Electrical and Electronics Engineering by E Keeda

INFOMATICA ACADEMY

CONTACT: 9821131002/9029004242

AC CIRCUITS  COMPARISON OF AC WITH DC: AC 1. AC stands for Alternating Current type supply. 2. Polarity of supply changes periodically.

DC 1. DC for Direct Current type supply. 2. Polarities are fixed.

3. AC Voltage or current not only varies its direction but magnitude is varying with time.

3. Both direction and magnitude are constant.

5. AC can be transformed from one voltage level to other voltage level with the help of transformer.

5. DC cannot be transformed.

6. AC has frequency from low to high, e.g. mains frequency is 50Hz.

6. Frequency of DC is zero.

7. Examples of AC voltage are Audio (signal from microphone), AC power line all oscillator O/P.

7. Dry cell, battery, Rectifier etc.

 ADVANTAGES OF AC: 1. The nature of sound is quite similar to that of AC supply and therefore its easy transmission as well as conversion of sound to electrical signal and electrical signal to sound with AC; is the main advantages of AC. 2. The transmitted signal easily received by the method of resonance, resonance is another specialty of AC. This method of receiving radio and TV signal with the help of resonance is known as „Tuning‟. 3. AC supply can be generated by simple motor without using another electrical energy. Generator is a common example of AC supply. In vehicles like scooter, motorcycle, AC magneto is also simple in construction. 4. DC supply can be generated by using AC. The quality of DC, generated by AC is not much superior, but that can be filtered by capacitor.

Electronics - I

AC Circuits

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 NATURE OF AC:

1. Amplitude (A): It is the magnitude of voltage or current at particular instant. 2. Period (T): It is the time taken by AC supply to complete one cycle. 3. Cycle: It is a set of magnitudes in one positive and one negative half cycle. 4. Frequency (f): It specifies the speed of rotating signal in terms of frequency. It is defined as “frequency is the number of cycle completed in a second”. 1 1 sec. f  Hz and period T  T f 5. Wavelength    : The distance covered by one cycle is called as its wavelength, it is denoted by  ( lambda), and it is measured in meters. The formula for  is C  f Where C = velocity of light = 3 x108 m/Sec. 6. Peak voltage: It is the maximum value of sine wave voltage either on positive or negative half cycle.  AC GENERATIOR: As explained earlier AC supply continuously varies in magnitude and periodically changes polarity. The fig. (a) Shows the idea of AC generated by a rotating loop and magnitude of voltage for different angular positions. The fig. (b) also shows how a rotary generator produces and AC voltage. The loop of conductor rotates through the magnetic field to generate AC voltage across its output terminals. The voltage generated by this method is known as induced voltage. Its magnitude depends on the angular position of the loop, at 00 it is parallel to magnetic field and at 900 , 2700 it is perpendicular. When it is parallel, the voltage is zero and at 900 it is maximum as shown in the fig. At 1800 , 2700 and 3600 it passes through same angle but the direction of current gets reversed because the terminals of the loop get interchanged.

Electronics - I

AC Circuits

INFOMATICA ACADEMY

CONTACT: 9821131002/9029004242

 RMS value of AC: Ac voltage is measured in terms of effective value which is called as “Root Mean Square‟ (RMS) Value. RMS value is the most correct value of sine wave voltage. Voltage Vrms  VP x 0.707orVrms 

Current I rms  I p x 0.707or I rms 

 Example 1: Calculate the peak and peak to peak voltage of 230V AC supply. Vrms  230V

Solution:

Vrms  Vp x 0.707

Vp 

Vrms 0.707



230 0.707

Vp  325.23V Vp  p  650.46V

Thus peak voltage is always more than RMS voltage.  Example 2: Calculate the RMS voltage of an AC supply whose Peak voltage is 450V. Solution:

Vrms  Vp x 0.707  450 x 0.707  318.15V

 REPRESENTATION OF SINE WAVE: If the magnitude of sine wave voltage is noted for 00 ,900 up to 3600 , it varies from 00  0,

900  Vp,

1800  0, 2700  Vp and 3600  0. It is very similar to a sine function

and therefore represented by a sine wave equation e = A sin ωt and i = A sin t ∴ e = A sin2 πft

But w = 2  f e = A sin t or e = A sin 2  ft

Where „e‟ is AC voltage, A = amplitude of the voltage it is also represented by Emax, f = the frequency of sine wave

Electronics - I

AC Circuits

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 Phase Angle: (i) If they are passing through the same points as shown in fig. (2.3) then it is said that these two AC signals are “in phase” or there is 00 phase shift.

(ii) If two AC signals are not passing through common points then they are said to be out of phase signal. The fig. (2.4) shows 900 and 1800 out of phase signals A & B. It shows that signal voltage A is leading to the signal voltage B. Vector diagram shows that lengths are equal because their amplitudes are equal.

 RESISTANCE IN AC CIRCUIT: Resistance behaves same as it behaves in DC circuit. Ohm‟s law also applicable. The current flowing through a resistance can be easily calculated by Ohm‟s law  Vrms  i   R  

Similarly, it does not affect phase between voltage and current. Therefore resistance has zero phase angles and it is therefore a non-reactive component.

Note that in AC circuits the resistance value of a resistor remains unchanged even if the frequency of AC supply gets changed. In case of capacitor or inductor it makes difference as explained in the following topic. Hence inductor and capacitor are known as reactive components.

Electronics - I

AC Circuits

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 CAPACITOR IN AC CIRCUITS: Capacitor is a reactive component it behaves in different way in AC circuit. When a capacitor is connected in AC circuit, it‟s resistance to AC, which is called as “reactance” it depends upon the frequency of AC voltage. Resistance in DC circuit and reactance in AC circuit, these two terms are different. Reactance is also measured in Ohms but it is not constant, it varies with frequency. “Xc” denotes capacitive reactance and it depends upon the value of capacitor as well as frequency of AC supply. Hence it is calculated by formula.

Xc 

1 2 fc



Similarly, capacitor produces 900 phase shift between voltage and current as shown in fig (2.8), current leads voltage by 900

Refer fig. (2.8) when capacitor is connected across a DC supply the lamp glows for a very short time and then it becomes OFF because current through the circuit becomes zero once capacitor is fully charged it is said that capacitor is DC open. In the next figure; when it is connected across an AC supply lamp glow due to charging and discharging, it is said that capacitor is AC short. When the frequency of AC supply is increased the lamp glows with more intensity. It shows that the capacitive reactance decreases with increase in frequency.  INDUCTOR IN AC CIRCUIT: Inductor is nothing but a coil it is also a reactive component, its opposition, to AC current is called as “inductive reactance” denoted by  X L  and it is calculated by formula,

X L  2 fL



Measured in Ohms. Inductive reactance is directly proportional to the value of inductance and frequency. Inductor also produces 900 phase shift between voltage and current but here current lags voltage by 900 as shown in fig (2.9).

Electronics - I

AC Circuits

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Refer fig. ( 2.9) when an inductor is connected across DC supply the lamp glows continuously because the inductor is acting as a wire, it is said that inductor is DC short, When it is connected across an AC supply lamp does not glow because it opposes AC current, it is said that inductor is AC open.  IMPEDANCE (Z): When circuit is complex, if capacitor, inductor and resistor all are present then total opposition is not the sum of Xc, X L and r but it is calculated by other method because there is an effect on phase shift. This effective total opposition is called as “impedance”. It is denoted by “Z”. When circuit is series circuit then impedance is calculated by Z  X 2  R2

Consider its vector diagram. If the phase of R is taken as a reference then the phase of reactance is 900 out of phase. In case of X L it makes 900 phase shift while in case of X C it makes phase shift of 900 .

Z  X 2  R2 While While tan  

X L XC or R R

  tan 1

XC XL or tan 1 R R

X is the total reactance and R is the total resistance of the circuit. If circuit is iin parallel then impedance is calculated by current equation

IT  I X2  I R2 Where I x  I L I C and therefor Z 

Electronics - I

V IT

AC Circuits

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Example 2: Calculate impedance of the circuit

Solution: As shown in given circuit diagram components are in parallel therefore first total current  IT  must be calculated.  IL 

V 100   4A X L 25

IC 

V 100   1A X C 100

IR 

V 100   4A R 25

(Voltage across each is 100V because they are in parallel)  I X  4 A  1A  3 A (Capacitive and Inductive currents are 1800 out of phase)

IT  I X2  I R2  32  42  5 A Now impedance Z 

V 100   20 IT 5

 COMPARISON OF XC AND XL : Capacitive Reactance ( X C )

Inductive Reactance ( X L )

1. It is the opposition of capacitance to 1. It is the opposition of inductance to AC AC current. current. 2. Formula : X C  1/ 2 f C

2. X L  2 f L

3. It increases for lower values of 3. It decreases capacitors. inductors.

for

lower

values

4. It increases for lower frequencies.

4. It increases for higher frequencies.

5. Voltage lags to the current by 900

5. Voltage leads to the current by 900

7. VC lags iC by900

7. VL leads iL by900

Electronics - I

AC Circuits

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1. If the value of peak voltage is 325.25 V find its RMS value Solution:

Vrms  Vp  0.707  325.25  0.707 Vrms  229.95 volt

  log 325.25  log 0.707   2.5122173  (0.155805)   2.3616   Anti log(2.3616)  229.5

2. Find the impedance when resistance of 30  and inductance of 40  are in series. Solution:

R  330

X  40 

ImpedanceZ X 2  R 2 

 40    30  2

 50

3. Resistance of 20 ohms, inductor having inductive reactance of 20 and capacitor having capacitive reactance of 60 . If they are connected in series find impedance and phase angle of the circuit. Solution:

Given R = 20, X L 20 , X C  60 Series circuit formula is Z  X 2  R 2 Where X  X C  X L  60  20  40 Z 

 40    20  2

 44.72

X L 20  1 R 20   tan 1 1  450 tan  

 RESONANCE (TUNED CIRCUIT): This is a useful term; it plays in an important role in radio, TV circuit to tune or to select p articular station signals. Resonance is defined, as “it is the moment in electrical circuit where circuit gives maximum response at particular frequency”. Response may be impedance or current in the circuit. The frequency at which circuit becomes resonant is called as „resonant‟ frequency and it is denoted by ( f r ) similarly the circuit, which shows resonance is known as tuned circuit. The main use of this circuit we find in radio and TV circuits. In electrical L-C-R circuit inductive reactance increases with frequency and capacitive reactance decreases. At a particular frequency X L becomes equal to X C they are equal and opposite this phenomenon is known as resonance. The frequency at which it happens is known as resonant frequency (fr.)

Electronics - I

AC Circuits

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There are two resonance circuits as follows 1. Series Resonance and 2. Parallel Resonance. 1. Series Resonance: A capacitor, inductor and AC voltage source when they are connected in series then they form series resonance circuit.

Fig. (2.11) shows the circuit diagram where „R‟ is an internal (DC) resistance of the inductor therefore it is also known as LCR circuit. The current meter shows the current for different frequencies of AC voltage source. The graph shows at resonant frequency „ f r ‟current is maximum and above „ f r ‟ or below „ f r ‟ current is rapidly dropping. Derivation: At Resonance X L  X C 2 frr L 

1 2 f r C

1 4 2 LC 1  fr  ......................Formula for Resonant frequency. 2 LC  f r2 

Electronics - I

AC Circuits

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 Characteristics Of Series Resonance: 1. Below resonant frequency; circuit is capacitive and current is small due to high XC 2. Above resonant frequency; circuit is inductive and current is small due to high X L . 3. At resonance circuit is resistive and maximum current flows through circuit because X L  X C  0. 4. At resonance voltage across capacitor is equal to the voltage across inductance but they are in 1800 out of phase. I x X C  I x X L EC   EL because series current I is same and at resonance X C  X L 5. At resonance, impedance of the circuit is very small. 6. The circuit is known as an acceptor circuit. Example: A series resonance circuit consists of 100 micro Henry inductance and 100pf capacitance. Find the resonant frequency. Solution: L = 100  H

C = 100 pf

Formula : f r  

1 2 LC 1 6





2 X 3.14 X 100 X 10 X 100 X 10 12 6.28 10 4 X 1018 1 1 1   14/2 14 6.28 10  6.28 X 107 6.28 10

 0.1592 X 107  1592 X 103 Hz or 1592 KHz  ‘Q’ of the Resonance: Illustrates the idea of q and its relationship with „Bandwidth‟. The bandwidth of resonance is important because it gives the idea about the band of frequencies for which resonance is effective. As shown in fig (2.13) bandwidth is calculated by formula, 10MHz. Bandwidth(f )   0.2MHzor 200 KHz 50 It shows 0.1 MHz below and 0.1 MHz above from 9.9 MHz-10 MHz-10.12 MHz actual resonances occur. Current for these frequencies is approximately close to the maximum value. Therefore this resonance circuit is called as „Band-pass-filter‟ it gives maximum output for a band of frequencies, For example all other frequencies above 10.1 MHz and below 9.9 MHz output is about zero, those frequencies are filtered. This series resonance is also known as an acceptor circuit.

Electronics - I

AC Circuits

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2. Parallel Resonance: This is another L-C-R circuit in which L-C are connected in parallel with input AC voltage. The basic difference between series and parallel resonance is, in parallel resonance impedance of circuit is maximum while in series resonance current is maximum. Resonant frequency is calculated by formula

fr 

1 2 LC

Because at this frequency capacitive reactance becomes equal to the inductive reactance ( X C  X L ).

 Characteristics Of Parallel Resonance: 1. Below resonant frequency; the circuit is inductive and impedance is small because X L is low. 2. Above resonant frequency; the circuit is capacitive and impedance is small because X C is low. 3. At resonance; the circuit is resistive and impedance is maximum because X L  X C 4. The circuit is known as rejecter circuit.  COMPARISON: Series Resonance

Parallel Resonance

1. L and C are in series

1. L and C are in parallel.

2. At resonance, current is maximum

2. At resonance, current is minimum

3. At resonance impedance in 3. At resonance impedance is maximum minimum 4. Circuit is capacitive below f r and 4. Inductive below f r and capacitive above fr . inductive above „ f r ‟ 5. Known as Band pass filter or 5. Known as band-stop filter or rejecter acceptor circuit circuit. 1 1 6. Formula : f r  6. Formula : f r  2 LC 2 LC

Electronics - I

AC Circuits

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1. The Resonant frequency of a LC circuit is 1 KHz calculate the value of inductance, if the capacitance is 50  f C = 50  f .

Solution: Given fr = 1 KHz

 fr 

1 2 LC

1X 103 

6.28 lx50 x106 squaring both sides 1 1 1 1 X 106    2 6 (6.28)  L  50 10 39.43  50 1971.92  5.07 104  0.5 103 H  0.5mH 2. A capacitor of 50  f , an inductance of 0.2025 H and a resistance of 21 are connected in series. At what frequency will the resonance occur? What will be the current at resonance if the supply voltage is 14 volts? Solution: (I)

fr 

2 LC 1 1  fr 2  2  2 4 LC 4(3.14)  0.2025  50 106 1   2.5042 103 106  2.5042 103 399.31106 f r2  2504.2 Hz  f r  50.04 Hz (II) At resonance circuit resistance is only the resistance of coil, which is given as 21  Current in this circuit is V 14 i    0.66 A R 21

Electronics - I

AC Circuits