L C LC Circuits 0 0 t V V C L t t U B U E

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 mean??   o sin  t A.It is an A.C. voltage source. Output voltage appears at the terminals and is sinusoidal in time with an angular frequency .   o sin  t R I(t)  Oscillating circuits have both AC voltage and current. Simple for resistors, but...

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 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.) 2 1 (primary) (secondary)   N N iron V2V2 V1V1 This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils.

Ideal Transformers (no load) 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: Therefore, N 2 > N 1  secondary V 2 is larger than primary V 1 (step-up) N 1 > N 2  secondary V 2 is smaller than primary V 1 (step-down) Note: “no load” means no current in secondary. The primary current, termed “the magnetizing current” is small! 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: No resistance losses All flux contained in iron Nothing connected on secondary   N 2 N 1 (primary) (secondary) iron V 2 V 1

Ideal Transformer (with a load) Power is dissipated only in the load resistor R.   N 2 N 1 (primary) (secondary) iron V 2 V 1 R Where did this power come from? It could come only from the voltage source in the primary: = Changing flux produced by primary coil induces an emf in secondary. When we connect a resistive load to secondary coil, emf in secondary  current I 2 in secondary

Lecture 18, ACT 1 The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary. –If V 1 = 120 V, what is the potential drop across the resistor R ? (a) 30 V (b) 120 V (c) 480 V 1A1B –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   N 2 N 1 (primary) (secondary) iron V 2 V 1 R

Lecture 18, ACT 1 The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary. –If V 1 = 120 V, what is the potential drop across the resistor R ? (a) 30 V (b) 120 V (c) 480 V 1A The ratio of turns, (N 2 /N 1 ) = (200/50) = 4 The ratio of secondary voltage to primary voltage is equal to the ratio of turns, (V 2 /V 1 ) = (N 2 /N 1 ) Therefore, V 2 = 480 V   N 2 N 1 (primary) (secondary) iron V 2 V 1 R

Lecture 18, ACT 1 The primary coil of an ideal transformer is connected to an AC voltage source as shown. There are 50 turns in the primary and 200 turns in the secondary. –If V 1 = 120 V, what is the potential drop across the resistor R ? (a) 30 V (b) 120 V (c) 480 V 1A The ratio of turns, (N 2 /N 1 ) = (200/50) = 4 The ratio (V 2 /V 1 ) = (N 2 /N 1 ). Therefore, V 2 = 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 1B Energy is conserved— 960 W should be produced in the primary P 1 = V 1 I 1 implies that 960W / 120V = 8 A   N 2 N 1 (primary) (secondary) iron V 2 V 1 R

An ideal transformer has N 1 = 4, N 2 = 6. Side 1 is connected to a generator with  = V max sin (  t ) 12 2) What is the maximum EMF on side 2? a) V 2max = 2   /3 b) V 2max =   c) V 2max = 3    Preflight 18:

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)I 1 < I 2 b)I 1 = I 2 c) I 1 > I 2

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 Where are we going? Oscillating circuits radio, TV, cell phone, ultrasound, clocks, computers, GPS

Energy in the Electric and Magnetic Fields … energy density... Energy stored in an inductor …. B Energy stored in a capacitor... … energy density E

LC Circuits Consider the RC and LC series circuits shown: 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! L C C R

RC/LC Circuits RC: current decays exponentially C R 0 t I 0 I Q L C LC: current oscillates I 0 0 t I Q

LC Oscillations (qualitative)    L C L C L C  L C

Alternate way to draw: L C V=0 V C = Q/C V L = L dI/dt V C +V L = 0 V C = -V L

LC Oscillations (qualitative) 0 I Q 0 t 0 dI dt t 0 V C 0 V L These voltages are opposite, since the cap and ind are traversed in “opposite” directions

Lecture 18, Act 2 At t=0, the capacitor in the LC circuit shown has a total charge Q 0. At t = t 1, the capacitor is uncharged. –What is the value of V ab =V b -V a, the voltage across the inductor at time t 1 ? (a) V ab < 0 (b) V ab = 0 (c) V ab > 0 (a) U L1 < U C1 (b) U L1 = U C1 (c) U L1 > U C1 – What is the relation between U L1, the energy stored in the inductor at t=t 1, and U C1, the energy stored in the capacitor at t=t 1 ? 2B 2A L C L C Q = 0 QQ = 0 t=0 t=t 1 a b

Lecture 18, Act 2 At t=0, the capacitor in the LC circuit shown has a total charge Q 0. At t = t 1, the capacitor is uncharged. –What is the value of V ab =V b -V a, the voltage across the inductor at time t 1 ? (a) V ab < 0 (b) V ab = 0 (c) V ab > 0 2A V ab 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! L C L C Q = 0 QQ = 0 t=0 t=t 1 a b

Lecture 18, Act 2 At t=0, the capacitor in the LC circuit shown has a total charge Q 0. At t = t 1, the capacitor is uncharged. (a) U L1 < U C1 (b) U L1 = U C1 (c) U L1 > U C1 2B At t=t 1, the charge on the capacitor is zero. At t=t 1, the current is a maximum. L C L C Q = 0 QQ = 0 t=0 t=t 1 a b –What is the relation between U L1, the energy stored in the inductor at t=t 1, and U C1, the energy stored in the capacitor at t=t 1 ?

At time t = 0 the capacitor is fully charged with Q max, and the current through the circuit is 0. 2) What is the potential difference across the inductor at t = 0? a) V L = 0 b) V L = Q max /C c) V L = Q max /2C 3) What is the potential difference across the inductor when the current is maximum? a) V L = 0 b) V L = Q max /C c) V L = Q max /2C Preflight 18:

LC Oscillations ( L with finite R ) If L has finite R, then –energy will be dissipated in R  the oscillations will be damped. R = 0 Q 0 t R  0 t 0 Q U max is max energy stored in the system  U is the energy dissipated in one cycle 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!]

Review of Voltage Drops Across Circuit Elements Voltage determined by integral of current and capacitance C I(t) Voltage determined by derivative of current and inductance L I(t)

LC Oscillations (quantitative, but only for R =0) Guess solution: (just harmonic oscillator!) where , Q 0 determined from initial conditions Procedure: differentiate above form for Q and substitute into loop equation to find  . Note: Dimensional analysis  L C I Q What is the oscillation frequency ω 0 ? Begin with the loop rule: remember:

LC Oscillations (quantitative) General solution: L C Differentiate: Substitute into loop eqn:  Therefore, which we could have determined from the mass on a spring result:

Lecture 18, Act 3 At t=0 the capacitor has charge Q 0 ; the resulting oscillations have frequency  0. The maximum current in the circuit during these oscillations has value I 0. –What is the relation between  0 and  2, the frequency of oscillations when the initial charge = 2Q 0 ? (a)  2 = 1/2  0 (b)  2 =  0 (c)  2 = 2  0 (a) I  = I  (b) I  = 2 I  (c) I  = 4 I  – What is the relation between I 0 and I 2, the maximum current in the circuit when the initial charge = 2Q 0 ? 3B 3A

Lecture 18, Act 3 At t=0 the capacitor has charge Q 0 ; the resulting oscillations have frequency  0. The maximum current in the circuit during these oscillations has value I 0. –What is the relation between  0 and  2, the frequency of oscillations when the initial charge = 2Q 0 ? (a)  2 = 1/2  0 (b)  2 =  0 (c)  2 = 2  0 3A Q 0 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 Q 0 ; the resulting oscillations have frequency  0. The maximum current in the circuit during these oscillations has value I 0. –What is the relation between I 0 and I 2, the maximum current in the circuit when the initial charge = 2Q 0 ? (a) I 2 = I 0 (b) I 2 = 2I 0 (c) I 2 = 4I 0 3B The initial charge determines the total energy in the circuit: U 0 = Q 0 2 /2C The maximum current occurs when Q=0 ! At this time, all the energy is in the inductor: U = 1/2 LI o 2 Therefore, doubling the initial charge quadruples the total energy. To quadruple the total energy, the max current must double!

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 Preflight 18:

LC Oscillations Energy Check The other unknowns ( Q 0,  ) are found from the initial conditions. E.g., in our original example we assumed initial values for the charge ( Q i ) and current ( 0 ). For these values: Q 0 = Q i,  = 0. Question: Does this solution conserve energy? Oscillation frequency has been found from the loop equation.

U E t 0 Energy Check U B 0 t Energy in Capacitor  Energy in Inductor Therefore, 4

Lecture 18, Act 4 At t=0 the current flowing through the circuit is 1/2 of its maximum value. – Which of the following plots best represents U B, the energy stored in the inductor as a function of time? 4 L C I Q (a) (b) (c) 0 0 UBUB time 0 0 UBUB 0 0 UBUB

Lecture 18, Act 4 At t=0 the current flowing through the circuit is 1/2 of its maximum value. – Which of the following plots best represents U B, the energy stored in the inductor as a function of time? 4 (a) (b) (c) 0 0 UBUB time 0 0 UBUB 0 0 UBUB 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!! L C I Q

Summary Quantitative description –Frequency of oscillations –Energy conservation 0 V C 0 V L Oscillating voltage and current Qualitative description Transformers used to step up/down voltage

Appendix: LCR Damping For your interest, we do not derive here, but only illustrate the following behavior t 0 Q 0 Q t L C In an LRC circuit,  depends also on R !