Physics 11b ecture #15 and ircuits A ircuits S&J hapter 3 & 33
Administravia Midterm # is Thursday If you can t take midterm, you MUST let us (me, arol and Shaun) know in writing before Wednesday noon Email will do opy to all three of us If you don t, you will not be given make-up exam even if you have a good reason If you can t take midterm due to emergencies, such as an illness, you must let us know as soon as possible You need to supply supporting documents (e.g. doctor s letter)
What We Did ast Time Inductance Mutual inductance Self inductance E For air-core solenoid Inductors in a circuit: di = M dt 1 E E1 di = dt µ = or N A = di = M dt Φ I B di V Vb Va dt = = V a V b circuit has a time constant τ = I Energy in an inductor U =, density u B I U B = = A µ
Today s Goals and circuits Oscillation in electricity Analogous to pendulum and mass-spring oscillator Introduce A circuits A voltage source Maximum and r.m.s. values of V and I Power dissipation in a resistor
ircuit An inductor and a capacitor are connected in series has initially a charge Q Switch closes at t = Step 1: define current I Also define charge Q Note Q and I are related: dq I dt =+ Pay attention to the sign! Step : apply Kirchhoff s loop rule +Q -Q I Q di dt = Q d Q + = dt dq dt = Q Have we seen this differential equation before?
Harmonic Oscillator Mass m is placed on a friction-free floor Spring pulls/pushes the mass m with force F = kx Hooke s aw m Newton s law: d x d x k F = m = dt x dt m Simple Harmonic Oscillation xt () = Acos( ωt+ φ) F F x ω = k m angular frequency x
ircuit An circuit is a simple harmonic oscillator et s assume the solution is Qt () = Acos( ωt+ φ) Plug into the differential equation dq dt Q = +Q -Q I HS = Aω cos( ωt+ φ) A HS = cos( ωt + φ) ω = 1 What about A and φ?
Initial onditions At t =, Q = Q and I = Starting from Qt () = Acos( ωt+ φ) Q() = Acosφ = Q dq I() = = Aωsin φ = dt t= +Q -Q I Final solution: with urrent is A = Q φ = Qt () = Qcosωt It () = Qω sinωt ω = 1
harge and urrent Q and I oscillates back and forth Their phases are off by 9 degrees Q lags behind I Period of oscillation π T = = π ω Frequency of oscillation 1 1 f = = T π Q Q Qω Qω Q= Q cosωt I = Qω sinωt ωt ωt
Why Oscillate? +Q discharges I = by flowing Q current thru I = Q ω harge gone, but keeps current flowing harge gone, but keeps current flowing I = +Q ω discharges by flowing current thru Q +Q I =
Mechanical Analogue
Energy U U Energy is exchanged between and Q = Q ( ω) Q 1 Q = = = At any moment Q I U = + Q cos ωt Qω sin ωt = + Q = U U +Q Q = I = U = Q U = Q I = Q ω U = Q = I = U = +Q I
Energy Energy is exchanged between and, but the total remains constant The situation is identical to the mass-spring oscillator Q Q U U m Q U U ωt
ircuit Ideal circuit oscillates forever eal ones don t Because there is always resistance Oscillation dies away et s analyze an circuit Step 1: define current I dq I = and harge Q dt Step : Kirchhoff di Q I dt = +Q I -Q dq + dq + Q = dt dt This equation may be a little too hard to solve Use solution given in textbook hapter 15
ircuit Solution Solution to dq + dq + Q = is dt dt t () = cos Qt Qe ω t exponential damping Angular frequency of oscillation is d oscillation +Q -Q I ω d 1 = Note for small Back to the circuit t e 1 ω d 1
Weakly-Damped ircuit Solution looks like Shrinking oscillation There is one condition: Q Qe t 1 1 ωd = Inside the square root must be positive > t critical resisance = Q i.e. the resistor must be small < = c 4 Qt () Qe cosω t d
Strong Damping If = c or larger, oscillation cannot happen Q ritically-damped Strongly-damped Weakly-damped ωt Solutions become exponential Time constant is shortest if critically damped, i.e., = c Q
Alternate-urrent ircuit A circuit is driven by an A voltage source ecall A generator in ecture #13 v V is the voltage amplitude ω is the angular frequency v= V ωt sin ω = π f = π T If we connect a resistor (e.g. a light bulb) to it i v V = = sinω = sinω i t I t This is trivial
Which Voltage? An A source is fully specified by V and ω ω may be given as, e.g. f = 6Hz ω = 377 rad/s Amplitude is often given by the root-mean-square ( v) = ( V ) sin ωt time average square root ( V ) V Vrms = = This is the A voltage we usually use Wall-outlet power is 11~1V r.m.s. V = V = 1V 17V rms ( V )
Power Dissipation et s consider the power now With connected to the A source v= V sinωt i = I sin ωt Power consumed by the resistor is Average over time (sin ωt 1/) You can also write V = I P = v i = V I ωt sin i VI V I Pave = = Vrms Irms Vrms = I rms = P ( V ) rms ave = = I rms
Summary circuit is a simple harmonic oscillator Qt () = Qcosωt ~ mass, ~ spring 1 ω = Energy passed between and circuit is a damped oscillator +Q -Q I t () = cos Qt Qe ω t Oscillates if A source < = c d 4 ω d 1 = v v= V ωt sin V = rms V