BAKISS HIYANA BT ABU BAKAR JKE,POLISAS

Transcription

1 BAKISS HIYANA BT ABU BAKAR JKE,POLISAS 1

2 1. Explain AC circuit concept and their analysis using AC circuit law. 2. Apply the knowledge of AC circuit in solving problem related to AC electrical circuit. 2

3 Resonance phenomenon & its functions Effect of changing the frequency RESONANCE Understand resonance in series and parallel circuits Graph of impedance vs frequency Resonance frequency equation Determine Q, BW, D 3

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5 Resonance is a condition in RLC circuit in which the capacitive and inductive reactance are equal in magnitude, thereby resulting in a purely resistive impedance. Z = R + j( ) ; note: = 0 X L = X C 5

6 Current will be maximum & offering minimum impedance. 6

7 Current will be minimum & offering maximum impedance. 7

8 Resonance circuit serves as stable frequency source. The frequency set by the tank circuit is solely dependent upon the value of L & C Resonance circuit serves as filter. Acting as a short of frequency filter to strain certain frequencies out of a mix of others 8

9 A series RLC circuit s reactance changes as you change the voltage source s frequency. Its total impedance also changes. 9

10 At low frequencies, Xc > XL and the circuit is primarily capacitive. At high frequencies, XL > Xc and the circuit is primarily inductive. 10

11 Reactance change as you change the voltage source s frequency. At low frequencies, XL < Xc and the circuit is primarily inductive. At high frequencies, Xc< XL and the circuit is primarily capacitive. 11

12 A series RLC circuit contains both inductive reactance (X L ) and capacitive reactance (Xc). Since X L and Xc have opposite phase angles, they tend to cancel each other out and the circuit s total reactance is smaller that either individual reactance: X T < X L & X T < Xc 12

13 The smaller reactance dominates, since a smaller reactance results in a larger branch current. 13

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16 (a) SERIES CIRCUIT: Q factor: - Q is the ratio of power stored to power dissipated in the circuit reactance and resistance. - Q is the ratio of its resonant frequency to its bandwidth. IF; 16

17 PARALLEL CIRCUIT: (a) Quality factor: the ratio of the circulating branch currents to the supply current. 17

18 (b) Frequency bandwidth, B = f2 f1: The difference between the two half-power frequencies. Bandwidth, Δf is measured between the 70.7% amplitude points of series resonant circuit. 18

19 Lower cut-off frequency, (ƒ L ): Upper cut-off frequency, (ƒ H ): 19

20 BW = f c /Q Where: f c = resonant frquency Q = quality factor 20

21 Parallel resonant response varies with Q. 21

22 In Figure above, the 100% current point is 50 ma. The 70.7% level is 0707(50 ma)=35.4 ma. The upper and lower band edges read from the curve are 291 Hz for f l and 355 Hz for f h. The bandwidth is 64 Hz, and the half power points are 32 Hz of the center resonant frequency BW = Δf = f h -f l = = 64 f l = f c - Δf/2 = = 291 f h = f c + Δf/2 = = 355 Since BW = f c /Q: Q = f c /BW = (323 Hz)/(64 Hz) = 5 22

23 (c) The dissipation factor, D: - The ratio of the power loss in a dielectric material to the total power transmitted through the dielectric. 23

24 CHARACTERISTIC SERIES CIRCUIT PARALLEL CIRCUIT Resonant frequency, fr Quality factor,q Bandwidth, BW Half power frequency, f L & f H & 24

25 A series resonance network consisting of a resistor of 30Ω, a capacitor of 2uF and an inductor of 20mH is connected across a sinusoidal supply voltage which has a constant output of 9 volts at all frequencies. Calculate, the resonant frequency, the current at resonance, the voltage across the inductor and capacitor at resonance, the quality factor and the bandwidth of the circuit. Also sketch the corresponding current waveform for all frequencies. 25

26 Resonant Frequency, ƒ r Bandwidth, BW Circuit Current at Resonance, I m The upper and lower -3dB frequency points, ƒ H and ƒ L Inductive Reactance at Resonance, X L Voltages across the inductor and the capacitor, V L, V C Current Waveform Quality factor, Q 26

27 A series circuit consists of a resistance of 4Ω, an inductance of 500mH and a variable capacitance connected across a 100V, 50Hz supply. Calculate the capacitance require to give series resonance and the voltages generated across both the inductor and the capacitor. Solution: Resonant Frequency, ƒ r Voltages across the inductor and the capacitor, V L, V C 27

28 A parallel resonance network consisting of a resistor of 60Ω, a capacitor of 120uF and an inductor of 200mH is connected across a sinusoidal supply voltage which has a constant output of 100 volts at all frequencies. Calculate, the resonant frequency, the quality factor and the bandwidth of the circuit, the circuit current at resonance and current magnification. 28

29 The upper and lower -3dB frequency points, ƒ H and ƒ L Resonant Frequency, ƒ r Inductive Reactance at Resonance, X L Quality factor, Q Circuit Current at Resonance, I T At resonance the dynamic impedance of the circuit is equal to R Bandwidth, BW Current Magnification, I mag We can check this value by calculating the current flowing through the inductor (or capacitor) at resonance. 29

30 For resonance to occur in any circuit it must have at least one inductor and one capacitor. Resonance is the result of oscillations in a circuit as stored energy is passed from the inductor to the capacitor. Resonance occurs when X L = X C and the imaginary part of the transfer function is zero. At resonance the impedance of the circuit is equal to the resistance value as Z = R. 30