Circuit Analyses in Complex PDF
Document Details
Uploaded by EngagingPlateau
Nnamdi Azikiwe University (UNIZIK)
Tags
Summary
This document provides a detailed analysis of circuits using complex algebra. It covers topics like voltage, current, impedance, and resonance in series RLC circuits. It includes examples and exercises in different contexts.
Full Transcript
# Circuit Analyses in Complex Recall for a signal, V=V<sub>m</sub>cos(wt-θ) ## Voltage Signal, V=V<sub>m</sub>cos(wt-θ) Complex algebra enables equations representing alternating vectors and currents and their phase relationships to be expressed in simple algebraic form. Pat V=500cos(2500t-20)...
# Circuit Analyses in Complex Recall for a signal, V=V<sub>m</sub>cos(wt-θ) ## Voltage Signal, V=V<sub>m</sub>cos(wt-θ) Complex algebra enables equations representing alternating vectors and currents and their phase relationships to be expressed in simple algebraic form. Pat V=500cos(2500t-20) Hence comparing the equations, w=2500 rad/sec Now, Xc=1/wc = 1/2500x10x10<sup>-6</sup> In complex algebra, resistance is called real quantities while reactance is called imaginary quantities. The symbol j is used to denote imaginary quantities. Z = R+jwL = R-jwC Z = 10+j10 Expressing in polar form, Z<sup>2</sup> = 10<sup>2</sup>-(10)<sup>2</sup> Z= √10<sup>2</sup>-(-100) Z= √102+(-100) Z= √14-j4-2 tanθ=-10/10=1 ## Resonance RLC Circuits ### Resonance in Series RLC Circuits Resonance of a series RLC circuit is said to occur when the inductive and capacitive reactances are equal in magnitude but they are 180° apart in phase. This implies that the impedance of the circuit would be minimum, as only the resistance will account for the impedance. Another implication is that the current will be maximum. Therefore, at resonance, Z=R. * Students' Concept: When the impedance (Z) is minimum, and this happens when the net reactance of the circuit is zero. Thus X<sub>L</sub>=X<sub>C</sub> or wL=1/wC hence, 2πf<sub>o</sub>L=1/2πf<sub>o</sub>C The frequency at which this resonance occurs is called the resonance frequency (f<sub>o</sub>). This is the frequency at which X<sub>L</sub>=X<sub>C</sub> 1.e. 2πf<sub>o</sub>L= 2πf<sub>o</sub>C f<sub>o</sub> = 1/2π√LC f<sub>o</sub> = 1/2π√LC Recall that impedance of the series RLC circuit is given by Z = R + X = R + (X<sub>L</sub>-X<sub>C</sub>) Z = R + (wL-1/wC) Squaring both sides Z<sup>2</sup> = R<sup>2</sup> + (wL - 1/wC)<sup>2</sup> Z = √R<sup>2</sup> + (wL - 1/wC)<sup>2</sup> The maximum current in the circuit at resonance: 1/2π√LC = 1/√LC 2πf<sub>o</sub> = 1/√LC w<sub>o</sub> = 1/√LC ### Application of Resonance In electronics, it is used in tuning circuits such as radios and TVs. It has an advantage of responding strongly to one particular frequency, while other frequencies have little effect. Hence, such a resonant circuit can select a definite frequency from a jumble of other signals available to. That resonant frequency responds to that of a particular incoming radio signal, which this happens, maximum current is obtained, and the distant radio station is clearly heard. i) The maximum current in the circuit at resonance. Solution! f<sub>o</sub> = 1/2π√LC = 1/2√1x10<sup>-3</sup>x10<sup>-6</sup> = 13 kHz ii) w<sub>o</sub> = 1/√LC = √1x10<sup>-3</sup>x10<sup>-6</sup> = 81.7 rad/s ### Example 2 A series circuit with R=5 ohms, L=1mH and C= 20μf, has an applied voltage V = 100V with a frequency of 1000 rad/sec. The value of the inductance is adjusted until the voltage across the resistor is maximum. Find the voltage across each element. Solution: The maximum voltage across the resistor occurs at resonance, when current is maximum (V=IR) at resonance X<sub>C</sub>=wL X<sub>C</sub>= 1000x20x10<sup>-6</sup> = 50 Ohms Since X<sub>C</sub>=X<sub>L</sub> X<sub>L</sub>=wL = 50 ohms Therefore, Z=R = 50 ohns I=V/Z=100/50=2A V<sub>L</sub> = IX<sub>L</sub> = (2)(50)=100° V<sub>C</sub> = IX<sub>C</sub> = (2)(50)=100° V<sub>C</sub> = 1Xc = (2)(50)(90) = 100290 Volts NB: Note that the voltage across the reactive elements are for larger than the applied voltage. A series resonant circuit is thus used as a voltage magnification circuit ### Exercise 1 A coil of resistance 2 ohms and inductance of 0.316 H is connected in series with a self capacitor of 6.316 μF across a 240V variable frequency supply. Calculate the value of the frequency at which resonance occurs; and for this condition, evaluate: i) The impedance of the complete circuit. ii) The phase angle of the complete circuit. iii) The current in the voltage across the capacitor and that across the coil. v) Sketch the vector-diagram ### Exercise 2 An air cored coil and a capacitor were connected in series across a 240V variable frequency supply. As the frequency was increased from zero, the r. m. s value of the supply current increased from zero to a maximum value of 204 mA at 25 Hz; and decreased to 5A at 50 Hz. Calculate the resistance, inductance, and capacitance of the circuit. ### Case 1 Recall a parallel RLC circuit such as we figure given below, The admittance of the circuit is given by Y = G+jB =G+j(B<sub>L</sub>-B<sub>C</sub>) =G+j(wC-1/wL) The circuit is in resonance when B=0, i.e; wC=1/wL w² =1/LC w=w<sub>o</sub>=1/√LC or f<sub>o</sub>=1/2π√LC where w<sub>o</sub> represents the resonant frequency. Also, since the admittance at resonance is seen as Y=G+j(w<sub>o</sub>C – 1/w<sub>o</sub>L) Y=G Y=1/R ∴ the impedance of the circuit is given as Z=1/Y Z= R+jW<sub>o</sub>L<sup>2</sup> Z= R+ 1/w<sub>o</sub>C<sup>2</sup> Substituting for w<sub>o</sub><sup>2</sup>=1/LC Z= R+1/1/LC – R ∴ Z= CR Consider the admittances of yet another type of parallel circuit below: R<sub>1</sub> Y L C Y = Y<sub>1</sub> + Y<sub>2</sub> = 1/R<sub>1</sub>+jX<sub>L</sub> + 1/R<sub>2</sub>-jX<sub>C</sub> Rationalizing; Y=(R<sub>1</sub>+jX<sub>L</sub>)(R<sub>1</sub>-jX<sub>L</sub>)/(R<sub>1</sub>+jX<sub>L</sub>)(R<sub>1</sub>-jX<sub>L</sub>) + (R<sub>2</sub>+jX<sub>C</sub>)(R<sub>2</sub>-jX<sub>C</sub>)/ (R<sub>2</sub>+jX<sub>C</sub>)(R<sub>2</sub>-jX<sub>C</sub>) = (R<sub>1</sub>+jX<sub>L</sub>)(R<sub>1</sub>-jX<sub>L</sub>) + (R<sub>2</sub>+jX<sub>C</sub>)(R<sub>2</sub>-jX<sub>C</sub>)/ (R<sub>1</sub><sup>2</sup>+X<sub>L</sub><sup>2</sup>) + (R<sub>2</sub><sup>2</sup>+X<sub>C</sub><sup>2</sup>) Y = R<sub>1</sub>-jX<sub>L</sub> + R<sub>2</sub>+jX<sub>C</sub> R<sub>1</sub><sup>2</sup>+X<sub>L</sub><sup>2</sup> + R<sub>2</sub><sup>2</sup>+X<sub>C</sub><sup>2</sup> Y= (R<sub>1</sub>+R<sub>2</sub>) + j(X<sub>C</sub>-X<sub>L</sub>) R<sub>1</sub><sup>2</sup>+X<sub>L</sub><sup>2</sup> + R<sub>2</sub><sup>2</sup>+X<sub>C</sub><sup>2</sup> Separating real and imaginary parts, Y<sub>R</sub> = (R<sub>1</sub>+R<sub>2</sub>)/R<sub>1</sub><sup>2</sup>+X<sub>L</sub><sup>2</sup> + R<sub>2</sub><sup>2</sup>+X<sub>C</sub><sup>2</sup> Y<sub>I</sub> = (X<sub>C</sub> - X<sub>L</sub>)/R<sub>1</sub><sup>2</sup>+X<sub>L</sub><sup>2</sup> + R<sub>2</sub><sup>2</sup>+X<sub>C</sub><sup>2</sup> The circuit is at resonance when the complex admittance is a real number. Thus: X<sub>L</sub>/R<sub>1</sub><sup>2</sup>+X<sub>L</sub><sup>2</sup> = X<sub>C</sub>/R<sub>2</sub><sup>2</sup>+X<sub>C</sub><sup>2</sup> (wL)/R<sub>1</sub><sup>2</sup>+(wL)<sup>2</sup> = (1/wC)/R<sub>2</sub><sup>2</sup>+(1/wC)<sup>2</sup> wC(R<sub>1</sub><sup>2</sup>+(wL)<sup>2</sup>) = wL( R<sub>2</sub><sup>2</sup>+(1/wC)<sup>2</sup>) wC(R<sub>1</sub><sup>2</sup>+w<sup>2</sup>L<sup>2</sup>) = wL( R<sub>2</sub><sup>2</sup>+1/w<sup>2</sup>C<sup>2</sup>) R<sub>1</sub><sup>2</sup>wC+w<sup>3</sup>LC<sup>2</sup> = wL( R<sub>2</sub><sup>2</sup>+1/w<sup>2</sup>C<sup>2</sup>) R<sub>1</sub><sup>2</sup>wC+w<sup>3</sup>LC<sup>2</sup> = R<sub>2</sub><sup>2</sup>wL+1/wC R<sub>1</sub><sup>2</sup>wC+w<sup>3</sup>LC<sup>2</sup> = R<sub>2</sub><sup>2</sup>wL+1/wC Rearranging and transposing, R<sub>2</sub><sup>2</sup>w<sup>2</sup>LC-w<sup>2</sup>L<sup>2</sup>=R<sub>1</sub><sup>2</sup>w<sup>2</sup>C-1/wC<sup>2</sup> w<sup>2</sup>= (R<sub>1</sub><sup>2</sup> - R<sub>2</sub><sup>2</sup>) w<sup>2</sup>LC w<sup>2</sup>= (R<sub>1</sub><sup>2</sup>-R<sub>2</sub><sup>2</sup>) w<sup>2</sup>C w<sup>2</sup>=(R<sub>1</sub><sup>2</sup>-R<sub>2</sub><sup>2</sup>)C = R<sub>1</sub><sup>2</sup>/(R<sub>1</sub><sup>2</sup>-R<sub>2</sub><sup>2</sup>)C = LC[(R<sub>1</sub><sup>2</sup>-L)/L] = LC(R<sub>1</sub><sup>2</sup>/L -1) w<sup>2</sup>= LC(R<sub>1</sub><sup>2</sup>-R<sub>2</sub><sup>2</sup>)/LC w<sub>o</sub>= √(R<sub>1</sub><sup>2</sup>-R<sub>2</sub><sup>2</sup>)/LC or f<sub>o</sub>= 1/2√LC(R<sub>1</sub><sup>2</sup>-R<sub>2</sub><sup>2</sup>)/LC ### Example 1 mH 10Ω 1μF For the circuit shown above, if the frequency is adjusted to the resonant frequency, calculate the supply current and the current in each branch. Assume V=10V. Solution: Recall Case 1 The supply current is given by 1/√CR I = 10√1x10<sup>-3</sup>x10<sup>-6</sup> = 10<sup>-1</sup>x10<sup>-3</sup>x10 = 0.01A = 10mA. For the circuit, w<sub>o</sub>= 1/√LC = 1/√1x10<sup>-3</sup>x10<sup>-6</sup> = 10<sup>3</sup>/√10<sup>-3</sup>x10<sup>-6</sup> = (10<sup>3</sup>)<sup>2</sup>/√(10<sup>-3</sup>)<sup>2</sup>(10<sup>-6</sup>)<sup>2</sup> = 9.95x10<sup>3</sup> rad/sec The current in the coil is I<sub>coil</sub> = V/Z<sub>coil</sub> = 10/√R<sup>2</sup>+(w<sub>o</sub>L)<sup>2</sup> = 10/√10<sup>2</sup> + (9.95x10<sup>3</sup>x1x10<sup>-3</sup>)<sup>2</sup> = 10/√10<sup>2</sup> + 99.5<sup>2</sup> = 10/√10000 + 99.5<sup>2</sup> = 10/√10000.25 = 10/99.5 I<sub>coil</sub>= 0.1 A. Current in the capacitor I<sub>C</sub>=V/X<sub>C</sub>=V/1/w<sub>o</sub>C =V<sub>1</sub>w<sub>o</sub>C =V<sub>1</sub>w<sub>o</sub>C Therefore, I<sub>C</sub>=10x99.5x10<sup>3</sup>x10<sup>-6</sup> = 99.5mA Note that the current in the branches is much greater than the circulating current. ### Exercise A Coil of resistance 20Ω and inductance 0.1 H is connected with a 5μF capacitor across a variable frequency supply. Calculate: i) The frequency at which resonance occurs. ii) The circuit impedance at resonance. iii) The admittance at resonance. iv) Sketch the vector diagram of the circuit.
