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Ac fundamentals and AC CIRCUITS

Q1. Explain and derive an expression for generation of AC quantity.

According to Faradays law of electromagnetic induction when a conductor is moving within a

magnetic field, an emf is induced in the conductor . such as that shown in Fig. 1, is rotated within a

magnetic field the induced emf can be measured between the ends of the conducting loop as a

potential difference. The magnitude of the emf induced depends on the strength of the magnetic field,

the length of the conductor forming the loop and the speed at which it is rotated. It also depends,

however, on the angle at which the conducting loop moves with respect to the direction of the

magnetic field. It can be shown that the induced emf is proportional to the sine of this angle. The emf

is a maximum when the conductor is moving at right angles to the direction of the magnetic field and

a minimum when it is moving in the direction of the field.

Due to the circular motion of the armature against the straight lines of force, a variable number of

lines of force will be cut even at a constant speed of the motion.

1.At zero degrees, the rectangular arm of the armature does not cut any lines of force, giving zero

voltage output. As the armature arm rotates at a constant speed toward the 90° position, more lines

are cut.

2.The lines of force are cut at most when the armature is at the 90° position, giving out the most

current on one direction. As it turns toward the 180° position, lesser number of lines of force are cut,

giving out lesser voltage until it becomes zero again at the 180° position.

3.The voltage starts to increase again as the armature heads to the opposite pole at the 270° position.

Toward this position, the current is generated on the opposite direction, giving out the maximum

voltage on the opposite side.

Mrs. V.S. Patki(M.E.Electrical)[email protected]

4. The voltage decrease again as it completes the full rotation(3600). In one rotation, the AC output is

produced with one complete cycle as represented in the sine wave.

According to flemings right hand rule we can find the direction of generated emf.

Q2. DIFINITIONS : Mrs. V.S. Patki(M.E.Electrical)[email protected]

https://en.wikipedia.org/wiki/Sine_wave

1. Cycle : One complete set of positive and negative values of alternating quantity is known as

cycle. Hence, diagram represents one complete cycle.

A cycle may also be sometimes specified in terms of angular measure. In that case, one complete

cycle is said to spread over 360º or 2π radians.

2. Time Period

The time taken by an alternating quantity to complete one cycle is called its time period T. For

example, a 50-Hz alternating current has a time period of 1/50 second.

T = 1/ f (sec) where f is frequency in Hz.

3. Frequency

The number of cycles/second is called the frequency of the alternating quantity. Its unit is hertz

(Hz).

It may be noted that the frequency is given by the reciprocal of the time period of the alternating

quantity.

f = 1/T or T = 1/f (Hz)

4. Amplitude: The maximum value, positive or negative, of an alternating quantity is known as its

amplitude. In above diagram Vmax is amplitude.

5. Angular frequency (ω): it is the angular distance covered in one second. It is also called as angular

velocity.

ω = 2 π f = 2 π / T ( rad/sec) where f is frequency in Hz. And T is time period.

6. Instantaneous value – Value at any instance is called instantaneous value. I-Im sinωt

Where i is instantaneous value that changes with t .

Alternating current and Alternating EMF


An alternating current is one whose magnitude changes sinusoidal with time .Thus alternating

current is given by

i=I m sin (ωt+Ф)

Where im=current amplitude or peak value of alternating current

The emf or voltage whose magnitude changes sinusoidal with time is known as alternating emf and

is represented by v=V m sin (ωt+Ф)

where Vm is the peak value of alternating current.

Q3. Define and derive the expression for RMS value of AC.

Root-Mean-Square (R.M.S.) Value The r.m.s. value of an alternating current is given by that steady (d.c.) current which whenflowing through a given circuit for a given time produces the same heat as produced by the alternating current when flowing through the same circuit for the same time.It is also known as the effective or virtual value of the alternating current.


Q4. Define and derive the expression for Average value of AC.

The average value Iavg of an alternating current is expressed by that steady current which transfers

across any circuit the same charge as is transferred by that alternating current during the same

time.

In the case of a symmetrical alternating current (i.e. one whose two half-cycles are exactly similar,

whether sinusoidal or non-sinusoidal), the average value over a complete cycle is zero. Hence, in

their case, the average value is obtained by adding or integrating the instantaneous values of current

over one half-cycle only. But in the case of an unsymmetrical alternating current (like half-wave

rectified current) the average value must always be taken over the whole cycle.


Phase Difference of a Sinusoidal Waveform

The current

waveform can also be said to be “lagging” behind

the voltage waveform by the phase angle, Φ if

current starts it zero and maximum value after

voltage. where, i lags v by angle Ө

Likewise, if the current, i has a positive value and crosses the reference axis reaching its maximum peak

and zero values at some time before the voltage, v then the current waveform

will be “leading” the voltage by some phase angle. Then the two waveforms

are said to have a Leading Phase Difference and the expression for both the

voltage and the current will be. where, i leads v by angle Ө


Two Sinusoidal Waveforms – “in-phase”

Voltage (vt) = Vm sin ωt

Current (it)= Im sin ωt

Both voltage and current reaches it zero and maximum value at same time called as in phase quntities.


A.C through pure resistor Figure below shows the circuit containing alternating voltage source V=V0sinω connected to a resistor of resistance R

Let at any instant of time ,i is the current in the circuit ,then from Kirchhoff’s loop ruleV0sinωtR or i= (V0/R) sinωtI = i0 sinωt Where, i0 = V0/R

From instantaneous values of alternating voltage and current ,we can conclude that in pure resistor ,the current is always in phase with applied voltage

Their relationship is graphically represented as

(7) A.C through pure inductor

Figure below shows the circuit in which voltage sourceV=Vmsinωt (10)is applied to pure inductor (zero resistance) coil of inductance L

As the current through the inductor varies and opposing induced emf is generated in it and is given by -Ldi/dt


From Kirchhoff’s loop rule

Where C is the constant of integration .This integration constant has dimensions of current and is independent of time. Since source has an emf which oscillates symmetrically about zero, the current it sustain also oscillates symmetrically about zero, so there is no time independent component of current that exists. Thus constant C=0

From instantaneous values of current and voltage (equation 11 and 10) we see that in pure

inductive circuit the current lags behind emf by a phase angle of π/2 This phase relationship is graphically shown below in the figure

From equation (12) peak value of current in the coil is i0=V0/ωLorV0=(ωL)i0

Comparing it with the ohm's law we find product ωL has dimension of resistance and it can be represented by


XL=ωLwhere XL is known as reactance of the coil which represents the effective opposition of the coil to the flow of alternating current

XL=ωL is zero for DC for which ω=0 and increases as the frequency of current increases

(8)AC through pure capacitor

Figure given below shows circuit containing alternating voltage source V=V0sinωtconnected to a capacitor of capacitance C

Suppose at any time t,q be the charge on the capacitor and i be the current in the circuit Since there is no resistance in the circuit, so the instantaneous potential drop q/C across the

capacitor must be equal to applied alternating voltagesoq/C=V0sinωt

Since i=dq/dt is the instantaneous current in the circuit so

is the peak value of current

Comparing equation (13) with V=V0sinωt ,we see that in a perfect capacitor current leads emf by a phase angle of π/2

This phase relationship is graphically shown below in the figure


Again comparing peak value of current with ohm's law ,we find that quantity 1/ωC has the dimension of the resistance

Thus the quantityXC=1/ωC =1/2πfC ---(14)is known as capacitive reactance

From equation (14) we see that capacitive reactance decreases with increasing frequency of current and in infinite for direct current for which frequency f=0

(9) Circuit containing inductance and resistance in series

Figure below shows pure inductor of inductance L connected in series with a resistor of resistance R through sinusoidal voltageV=V0sin(ωt+φ)

An alternating current I flowing in the circuit gives rise to voltage drop VR across the resistor and voltage drop VL across the coil

Voltage drop VR across R would be in phase with current but voltage drop across the inductor will lead the current by a phase factor π/2

Now voltage drop across the resistor R isVR=IRand across inductorVL=I(ωL)where I is the value of current in the circuit at a given instant of time

So voltage phasors diagram is


In figure (10) we have taken current as a reference quantity because same amount of current flows through both the components. Thus fro phasors diagram

is known as impedance of the circuit

Current in steady state is

and it lags behind applied voltage by an angle φ such thattanφ=ωL/R ---(16)

(10) Circuit containing capacitance and resistance in series

Figure below shows a circuit containing capacitor and resistor connected in series through a sinusoidal voltage source of voltageV=V0sin(ωt+φ)

In this case instantaneous P.D across R isVR=IRand across the capacitor C isVC=I/ωC


In this case VR is in phase with current i and VC lags behind i by a phase angle 900

Figure 11(b) shows the phasors diagram where vector OA represent the resultant of VR and VC which is the applied Voltage thus

is called the impedance of the circuit

Again from the phasors diagram applied voltage lags behind the current by a phase angle φ given bytanφ= VC/ VR=1/ωCR ---(18)

(11) LCR series circuit

Figure below shows a circuit containing a capacitor ,resistor and inductor connected in series through an alternating voltage source

Same amount of current will flow in all the three circuit components and vector sum of potential drop across each component would be equal to the applied voltage

If i be the amount of current in the circuit at any time and VL,VC and VR the potential drop across L,C and R respectively thenVR=iR ⇒ Voltage is in phase with iVL=iωL ⇒ Voltage is leading i by 900

VC=i/ωC ⇒ Voltage is lagging behind i by 900

Since VL is ahead of i by 90 and VC is behind by 90 so that phase difference between VL and VC is 180 and they are in direct opposition to each other as shown in the figure 12(b)


In figure 12(b) we have assumed that VL is greater than VC which makes i lags behind V.If VC > VL then i lead V

In this phasors diagram OA represent VR,AD represent VC and AC represent VL.So in this case as we have assumed that VL > VC ,there resultant will be (VL -VC) represented by vector AD

Vector OB represent resultant of vectors VR and (VL -VC) and this vector OB is the resultant of all the three ,which is equal to applied voltage V,thus

is called impedance of the circuit

From phasors diagram 12(b),current i lag behind resultant voltage V by an phase angle given by,

From equation (20) three cases arises

(i) When ωL > 1/ωC then tanφ is positive i.e. φ is positive and voltage leads the current i(ii) When ωL < 1/ωC,then tanφ is negative i.e. φ is negative and voltage lags behind the current i(iii) When ωL = 1/ωC ,then tanφ is zero i.e. φ is zero and voltage and current are in phase

Again considering case (iii) where ωL = 1/ωC,we have


which is the minimum value Z can have.

This is the case where XL=XC,the circuit is said to be in electric resonance where the impedance is purely resistive and minimum and currents has its maximum value

Hence at resonanceωL = 1/ωCor ω=1/√LC ---(21)But ω=2πf where f is the frequency of applied voltage .Thereforef0=1/2π√LC ---(22)This frequency is called resonant frequency of the circuit and peak current in this case isi0=V0/Rand reactance is zero

We will now define resonance curves which shows the variation in circuit current (peak current i0) with change in frequency of the applied voltage

Figure below shows the shape of resonance curve for various values of resistance R

for small value of R,the resonance is sharp which means that if applied frequency is lesser to resonant frequency f0,the current is high otherwise

For large values of R,the curve is broad sided which means that those is limited change in current for resonance and non -resonance conditions


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