Unit 3

Basic Electrical

When a coil is placed in uniform magnetic field. When coil rotates in opposite direction at constant angular velocity the emf is induced in the coil. The magnitude of emf induced depends on the flux cut by the conductor. The magnitude of induced emf becomes maximum whenconductor is perpendicular to the lines of forces. The direction of emf induced is determined by Flemings right hand rule.

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.

The function which repeats itself after a particular interval of time is called as periodic function. For example sinx

The alternating quantities (voltages and currents) in practice are represented by straight lines having definite direction and length. Such lines are called the phasors and the diagrams in which phasors represent currents, voltages and their phase difference are known as phasor diagrams.

Though phasor diagrams can be drawn to represent either maximum or effective values of voltages and currents but since effective values are of much more importance, phasor diagrams are mostly drawn to represent effective values.

In order to achieve consistent and accurate results it is essential to follow certain conventions.

The above figure shows that OA is the maximum ac quantity present. The OA represents emf on vertical axis. This phasor of OA in ACW direction represents sinusoidal voltage or current. Its angular velocity is one revolution complete in same time as by the alternating voltage or current.

The ACW is taken positive for phasor. In series circuit current phasor is taken as reference as the current is same. In parallel circuit voltage is taken as reference as it is same throughout.

The series RLC circuit is shown below. Let current i(t) be sinusoidal. The value of instantaneous voltage across R is in phase with current. The instantaneous voltage across Lleads current by 900. The instantaneous voltage across C lags current by 90o.

Writing the loop equations, we get

Vs-VR-VL-VC=0

VS-IR-L- = 0

VS= IR + L +

The voltage triangle will be

Vs=

VR = iRsint

VL= iXLsint+90)

VC= iXcsint-90)

Z = =

Impedance Z=

Que) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate total impedance?

Sol: Impedance Z=

Z= = 18.27ohm

Que) For a series RLC circuit having R=12ohms, L= 0.2H, C=60F. They are connected across 100v 50Hz supply. Calculate circuit current?

Sol: I=

Z=

Z= = 13.89ohm

I = 100/13.89 =7.2A

Que) For a series RLC circuit having R=10ohms, L= 0.15H, C=100F. They are connected across 100v 50Hz supply. Calculate power factor?

Sol: cosφ =

Impedance Z=

Z= = 18.27ohm

Cosφ = =

φ = 56.81o lagging

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.

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 =3ohm

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-j0.022siemens

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 = 0.8cosφ = 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

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 ωt①

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 =

Pavg = VrmsIrms

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

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.

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)

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

Phase system only

- 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 = VPhIPh Cos Ø

For 3 Ø total power is

PT= 3 VPhIPh 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 = VPhand 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)