A.C. Through Pure Ohmic Resistance Alone
The electric circuit is made up of the basic components namely resistor, inductors, and capacitors. Their effect on the flow of current in the circuit is called as resistance(R), reluctance (L), and capacitance(C) respectively.
These RLC components can be connected in the circuit in many possible series parallel combinations like parallel, series-parallel, parallel-series, and other complex combinations.
According to Ohm’s law,
I= V/R ….. Eq.1
Where I= Current
Instantaneous AC current is given by
By replacing equation 1, we get
Now we know that P= VI
Where P= power
p=(Vmax Sin ωt)(Imax Sin ωt)
p= VmaxImax((1-Cos2ωt) /2) as Sin2wt = (1- Cos2ωt)/2
p= VmaxImax/2 – VmaxImaxCos2ωt/2
For a complete cycle 2cos 2 wt is zero.
Hence, power for the complete cycle
p= (Vmax/√2) (Imax/√2)
whereVrms = R.M.S. value of applied voltage, and
Irms = R.M.S. value of the current.
Thus from the above equation, we derived that RMS values are used for power calculation.
It can be seen from the Waveform graph that power in a cycle can never become negative, in other words, power in a purely resistive circuit can never become zero
And current and voltage are always in phase with each other.
A.C. through pure inductance alone
The above diagram depicts the L. Henry circuit of pure Inductance
Where V= VmaxSin ωt
An electromagnetic flux would be produced because of the self-inductance of the coil, whenever an AC voltage is applied to the pure inductive coil. Now, because of lack of ohmic drop in such case, the applied current is ought to overcome the induced emf.
The above figure depicts AC through pure inductance alone, where resultant power continues to be zero.
Now in the above equation, the negative sign depicts that emf opposes any change that may occur in current. And to maintain the current, external agent has to interrupt.
u= L di/dt
Replacing ‘u’ by ‘VmaxSinωt’, we may write it as
On integrating both the sides, we get
On cancelling out d from both the sides, we obtain
i= Vmax / XL Sin(ωt- π/2)
Now we know that the value of current will be maximum when Sin(ωt- π/2)=1
so we conclude that,
Now, we know that in pure inductive circuit, current leads the voltage by 90°. In simple words, the phase difference between current and voltage will be 90°.
So understand that if current and voltage are 90° out of phase, the power will become zero.
The reason for the same would be
Power in AC circuit= V I CosΦ
If we take angle between curent and voltage as 90°, then Cos Φ= Cos 90°
Power (P)= V I Cos (90)= 0
So now we know that if we put Cos (90)= 0, then the corresponding obtained power will be zero.
A.C through pure capacitance alone
Below diagram shows a circuit having a pure capacitor of capacitance (C) connected to an A>C supply voltage given by
Now considering that the current in the circuit at any particular instant be,
i= D (Cv)/dt
The value of current will be maximum when SIN (ωt+ 90°)=1
= ωCVm Cos ωt =ωCVm SIN (ωt+ 90°)
i= ImSIN (ωt+ 90°)
Power: To understand this you must notice, Fig, 26 (c), the whole figure is notably divided into various quarters representing the graph of resultant power in AC through pure capacitance alone. You would notice that in first quarter Capacitor plates electric field is storing all the power or energy been supplied by the source.
Moving on to the next quarter cycle, you would notice that, this electric field between the capacitor plates collapses and whatever power or energy was stored in this field returns to the source.
The entire process by both quarters of the cycle repeats itself continuously, and in the whole process, no power or energy is been consumed. This expression could be mathematically expressed as
Power for the whole cycle = Vmax / √2 . Imax / √2 ʃ2π0 sin 2 ωt = 0
So from the above explanation of power in AC through pure capacitance only, we conclude that power consumed by the circuit will always remain zero and the current always leads the applied voltage by 90°.
Variation in frequency (f) and capacitive reactance (Xc)
We know that the formula for capacitive reactance is Xc= 1/ 2πfl. And the below given figure is the graphical representation of the same.
The points to be understood about Xc and f are:
- When f=0, Xcbecomes infinity and behaves as an open
- And when inverse of this happens that is when the frequency becomes infinite, at that point Xc decreases and becomes zero.
A physical quantity can be either a scalar or a vector. A scalar is a physical quantity that requires only magnitude and no other characteristics for its description, whereas vectors or Phasors need a magnitude as well as a direction for describing themselves.
Phasors : A phasor is a complex number representing a sinusoidal function whose amplitude (A), angular frequency(ω) and initial phase(θ) are time invariant.
Complex numbers: Apart from real numbers there are other numbers that can be used while dealing with such frequencies. Complex numbers are introduced to allow of complex number equations which happen to be square roots of negative numbers. This category of number is known as “imaginary number”. Imaginary numbers are represented as multiples if and it is denoted by the letter ‘j’. This j = and is known as the j operator.
Vector rotation of the j operator:
The j operator isused for indicating anticlockwise rotation of vector, each power of j i.e.: j ,j2,j3etc shows the rotation of the vector by 90o anticlockwise(as in fig. below)
This means multiplying an imaginary number by j2 will help in rotating vector by 180o anticlockwise, by j3 rotates it 270o and so on. However these multiplications have no effect on the magnitude of vector.
There are three ways to show phasor:
- Rectangular or Cartesian form
- Polar form
- Exponential form
THE RECTANGULAR OR CARTESIAN FORM:
In this form complex number is shown as a point on a two-dimensional plane called s-plane or complex plane. Its generalized form is:
Z = x + jy
Z => the complex number showing vector
x => the real part
y => the imaginary part
Graphical representation of rectangular form:
This kind of graphical representation of theform of a rectangle as Argand diagram. The horizontal axis represents all real number and the vertical axis represents the imaginary number. In the 1st quadrant, both the real and imaginary number has a positive value. In the 2nd quadrant, the real number is negative whereas the imaginary number is positive. In the 3rd quadrant, both the real and the imaginary number have a negative value. And, in the 4th quadrant, the real number is positive whereas the imaginary number has a negative value.
The Complex Conjugate
Complex Conjugate or conjugate of a complex number is obtained by reverting algebraic sign of imaginary part of the complex number. Complex conjugate of a complex number U is denoted by Ū. So if U= a + jb, Ū = a-jb. For example, if U= 5 + j6; Ū = 5 – j6.
The addition of complex number and its conjugate is always a real number while the subtraction will always yield an imaginary number. When a complex number is multiplied by its conjugate, i.e. U x Ū = (a+jb)(a-jb) = a2 – j2 b2 = a2 + b2 , the quadrature component gets removed.
THE POALR FORM
In the rectangular form points are plotted on the complex plane, but in the polar forma complex number is written in terms of its magnitude and angle . Thus in polar form a vector is represented as:
Z = A∠±θ
Z => complex number representing the vector
A => the magnitude of the vector
θ => angle of A which can be positive or negative
Polar form representation of a complex number
As the polar form representation is based on the triangular form simple trigonometry or geometry of triangles can be used to find magnitude and angle of complex number.
So by Pythagoras’s theorem:
A2 = x2 + y2
- A = (x2 + y2)
Also, x = A.cosθ , y = A.sinθ
Using trigonometry: θ = tan-1 (y/x)
The Complex Conjugate
To obtain the Conjugate of a complex number in polar form, the sign of the angle is reversed while the magnitude remains as earlier. So in polar form if U = A ∠θ; Ū = A∠-θ.
THE EXPONENTIAL FORM
This mode of showing a complex number is similar to polar format and corresponds to magnitude and phase angle, but uses base of base logarithm, e= 2.71 to find the value of the complex number. The Exponential form uses both sine and cosine details of right angled triangle to state a complex number in a complex plane. Thus in Exponential form :
Z = A ejθ
Or Z = A(cosθ + jsinθ)
So, Exponential form can be represented as:
Converting rectangular form to polar form:
To convert: Z = 4 + j3
In polar form: Z= A∠θ
So, A∠θ = 4 + j3
- A = (42 + 32)
- A = 5
And θ = tan-1(3/4) = 37o
Hence, 4 + j3 = 5∠37o
Converting polar to rectangular form:
To convert: Z = 6∠30o
In rectangular form: Z = a + jb
Now a= Acosθ and b= Asinθ
Therefore, 6∠30o = 6*cos 30o + j(6*sin 30o)
= 6*0.866 + j(6*0.5)
= 5.2 +j3
Thus, 6∠30o = 5.2 + j3
ADDITION AND SUBTRACTION OF VECTOR:
The addition and subtraction of phasors is easier to perform if the phasors are written in the rectangular form. So if two complex numbers, that are to be added or subtracted, are written in the polar form they must be first converted to rectangular form so as to perform addition or subtraction over them. Once they are converted to the rectangular form, addition or subtraction is carried over as follows:
Addition: If A and B are two complex numbers, such that:
A = 4+j1 and B= 2 + j3
A + B = (4+j1) + (2+j3)
A + B = (4+2)+ j(1+3)
Thus, A + B = 6 + j4
Subtraction: If A and B are two complex numbers, such that:
A = 4+j1 and B= 2 + j3
A + B = (4+j1) – (2+j3)
A + B = (4-2)+ j(1-3)
Thus, A + B = 2 – j2
MULTIPLICATION AND DIVISION OF VECTORS:
The multiplication and division of phasors is easier to perform if the phasors are written in the polar form. So if two complex numbers, that are to be multiplied or divide, are written in the polar form they must be first converted to rectangular form so as to perform multiplication or division over them. Once they are converted to the rectangular form, multiplication or division is carried over as follows:
Multiplication in polar form:
X * Y = A * B ∠θ1 + θ2
Multiplying together 6∠30O and 8∠-45O in polar form gives us.
X * Y = 6 * 8∠30O+(-45O) = 48∠-15O
Division in polar form:
When two vectors in polar form are to be divided together, the two modulus should be divided and then angles subtracted.
X/Y = (A/B)∠ θ1 –θ2
X/Y = (6/8)∠30O-(-45O) = 0.75∠75O
AC series circuits
It is already discussed that an electric circuit is made of three components namely, resistor, capacitor, and inductor. All these three are denoted as R for resistance, l for reluctance, and C for capacitance.
In the forthcoming section, let’s discuss various possible combinations of these RLC components. These combinations could be R-L, R-C, and R-L-C.
R-L circuit that is resistance and inductance in a series.
From the above diagram, it could be gathered that
The Voltage drop across R(Vr) and the Voltage drop across the coil (Vl) can be taken as the reference vector in the phasor diagram. And the vector sum of Vr and VL is represented by vector OB in the diagram.
Impedance of the circuit
Z= R2+ Xl2
Where R= resistance and Xl2= total opposition offered to the AC circuit by R-L series circuit
And the above-mentioned equation of impedance falls true for the impedance triangle ACB.
Now if you observe figure (d), where I lag V by angle Φ, the following equation could be obtained.
Power factor Cos Φ= R/Z
And if we take voltage as V= VmaxSinωt, then the equation for figure (d) would become
Thus obtained is
Now let’s move to figure (e), where the resolution of I is been shown into two components, which are I Cos Φ along the V (voltage) and I Sin Φ which is lying perpendicular to the voltage.
So now from the above figure, Average consumed by the circuit can be written as
V I CosΦ, for the first component
i.e. P= V IcosΦ
The term cosΦ is called the power factor of the circuit
Also, it must be noted that,
True power (Watts) = volt-ampere* power factor
i.e. Watts= VA * Cos Φ
Also, pure inductance consumes no power and the power been consumed would be by ohmic resistance only, hence
p = VI cosΦ
The above notion shows that power is been consumed by resistance alone and not by the inductor.
Now moving on to last figure (f), this is a graphical representation of power consumed in R-L circuits
So summing up that R-L circuit has current, Impedance, power factor and power consumed.
Apparent, active, and reactive power
The combination of reactive power and the active power will be apparent power. The unit of active power is watts, and the apparent power could be calculated in terms of volt-amperes,i.e., V*I. Similarly, reactive power could be expressed as volt-ampere reactive (var).
The active power which is also known as actual power does the useful work. Reactive power is used by magnets in creating the magnetic flux. and lastly, apparent power is simply the vectorial summation of KVAR and KW.
From the above figures, we get
And the further formulas have been derived using the above figure being
- actual power (P)= VI CosΦ
- Apparent power = rms * circuit resistance
VI= (I * Z)
I= I2Z volt-ampere
Reactive components develop an amount of unused power which is termed as reactive power, such as inductor and capacitor in any AC circuit that dissipates zero power. And these powers that keep traveling back and forth or react upon themselves are called reactive power. This power is rightly termed as ‘phantom power.’
All the three powers are shown in the above figure,
W= VA cosΦ
And VAR= VA SinΦ
VA = W/CosΦ
Where W= power factor
VA= True power
R-C circuit (resistance and capacitance in a series)
A resistor-capacitor circuit or RC network which is made of resistor and capacitor been driven by a voltage.
It is considered to be a good combination in the series circuit because the capacitor can store the energy and the resistor can control the rate at which it charges. RC has an exponential dimension of the time constant. From which you could guess that RC’s time dimension = 1 ohm * 1 farad= 1 second.
So if we take V= voltage, and I= resultant current then
Vr=Ir, When voltage drops across R
And Vc= IXC , When voltage drops across C
Now, the figure (c) in above diagram shows the impedance triangle, ABC, where
Where Z= Impedance, R= resistance, and X= Capacitive reactance
Now moving on to graphical representation of figure (d) in the above diagram,
Power factor, cos Φ= R/Z
And instantaneous power, p= V max Sin rot X/ V max SIN (ωt+Φ)
And lastly, an average power consumed in a circuit in an entire cycle,
P= average of Vmax/2. I max/2 cosΦ- an average of Vmax/2.Imax /2 cos (ωt+Φ)
Thus in R-C circuit, we have impedance, current, power factor and power consumed.
An RLC circuit consists of resistor, capacitor and inductor, all in one series.
The series RLC circuit in the above diagram has a single loop having instantaneous current flowing through that loop. Now the inductance and capacitance, Xl and XC respectively, are supply frequency function, the sinusoidal response of the RLC circuit will, therefore, vary with the frequency, f. thus in this type of circuit,
Impedance will be defined as Z= R2+ (XL+Xc)
Current would remain at I= V/Z
Power consumed would become as CosΦ=R/Z
And power consumed would be VI Cos Φ= I2R
Resonance in RLC circuits
The resonance in RLC series occurs when inductive and capacitive reactance are equal magnitude wise
And but still, cancels each other as they are 180 degrees apart in phase.
Resonance occurs when Xl=Xc, then the frequency can be calculated as,
f= 1/2π √LC)
Half power frequencies, Bandwidth, and Selectivity
Half power frequency
The half power frequency is a point at which the output power drops to half of its mid
band value. It is commonly used with cutoff frequency.
Bandwidth basically means a range of frequencies within any given band. Selectivity of
the circuit is this ratio between the bandwidth and the resonance frequency.
In this RLC circuit, when you vary the frequency, having Q as the quality of the resonance
circuit, the equation thus obtained becomes,
Q-factor of a resonant series circuit
Quality factor in simplest terms is a measure of the quality of the resonant circuit. The desirable Q-factor is the higher one as it indicates narrower bandwidth.
Q= Pstored/Pdissipited , the Quality factor is the measure of the ratio of power store to the power dissipated.
The formula explained is applicable to the series resonant circuit as well as a parallel resonance circuit.
This Q-factor is generally dimensionless. A higher Q-factor in sinusoidally driven resonant with greater amplitude but smaller frequencies around the frequency for which they resonate.
A.C parallel circuits
These days there are a lot of transmission and distribution systems have come up. But out of all parallel circuit has proved its reliability the most and we come across this system of power transmission the most.
Mostly power circuits are made up of constant voltage circuits with parallel connected loads, which can be solved in three ways
- Admittance method
- Vector algebra or J method
- Phasor/vector method
It is the most common method that is been used in solving A.C. parallel circuits. Admittance of circuit is basically the reciprocal of impedance. Therefore, admittance can be said to be the effective ability of the circuit which allows the A.C. current to flow through it. The newly recognized unit of admittance is Siemens. So any circuit with 1 ohm impedance will have admittance as Siemens. Admittance is generally denoted as Y.
So mathematically it can be represented as,
Where Z= impedance
An admittance triangle is also represented as impedance triangle with two important components
Where Conductance (g) = YCosɸ=1/Z . R/Z=R/Z2=R/R2+Z2
Even if circuit parameters changes, the value of conductance continues to be positive.
Susceptance is the perpendicular in the admittance triangle
Susceptance (b) = Y Sinɸ= 1/Z . X/Z= X/Z2= X/R2+Z2
To solve parallel circuits, all the branches are first arranged in parallel form. Each branch contain components like resistance, capacitance, or inductance and each circuit is separately analyzed. After analyzing it carefully, the effects of all the branches are combined together to procure the desired result.
In the above figure, in order to solve AC parallel circuit, first of all find the impedance of each branch of the circuit independently.
Now determine the magnitude value of current and phase angle with the voltage,
Now draw the Phasor diagram as per various branches, by the method of components
So the current I will be Ixx+IYY
Now find the phase angle between the current and total voltage
ɸ= tan -1 IXX/IYY
Series parallel circuit
There are two ways of solving series parallel circuits, which are
- Admittance method
- Symbolic method
In a series-parallel circuit admittance method is the best approach to solve or find out the impedance of the circuit. In this method the parallel branch are first converted into its equivalent series form and then it is connected to the remaining circuit and then the total impedance is calculated.
Resonance in parallel circuits
The primary condition for the parallel circuit being in resonance is that power factor of the circuit becomes 1 or unity. This happens when the impedance due to capacitance and impedance due to inductance becomes equal in the circuit. So if the current through the inductor and the current through the capacitor is equal, then voltage and current are in the same phase and leads to resonance.
Links of Previous Main Topic:-
- Current Electricity Basic Concepts
- Introduction to Alternating Current
- Generation and Equation of Altering Voltages and Currents
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