Lecture 4. Impedance of AC Circuits. Don t forget to complete course evaluations: https://sakai.rutgers.edu/portal/site/sirs Post-test. You are required to attend one of the lectures on Thursday, Dec. 7 when the post test will be given. If you do not take the post test, you will get a zero for % of your grade. Post-on-line survey. Please take the end-of-semester on-line survey at your earliest convenience but not later than midnight on Friday, December 5th 07. The link is https://rutgers.qualtrics.com/jfe/form/sv_aeqnv87qfuwyzx. We ask you to take this survey seriously because it helps us make improvements to this course and future courses.
Response of L-C-R Circuits to Alternating Currents So far we have considered transient processes in R-C and R-L circuits: the approach to the stationary (time-independent) state after some perturbation (e.g., switch on / off). Today we ll discuss the AC (cos ) driven circuits, e.g. the circuits driven by an AC power source. We disregard all transient processes and instead consider the steady-state AC currents: currents and voltages vary with time as ccc + φ, but their amplitudes are t-independent.
Amplitudes, rms Values, and Average Power in AC Circuits V t I t RLC Instantaneous values: Instantaneous power: V t = V 0 cos I t = I 0 cos + φ P t = V t I t Currents and voltages are NOT necessarily in phase, φ is the phase shift between V and I (the phase angle ). P t = V 0 cos I 0 cos + φ = V 0I 0 cos + φ + cos φ Average power: P aa P t = V 0I 0 cos φ Root mean square (rms): the square root of the average of the square of the quantity: a rrr = a t V rrr = V 0 I rrr = I 0 P aa = V rrr I rrr cos φ ccc φ - the power factor 3
Examples Average power: P aa P t = V 0I 0 cos φ Resistive loads φ = 0 I t V t R Capacitive and inductive loads φ = ± π Active (average) power: I V P (see below) P = RR S = V rrr I rrr cos φ = 0 φ = ± π π π phase ωt 4
Phasors / Complex Number Representation Problem: To find current/voltage in R-L-C circuits, we need to solve differential equations. Solution: The use of complex numbers / phasors allows us to replace linear differential equations with algebraic ones, and reduce trigonometry to algebra. Instantaneous value phasor We represent voltages and currents in the R-L-C circuits as the phase vectors (phasors) on the D plane. Quantity: A t = A 0 cos ωt. Corresponding phasor: a vector of length A 0 rotating counterclockwise with the angular frequency ω. Instantaneous value of A t is the projection of the phasor onto the horizontal axis. φ φ If all the quantities oscillate with the same ω, we can get rid of the term by using the rotating ( merry-goaround ) reference frame. We ll consider the steady state of AC circuits, when all amplitudes (the phasor lengths) are t-independent, and the only time dependence remaining is in the single frequency sinusoidal oscillation of voltages and currents. The t-independent angle between different phasors represents their relative phase. 5
b b φ φ Z a + ii = RR Z + i II Z RR Z = r cccφ II Z = r sssφ a Z = re iφ Z a ii = RR Z i II Z complex conjugate of Z Complex Numbers, Phasors Imaginary unit: Z Z Z i e iφ = cos φ + i sin φ e iπ = i A ccc ωt + φ = A A sss ωt + φ = A e iπ = i Z = i = ei +φ + ei +φ i = i Euler s relationship i +φ e e i i +φ The absolute value (or modulus or magnitude): = RR A ei +φ = II A ei +φ r cccc + i ssss r cccc i ssss = r ccc φ + sss φ = r Phasor: refer to either A e i +φ or just A e i φ. In the latter case, it is understood to be a shorthand notation: all phasors rotate with the same angular frequency and the time-independent angle φ between different phasors represents their relative phases. 6
Complex Numbers, Phasors (cont d) The use of complex numbers / phasors allows us to replace linear differential equations with algebraic ones, and reduce trigonometry to algebra: Addition: a + ii + c + ii = a + c + i b + d Multiplication: Differentiation: Ae iω t Be iω t = AAe i ω +ω t d dd AAiii = iωaa iii 7
V t Complex Power, Active Power and Reactive Power I t P t = V 0 cos I 0 cos + φ = V 0I 0 cos + φ + cos φ RLC P aa = V rrr I rrr cos φ Complex power: S = V t I t = V rrr e iii i +φ I rrr e = V rrr I rrr e ii = V rrr I rrr cos φ i sin φ = P + ii Average (active, real) power Reactive power P = RR S = V rrr I rrr ccc φ Apparent power: S = P + Q cos φ = P S 8
V t I t Resistor AC current through a resistor and AC voltage across the resistor are always in phase. Power dissipated in a resistor: P aa = V 0I 0 cccc = V 0I 0 = V rrr I rrr φ = 0 Active (average) power: P = RR S = V rrr I rrr cos φ = V rrr I rrr φ = 0 Q = II S = 0 If the load is purely resistive, both the current and voltage reverse their polarity at the same time. At every instant the product of voltage and current is positive or zero, with the result that the direction of energy flow (recall Poynting vector) does not reverse: only the active power is transferred from the power source to the load. 9
V t I t RLC Reactance and Impedance Both Reactance and Impedance are measures of how much the circuit impedes the flow of current. Units: Ohms. Impedance relates the time-independent phasors that represent the instantaneous values of voltage and current in the circuit: V rrr e iωt i φ = I rrr e Z V t = I t Z V rrr = I rrr X e ii all terms are real! Z = Z e ii The impedance is a complex number; it is a time-independent phasor, in contrast to V(t) and I(t). Reactance is the length of the impedance phasor, it is a real number. X Z V t I t V rrr = I rrr e ii Z e ii II I φ φ V RR Z V is the reference phasor 0
Reference Phasors II I φ φ V RR Z V is the reference phasor V rrr e iωt i +φ = I rrr e Z V rrr = I rrr e ii Z e iφ e iφ Since V rrr is real, φ must be the same as φ. II φ φ Z I RR V I is the reference phasor V rrr = I rrr e ii Z e ii I rrr = V rrre ii Z e ii = V rrre ii Z eii
I t Resistor Reactance and Impedance V t Resistor: V = IR = IX R Z R = X R = R Impedance and Reactance for a resistor are the same as its resistance, they are frequencyindependent.
V t I t Phase Shift in Capacitive Circuits V t V t = V 0 e iii i +φ I t = I 0 e Q t C = 0 dv t dd I t = dq t dd = C dd t dd = ii V 0 e iii I 0 e i(+φ) = iicv 0 e iii i = e iπ φ = π/ For a capacitor, voltage LAGS current by 90 0. V t I t Current (reference phasor) 3
I t Phase Shift in Capacitive Circuit (cont d) V t V t I t Active (average) power: P = RR S = V rrr I rrr cos φ = 0 Current (reference phasor) φ = π π I V P π phase ωt If the loads are purely reactive (capacitors and/or inductors, no resistive elements), then the voltage and current are 90 0 out of phase. For half of each cycle, V I is positive, but on the other half of the cycle, the product is negative - on average, exactly as much energy flows toward the load as flows back. There is no net transfer of energy to the load - only reactive power flows back and forth. 4
Capacitor Reactance and Impedance I t V t V t I t Current (reference phasor) V t = I t Z Z = Z e ii V t I t iicv 0 e iii = I 0 e i +π V 0 e iii = I 0 e i +π iic Z C = i = i V 0 = I 0 V rrr = I rrr X C = 5
V t I t Phase Shift in Inductive Circuits V t L V t = V 0 e iii i +φ I t = I 0 e dd t dd dd t dd = 0 V t = L i +φ = ii I 0 e dd t dd V 0 e iii i +φ = iωli 0 e i = e iπ V 0 e iii = ω LL 0 e i +φ+π φ = π/ In an inductor, current LAGS voltage by 90 0 6
I t Phase Shift in Inductive Circuit (cont d) V t Active (average) power: P = RR S = V rrr I rrr cos φ = 0 φ = π π I V P π time t phase ωt Again, the power IS NOT dissipated in an inductor: it is stored in the magnetic field of the inductor for half a period, and returned to the power source for another half. 7
I t Inductor Reactance and Impedance V t V t = I t Z Z = Z e ii V t I t V 0 e iii = iωl I 0 e i π Z L = i V 0 = ωl I 0 V rrr = I rrr V rrr = I rrr X L X L = ωl 8
Reactances and Impedances: Summary Resistor Z R = X R = R V rrr = I rrr R V t = I t R Capacitor X C = Z C = i = i V rrr = I rrr X C V t = I t Z C Inductor X L = ωl Z L = i V rrr = I rrr X L V t = I t Z L The impedance reflects the phase shift between V and I. In general, the impedance is a complex number (in contrast to reactance, which is real). 9
Impedance of Series and Parallel circuits I For R, C, and L in series: Z = Z + Z + Z 3 + = R + ix L ix c = R + i For R, C, and L in parallel: Z = Z + Z + Z 3 + = R + i
Low-Pass Filter Goal: to suppress high-frequency (f > f 0 ) components in the spectrum of a signal. Spectral power 0 3 f, khz 0 3 f, khz Qualitatively: V ii drops across R and C, current is the same through both R and C. At high ω the reactance of the capacitor is small, and the rms value of the output voltage will also be small. Spectral power High-Pass Filter: Spectral power 0 3 f, khz 0 3 f, khz Spectral power 3
Low-Pass Filter V ii = I R ix c Total impedance: Z = Z R + Z C = R ix c Z = R + X c V ii0 = I 0 R + X c V ooo = I ix c V ooo0 = I 0 X c I V R V ii Z V C = V ooo V ooo0 V ii0 = X c R + X c = ωc R + = ωrc + = ωτ RR + Output power: V ooo0 V ii0 = ωτ RR + ωτ RR ω τ RR ω τ RR / Cutoff frequency: ω 0 = πf 0 = RR We want to suppress the high-frequency (f > 0kkk) components in the output of an audio amplifier with the output resistance 00 Ω. What capacitance do you need? C = πf 0 R = π0 4 00 F = 60nn 4
Problem : R-C circuit II I φ φ φ V R R V ii V C = V ooo Z Z RR A voltage V ii = V 0 ccc from an AC power supply with ω = 0 3 rrr/s is applied across a resistor with R = 00 Ω in series with a capacitor with C = 0. μμ. The current in the circuit is I = I 0 ccc + φ. The phase angle φ is given by ttt φ = A. 50 B. 50 C. 0 6 D. 0 6 tan φ = V ii t = I t Z = I 0 e iφ Z e iφ Z Z e ii = R i φ = aaaaaa III RRR = aaaaaa X C R R = 0 3 = 50 00 0. 0 6 Note that φ is positive (as it should be for the RC circuits, CIVIL). 5
I Impedance of Series R-L-C Circuits For R, C, and L in series: Z = R + ix L ix c = R + i V V t = I t R + i V L = iωll Z = Z Z = R + V R V I t Current (reference phasor) V rrr = I rrr R + V C = i I 6
Resonance in the L-C I I(ii L ) II I I( ii C ) Condition for the Resonance: reactances for L and C are of the same magnitude: X C = = X L = ωl f 0 = ω 0 π ω 0 C = ω 0L ω 0 = LL At ω =ω 0 minimum (real) impedance max current. 7
I Resonance in Series R-L-C circuits For R, C, and L in series: Z = Z Z = R + I = V Z = V Z = R + ix L ix c = R + i R + Resonance condition: = ω 0 = LL - resonance frequency At ω =ω 0 minimum (real) impedance, max current. Note that at ω =ω 0, V C and V L can be greater than V. ω > ω 0 ω < ω 0 ω = ω 0 8
Problem : Series R-L-C circuits V L = iωll V rrr L =. 80 = 76V V R V rrr R =. 40 = 88V V V rrr C =. 0 = 4V V C = i I V rrr = I rrr Z = I rrr R + V rrr =. 40 + 80 0 = 0V 9
Next time: post-test! Email me your questions for the review session on Monday, //07! 30
Problem 3: Series R-L-C circuits An R-L-C series circuit with an inductance of 0.9H, a resistance of 44 Ω, and a capacitance of 7.7 µf carries an rms current of 0.446A with a frequency of 39Hz.. What is the impedance of the circuit? ω = 455 rrr/s Z = R + ix L ix c = R + i. What is the phase angle? Z 0 = R + = 339Ω II φ φ Z V RR tan φ = ωl ωc R = 455 0.9 455 7.7 0 6 44 0.97 arctan 0.97 0.77 rrr I 3. What is the rms voltage of the source? V rrr = I rrr Z 0 = 0.446A 339Ω = 5V 4. What average power is delivered by the source? cos 0.77 = 0.7 - power factor for this circuit P aa = V rrr I rrr cos φ = 5V 0.446A 0.7 = 48.6W - average rate at which electrical energy is converted to thermal energy in the resistor 3
Appendix : R-L-C circuits P aa = V rrr I rrr cos φ = V 0 I 0 cos φ cos φ = P aa V 0 I 0 = arccos 0.6 0.93 30W 00V 0.5A = 0.6 3
Appendix : Transformer Φ B - the flux per turn ℇ p = N p dφ B dd ℇ s = N s dφ B dd For an ideal transformer (R s = R p = 0): Energy conservation: ℇ p ℇ s = N p N s V p V s = N p N s V p I p = V s I s I s I p = N p N s V p V s I p = N p N s V s I s V p I p = N p N s R - as if the source had been connected directly to a resistance N p N s R impedance transformation Using mutual inductance M = L p L s : M Φ s I p = N sφ B I p M di p dd = N dφ B s dd ℇ s = M di p dd 33
Sloppy formulation Example ℇ s = M di p dd = I sr s M = I sr s di p dd = 0.4 cos 377t 377 5 cos 377t =.55mm ℇ s = M di p dd = M 6t = 0 0 3 6 3 = 0.8V 34
Parallel R-L-C Circuit: Example P aa = V rrr I rrr cos φ I rrr = V rrr Z Z = R + i Z = R + tan φ = /R = 3 4 II φ φ I Z V RR I rrr = V rrr Z = V rrr R + cos φ = = 0 + ttt φ = 4 5 0.5 + 6 3 = 50 A P aa = V rrr I rrr cos φ = 0 50 4 5 = 00W 35
I Parallel Resonance in the R-L-C circuits Z = R + iii + ωc i = R i + i = X L X C R + i Z = R + ωc ωl At the resonance frequency ω 0 = LL Z is at its minimum ω 0 L is a short ω C is a short I = V Z = V R + min at = or ω 0 = LL Note that at ω =ω 0, I C and I L can be greater than I. R = Ω, C = F, L = H, and V = V 36