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0, r1 .. r1 n VC 1.01 1 UE 0 f( x ) 0 C 1.01 1 0 1.01 1 0 2 VL 4 x L t 6 6.28 LC Circuits 0 f( x ) 0 1.01 1 0 0 2 tx 4 6 6.28 UB t Today... • Oscillating voltage and current • Transformers • Qualitative descriptions: • LC circuits (ideal inductor) • LC circuits (L with finite R) • Quantitative descriptions: • LC circuits (ideal inductor) • Frequency of oscillations • Energy conservation Oscillating Current and Voltage Q. What does eosinwt ~ mean?? A. It is an A.C. voltage source. Output voltage appears at the terminals and is sinusoidal in time with an angular frequency w. Oscillating circuits have both AC voltage and current. I(t) Simple for resistors, but... eosinwt ~ R εo I(t) sin ωt R Transformers • AC voltages can be stepped up or stepped down by the use of transformers. • The AC current in the primary circuit creates a time-varying magnetic field in the iron e • This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils. iron ~ V1 V2 N1 (primary) N2 (secondary) • The iron is used to maximize the mutual inductance. We assume that the entire flux produced by each turn of the primary is trapped in the iron. (Recall from magnetism lab how the ferromagnet “sucks in” the B-field.) Ideal No resistance losses Transformers (no load) All flux contained in iron Nothing connected on secondary • The primary circuit is just an AC voltage source in series with an inductor. The change in flux produced in each turn is given by: iron e ~ dturn V1 dt N1 • The change in flux per turn in the secondary coil is the same as the change in flux per turn in the primary coil (ideal case). The induced voltage appearing across the secondary coil is given by: V2 N 2 V1 V2 N1 N2 (primary) (secondary) dturn N 2 V1 dt N1 • Therefore, • N2 > N1 secondary V2 is larger than primary V1 (step-up) • N1 > N2 secondary V2 is smaller than primary V1 (step-down) • Note: “no load” means no current in secondary. The primary current, termed “the magnetizing current” is small! Ideal Transformer (with a load) • Changing flux produced by primary coil induces an emf in secondary. When we connect a resistive load to secondary coil, e ~ emf in secondary current I2 in secondary Power is dissipated only in the load resistor R. 2 Pdissipated iron V1 V2 N1 V2 2 I2 R V2 I 2 R (primary) N2 (secondary) Where did this power come from? It could come only from the voltage source in the primary: Pgenerated V1I1 I1 V2 I 2 V1 = N2 N1 V1 V1 N2 N1 I1 V1I1 V2 I 2 N2 I2 N1 1 R Lecture 18, ACT 1 • The primary coil of an ideal transformer is connected to an AC voltage source as shown. e ~ There are 50 turns in the primary and 200 turns in the secondary. 1A – If V1 = 120 V, what is the potential drop across the resistor R ? iron V1 V2 N1 (primary) (a) 30 V 1B (b) 120 V N2 (secondary) (c) 480 V – If 960 W are dissipated in the resistor R, what is the current in the primary ? (a) 8 A (b) 16 A (c) 32 A R Lecture 18, ACT 1 • The primary coil of an ideal transformer is connected to an AC voltage source as shown. e ~ There are 50 turns in the primary and 200 turns in the secondary. 1A – If V1 = 120 V, what is the potential drop across the resistor R ? iron V1 V2 N1 (primary) (a) 30 V (b) 120 V N2 (secondary) (c) 480 V The ratio of turns, (N2/N1) = (200/50) = 4 The ratio of secondary voltage to primary voltage is equal to the ratio of turns, (V2/V1) = (N2/N1) Therefore, V2= 480 V R iron Lecture 18, ACT 1 • The primary coil of an ideal transformer is connected to an AC voltage source as shown.e ~ There are 50 turns in the primary and 200 turns in the secondary. – If V1 = 120 V, what is the potential drop across the resistor R ? 1A (a) 30 V (b) 120 V V1 V2 N1 (primary) N2 (secondary) (c) 480 V The ratio of turns, (N2/N1) = (200/50) = 4 The ratio (V2/V1) = (N2/N1). Therefore, V2= 480 V 1B – If 960 W are dissipated in the resistor R, what is the current in the primary ? (a) 8 A (b) 16 A (c) 32 A Energy is conserved—960 W should be produced in the primary P1 = V1 I1 implies that 960W / 120V = 8 A R Preflight 18: An ideal transformer has N1 = 4, N2 = 6. Side 1 is connected to a generator with e = Vmax sin ( w t ) 2) What is the maximum EMF on side 2? a) V2max = 2e1/3 b) V2max = e1 c) V2max = 3e1/2 1 2 An ideal transformer steps down the voltage in the secondary circuit. The number of loops on each side is unknown. 4) Compare the currents in the primary and secondary circuits a) I1 < I2 b) I1 = I2 c) I1 > I2 Where are we going? • Oscillating circuits • radio, TV, cell phone, ultrasound, clocks, computers, GPS What’s Next? • Why and how do oscillations occur in circuits containing capacitors and inductors? • naturally occurring, not driven for now • stored energy • capacitive <-> inductive Energy in the Electric and Magnetic Fields Energy stored in a capacitor ... U 1 CV 2 2 … energy density ... Energy stored in an inductor …. +++ +++ --- --- 1 uelectric e 0 E 2 2 B 1 2 U LI 2 … energy density ... E umagnetic 1 B2 2 0 LC Circuits • Consider the RC and LC series circuits shown: ++++ ---- C R ++++ ---- C L • Suppose that the circuits are formed at t=0 with the capacitor charged to value Q. There is a qualitative difference in the time development of the currents produced in these two cases. Why?? • Consider from point of view of energy! • In the RC circuit, any current developed will cause energy to be dissipated in the resistor. • In the LC circuit, there is NO mechanism for energy dissipation; energy can be stored both in the capacitor and the inductor! RC/LC Circuits I Q+++ --- I Q+++ --- C R I I 0 0 t L LC: current oscillates RC: current decays exponentially 0 C 0 t LC Oscillations (qualitative) I 0 + + - - C I I0 L C Q0 Q Q0 I 0 I I0 C Q0 L L - - + + C L Q Q0 Alternate way to draw: C VC = Q/C V=0 L VL = L dI/dt VC+VL = 0 VC = -VL r1 0, .. r1 LC Oscillations n 1.01 1 r1 x 0 , .. r1 (qualitative) n Q 0 VC f( x ) 0 1.01 1 1.01 1 1.01 1 0 0 f( x ) 0 0 2 4 x I 6 6.28 1.01 1 0 1.01 1 0 0 f( x ) 0 2 VL 1.01 1 1.01 1 0 0 dI dt 2 4 x 0 f( x ) 0 These voltages are f( x ) 0 0 opposite, since the 6.28 cap and ind are traversed in “opposite” 1.01 1 directions 0 1.01 1 0 2 tx 4 x 6 6.28 6 0 0 4 6 6.28 2 tx 4 6 6.28 2 Lecture 18, Act 2 t=0 • At t=0, the capacitor in the LC circuit shown has a total charge + + Q0. At t = t1, the capacitor is = Q0 Q uncharged. - C – What is the value of Vab=Vb-Va, 2A the voltage across the inductor at time t1? (a) Vab < 0 (b) Vab = 0 t=t 1 a L Q =0 C L b (c) Vab > 0 2B – What is the relation between UL1, the energy stored in the inductor at t=t1, and UC1, the energy stored in the capacitor at t=t1? (a) UL1 < UC1 (b) UL1 = UC1 (c) UL1 > UC1 Lecture 18, Act 2 t=0 • At t=0, the capacitor in the LC circuit shown has a total charge + + Q0. At t = t1, the capacitor is = Q0 Q uncharged. - C – What is the value of Vab=Vb-Va, 2A the voltage across the inductor at time t1? (a) Vab < 0 (b) Vab = 0 t=t 1 a L Q =0 C L b (c) Vab > 0 • Vab is the voltage across the inductor, but it is also (minus) the voltage across the capacitor! • Since the charge on the capacitor is zero, the voltage across the capacitor is zero! Lecture 18, Actt=02 • At t=0, the capacitor in the LC circuit shown has a total charge + + Q0. At t = t1, the capacitor is = Q0 Q uncharged. - C – What is the relation between UL1, 2B the energy stored in the inductor at t=t1, and UC1, the energy stored in the capacitor at t=t1? (a) UL1 < UC1 (b) UL1 = UC1 • At t=t1, the charge on the capacitor is zero. Q12 U C1 0 2C t=t 1 a L Q =0 C L b (c) UL1 > UC1 • At t=t1, the current is a maximum. 1 2 Q02 U L1 LI1 0 2 2C Preflight 18: At time t = 0 the capacitor is fully charged with Qmax, and the current through the circuit is 0. 2) What is the potential difference across the inductor at t = 0? a) VL = 0 b) VL = Qmax/C c) VL = Qmax/2C 3) What is the potential difference across the inductor when the current is maximum? a) VL = 0 b) VL = Qmax/C c) VL = Qmax/2C LC Oscillations (L with finite R) • If L has finite R, then 0, will be damped. r1– energy will be dissipated in R r1the oscillations r1 10 n 100 .. r1 x 0 , .. r1 n n R=0 R0 1 1 Q Q f( x ) 00 f( x ) 00 1 0 t 5 x 1 10 0 t5 10 x • The number of oscillations is described by the “Q” of the oscillator (we will return to this in Lect. 20) [NOTE: Q here is not charge!] Q 2 U max DU Umax is max energy stored in the system DU is the energy dissipated in one cycle Review of Voltage Drops Across Circuit Elements I(t) C Q Id t V C C Voltage determined by integral of current and capacitance I(t) 2 L dI d Q V L L 2 dt dt Voltage determined by derivative of current and inductance LC Oscillations (quantitative, but only for R=0) I • What is the oscillation frequency ω0? • Begin with the loop rule: Q d 2Q Q L 2 0 dt C + + - - C L • Guess solution: (just harmonic oscillator!) Q Q0 cos(w0t ) where remember: , Q0 determined from initial conditions d2x m 2 kx 0 dt • Procedure: differentiate above form for Q and substitute into loop equation to find w0. 1 • Note: Dimensional analysis w0 LC LC Oscillations (quantitative) • General solution: Q Q0 cos(w0t ) + + - - C L • Differentiate: dQ w 0Q0 sin( w 0t ) dt d 2Q Q L 2 0 dt C d 2Q 2 w 0 Q0 cos(w 0t ) 2 dt • Substitute into loop eqn: 1 1 2 w L 0 L w Q0 cos(w 0t ) Q0 cos(w 0t ) 0 0 C C 2 0 Therefore, 1 w0 LC which we could have determined from the mass on a spring result: k 1/ C 1 w0 m L LC 3 Lecture 18, Act 3 t=0 • At t=0 the capacitor has charge Q0; the resulting oscillations have frequency w0. The maximum current in the circuit during these oscillations has value I0. 3A – What is the relation between w0 and w2, the frequency of oscillations when the initial charge = 2Q0? (a) w2 = 1/2 w0 (b) w2 = w0 + + Q Q0 - - C L (c) w2 = 2w0 3B – What is the relation between I0 and I2, the maximum current in the circuit when the initial charge = 2Q0? (a) I2 = I0 (b) I2 = 2I0 (c) I2 = 4I0 Lecture 18, Act 3 t=0 • At t=0 the capacitor has charge Q0; the resulting oscillations have frequency w0. The maximum current in the circuit during these oscillations has value I0. 3A – What is the relation between w0 and w2, the frequency of oscillations when the initial charge = 2Q0? (a) w2 = 1/2 w0 (b) w2 = w0 + + Q Q0 - - C L (c) w2 = 2w0 • Q0 determines the amplitude of the oscillations (initial condition) • The frequency of the oscillations is determined by the circuit parameters (L, C), just as the frequency of oscillations of a mass on a spring was determined by the physical parameters (k, m)! Lecture 18, Act 3 • At t=0 the capacitor has charge Q0; the resulting oscillations have frequency w0. The maximum current in the circuit during these oscillations has value I0. 3B – What is the relation between I0 and I2, the maximum current in the circuit when the initial charge = 2Q0? (a) I2 = I0 (b) I2 = 2I0 t=0 + + Q Q0 - - C L (c) I2 = 4I0 • The initial charge determines the total energy in the circuit: U0 = Q02/2C • The maximum current occurs when Q=0! • At this time, all the energy is in the inductor: U = 1/2 LIo2 • Therefore, doubling the initial charge quadruples the total energy. • To quadruple the total energy, the max current must double! Preflight 18: The current in a LC circuit is a sinusoidal oscillation, with frequency ω. 5) If the inductance of the circuit is increased, what will happen to the frequency ω? a) increase b) decrease c) doesn’t change 6) If the capacitance of the circuit is increased, what will happen to the frequency? a) increase b) decrease c) doesn’t change LC Oscillations Energy Check • Oscillation frequency w 0 1 has been found from the LC loop equation. • The other unknowns ( Q0, ) are found from the initial conditions. E.g., in our original example we assumed initial values for the charge (Qi) and current (0). For these values: Q0 = Qi, = 0. • Question: Does this solution conserve energy? 1 Q 2 (t ) 1 2 U E (t ) Q0 cos 2 (w 0t ) 2 C 2C 1 1 U B (t ) Li 2 (t ) Lw 02Q02 sin 2 (w 0t ) 2 2 r1 Energyx Check 0 , .. r1 n Energy in Capacitor U E (t ) 1 2 Q0 cos 2 (w 0t ) 2C UE 1 f( x ) 0.5 Energy in Inductor LC U B (t ) 1 U B (t ) Lw 02Q02 sin 2 (w 0t ) x 04, r1 ..00r1 0 2 n 1 1 w0 1 2 Q0 sin 2 (w 0t ) 2C Therefore, Q02 U E (t ) U B (t ) 2C 2 t 4 6 4 6 x UB f( x ) 0.5 00 0 2 t Lecture 18, Act 4 I • At t=0 the current flowing through the circuit is 1/2 of its maximum value. – Which of the following plots best represents UB, the energy stored in the inductor as a 4 function of time? (a) x 0, r1 x (b) .. r1 n 0, r1 n .. r1 1 1 1 UBf( x) 0.5 Uf(Bx ) 0.5 UBf( x) 0.5 00 0 0 2 4 x time 6 00 0 0 2 4 x time 6 Q 00 00 + + - - C L (c) 2 4 x time 6 I Lecture 18, Act 4 • At t=0 the current flowing through the circuit is 1/2 + + of its maximum value. Q - – Which of the following plots best represents UB, the energy stored in the inductor as a function of 4 time? r1 (a) x 0, r1 x (b) .. r1 n 0, n .. r1 1 1 1 UBf( x) 0.5 Uf(Bx ) 0.5 UBf( x) 0.5 00 0 0 2 4 x time 6 00 0 0 2 4 x time 6 00 00 C L (c) 2 4 6 x time • The key here is to realize that the energy stored in the inductor is proportional to the CURRENT SQUARED. • Therefore, if the current at t=0 is 1/2 its maximum value, the energy stored in the inductor will be 1/4 of its maximum value!! Summary • Transformers used to step up/down voltage V2 N2 V1 N1 I1V1 I 2V2 I1 N2 I2 N1 • Oscillating voltage and current x 0, •.. r1 Qualitative description VC V r1 n 1.01 1 1.01 1 L 0 0 f( x ) 0 f( x ) 0 1.01 1 0 • 0 2 4 x 6 6.28 1.01 1 0 2 4 Quantitative description 0 x 6 6.28 – Frequency of oscillations – Energy conservation w0 1 LC Appendix: LCR Damping r1 r1 10 n 100 For your interest, we do not derive but only illustrate the x 0 , here, .. r1 n following behavior 1 R + + - - C L 0 f( x ) 0 Q Q0e t cos(w 'o t ) R R0 Q R 2L t 1 0 r1circuit, r1depends 10 5 nalso 100on10R ! In anx LRC w 0 , .. r1 x n 1 Q 1 R w 'o 2 LC 4 L 2 R R0 4 f( x ) 0 0 1 0 t5 10