system.

Phase sequence:

The sequence in which the three phases reach their maximum positive values. Sequence is R-Y-B. Three colours used to denote three faces are red ,yellow and blue.

The direction of rotation of 3Φ machines depends on phase sequence. If a sequence is changed i.e. R-B-Y then the direction of rotation will be reversed.

Types of loads

Star connection of load
Delta connection of load

Balanced load:

Balanced load is that in which magnitudes of all impedances connected in the load are are equal and the phase angles of them are also equal.

i.e.

If. ≠ ≠ then it is unbalanced load

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

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Cos 300 =

Complete phases diagram for delta connected balanced Inductive load.

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

Q3) What is Series Resonance Frequency? How it is determined?A3)

In a series resonant circuit, the resonant frequency, ƒr point can be calculated as follows.

XL = Xc --- 2 π f L = 1/ 2 π f C

f 2 = 1/ 2 π L x 2 π C = 1/ 4 π 2 L C

f = [1/ 4 π 2 L C ] ½

fr = 1/ 2 π

In the circuit high quality factor is required to ensure low energy dissipation and low oscillation damping but high-quality factor can only be achieved by bandwidth. Therefore, it is required to maintain balance between these two

Q = fr/BW

Fr = 1/ 2 π

Q = 1/R

BW = R/L

Q4) Derive series magnetic current with air gap. A4)

In series magnetic circuit flux is same in each part of circuit

Consider a composite magnetic circuit made from different material of lengths ɻ1,ɻ2, and ɻ3 cross sectional area a1, a2 And a3 and relative permeability’s 1 , 2 , 3 resp. with an airgap of length ɻɡ

Total reluctance S = S1 + S2+ S3 + Sg

We know that S =

+ + +

But S = and mmf = S x Ø

mmf = Ø + + + for air

Now B =

mmȴ = + + +

but B=μH

H = =

Total mmȴ = H1 ɻ1+ H2 ɻ2 + H3 ɻ3 + Hƪɻƪ

A magnetic circuit which has more than one path for flux called parallel magnetic circuit

Mean length ABCDA = ɻ₁, and ADEFA = ɻ₂

Mean length path for central limb = ɻc

Reluctance ABCDA = S1, ADEFA = S2

And central limb = SC

Now total mmȴ = N

i.e. mmf =

For path ABC AD

N =

Where S1 = , S2 = , and Sc =

Where area of cross selection

for parallel ekt.

Total mmȴ = mmȴ by central limb + mmȴ required by any one of outer limb

= Ø sc + [Ø1S1 or Ø2 S2]

As mmf across parallel branch is same

Ø1 S1 = Ø2 S2

Hence while calculating total mmf, the mmf of only one of 2 parallel branches must be considered.

Q5) Explain Apparent, active & reactive power.A5)

Apparent power: (S):- it is defined as product of rms value of voltage (v) and current (I), or it is the total power/maximum power

S= V × I

Unit - Volte- Ampere (VA)

In kilo – KVA

2. Real power/ True power/Active power/Useful power: (P) it is defined as the product of rms value of voltage and current and the active component or it is the average or actual power consumed by the resistive path (R) in the given combinational circuit.

It is measured in watts

P = VI Φ watts / KW, where Φ is the power factor angle.

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

It is defined as the product of voltage, current and sine B and I

Therefore,

Q= V.I Φ

Unit –VA R

In kilo- KVAR

Represent Sinusoidal waveforms – Average and effective values.

Peat to peak value:

The value of an alternating quantity from its positive peak to negative peak

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.

Q6) What is RMS value: Root mean square value?A6)

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.

Q7) Explain Peak or krest factor (kp).A7)

It is the ratio of maximum value to rms value of given alternating quantity

Kp =

Kp = 1.414

Form factor (Kf): For numerical “It is the ratio of RMS value to average value of given alternating quality”.

Q8) Represent in various forms.A8)

The instantaneous voltage equation

V(t) = V m sin (w t + Ø)

Polar Form

The instantaneous voltage equation is given by

Vt= vm sin (w t +Ø)

which can be represented by polar form

vt = L Ø

where = peak value

e.g. vt =30 sin (w t + 90)

polar form < 90

polar form is suitable for multiplication and division of phases.

2. Rectangular Form:

The instantaneous voltage equation is given by

Vt = v m sin (w t +Ø)

which can be represented by Rectangular Form

vt =

where X = or Vm cos Ø

Y =or Vm sin Ø

Vt = v m cos Ø + i vm sin Ø

e.g. 30 sin (w t + 90)

Rectangular form vt = 30 cos 90 + i 30 sin 90

Rectangular Form is suitable for addition and subtraction of Phases.

Trigonometric Form (Polar to Rectangular)

If the phases are given in polar form from L Ø then it can be represented in

rectangular form by expressing X and Y component in form of and Ø.

Polar: Vt = L Ø

Rectangular: vt= + i where

𝑥 = cos Ø

𝑦 = sin Ø

Exponential Form (R-P)

Given equation Rect : v(+) =𝑥 + i𝑦

Polar: v(+) =

Where magnetite

And

2 Phase added and substrate using Rect. Form

Let V1 = 𝑥1 + i𝑦1

V2 = 𝑥2 + i𝑦2

(V1 + V2) = (𝑥1 + i𝑦1) + (𝑥2 + i𝑦2)

= (𝑥1 + 𝑥2) + i(𝑦1 + 𝑦2)

(V1 – V2) = (𝑥1 – 𝑥2) + i(𝑦1 - 𝑦2)

For add or subst. If eqtn. Is given in polar form, we have to connect into Rect. Form and then add/ subtract.

Two phases divide/ multiply by polar

Let V1 = π1 L Ø1

Let V2 = π2 L Ø2

(V1 V2) = (π1 L Ø1) (π2 L Ø2)

(V1 X V2) = (π1 x π2) L (Ø1 x Ø2)

For dividing

= L 1 - 2

Q9) Write a note on: Reactance and Impedance.A9)

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

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