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Physics 1B Electricity & Magne4sm Frank Wuerthwein (Prof) Edward Ronan (TA) UCSD Some Details on Final Exam • 20 Ques4ons • Will make available a prac4ce ﬁnal late night on Monday, March 19th on website. • Most of the ques4ons for ﬁnal will be taken from: – Homework – Quiz 1,2,3,4 – Prac4ce Final – Examples worked in class Outline of today • Addi4onal material on AC currents • Chapter 24: Electromagne4c waves " " AC Circuits with Ind. Consider a simple circuit containing an inductor and an AC source. Ini4ally the current in hindered by the inductor s induc4on. Even though the initial value of current will be a maximum, the voltage drop will oppose it. When the time rate of change of the current is maximum then the voltage drop across the inductor is a maximum. AC Circuits with Ind. " " " " Also, when the time rate of change of the current is zero then the voltage drop across the inductor is also zero. Giving us the following: As time moves on, we would first see a maximum voltage drop, then a maximum current. The voltage across the inductor, ΔVL, leads the current by 90o. AC Circuits with Ind. " " " " What determines how much current the inductor gets? The inductance, L, of the inductor and the frequency, f, of the AC current. Because of this we define a new variable which basically measures how much the inductor resists current for a given AC source. This variable is called inductive reactance, XL, and is given by: X L = 2"fL = #L " where f is measured in Hz and L is in H, this means that XL will be in Ω. AC Circuits with Ind. " Since inductive reactance measures how an inductor resists current in an AC circuit, we can say that Ohm s Law becomes: "VL,rms = Irms X L " The generator in a purely inductive AC circuit has an angular frequency of 120π rad/s. If Vmax = 140V and L = 0.100H, !what is the rms current in the circuit? Imax "Vmax "Vmax = = XL #L Imax 140V = = 3.71A 120" rad s (0.100H) ( ) Imax 3.71A Irms = = = 2.63A 2 2 RLC Circuit " " " " When we examine the different graphs of the voltage across circuit elements we compare them to the current graph. The instantaneous voltage across resistor is in phase with the current. The instantaneous voltage across the inductor leads the current by 90o. The instantaneous voltage across the capacitor lags the current by 90o. RLC Circuit " " When you want to find the sum of the voltage drops (ΔVmax) across the RLC circuit you simply cannot add the separate maximum voltage values (they are not all in phase). You have to add them as vectors (called phasors). Note that ΔVL and ΔVC are on the same line (180o difference) and that ΔVR is 90o from both of them. (ΔVL – ΔVC) comes from 180o difference. φ RLC Circuit " In order to find ΔVmax, use Pythagorean theorem: 2 R "Vmax = "V + ("VL # "VC ) " " 2 We define the angle between ΔVmax and the current as the phase angle, φ. ! phase angle is given The by: #VL $ #VC ) ( tan " = #VR φ " RLC Circuit " " We can also define an overall RLC circuit resistance to current flow that is based on the resistor, inductor, and capacitor values. We define this term to be impedance, Z, (which will have units of Ω) in terms of R, XL and XC. We can also define the phase angle this way: RLC Circuit " For an RLC circuit, Ohm s Law becomes: "Vmax = Imax Z " " No power losses are associated with pure capacitors and pure inductors in an AC circuit. Thus, ! the average power delivered to the generator is converted to internal energy in the resistor. Pavg = Irms"VR = Irms"Vrms cos # " cos is called the power factor of the circuit. " Phase shifts can be used to maximize power outputs. RLC Circuit " " " " " Resonance occurs where you have the maximum current value for a given circuit. This will maximize the power output as well (need cos to be 1). You need to minimize impedance for this to occur. This occurs when XC=XL. The frequency for this to occur is, fo: RLC Circuits • Example • In a series RLC circuit, suppose R = 300Ω, L = 60mH, C = 0.50μF, Vmax = 50V, and ω = 10,000rad/s. Find the impedance Z of the circuit and the phase angle, ϕ. Answer " First, we should calculate the reactances, XL and XC, and then the impedance. " RLC Circuits " " Answer The reactances are: X L = "L = (10,000 ! " rad The impedance will be given by: 2 Z = R + ( X L " XC ) = (300#) Z = 500" " H) = 600% 1 1 XC = = = 200% $6 "C (10,000 rad sec)(0.50 #10 F) 2 ! )(60 #10 sec $3 To find the phase angle use: 2 + (600# " 200#) 2 " " RLC Circuits Answer This gives us: % 600$ # 200$ ( #1% 400$ ( " = tan ' * = tan ' * & ) & 300$ ) 300$ #1 " = 53° " " " " The ! positive angle means that the circuit is more inductive than capacitive. So, the voltage leads the current. Now, from these pieces of information observable ! can be determined. information For example, what if the question also asked for the current amplitude in the circuit and the voltage amplitude across each circuit element? " " RLC Circuits Answer Since we know the impedance and the maximum voltage we can easily compute: Imax " Vmax 50V = = = 0.10A Z 500" Which means that for the circuit elements: "VR ,max = Imax R = (0.10A)( 300#) = 30V ! "VL,max = Imax X L = (0.10A)(600#) = 60V "VC ,max = Imax X C = (0.10A)(200#) = 20V Chapter 24 Electromagne4c Waves Plane Electromagnetic waves - description E & B fields are the same everywhere within the plane. E-field B-field Wave fronts (infinite & parallel) E & B fields always orthogonal. E & B fields always orthogonal to wave velocity. E & B ﬁelds vs x,t E(x,t) = E sin(kx - ωt) in y-direction B(x,t) = B sin(kx - ωt) in z-direction E & B are in phase and orthogonal in space. Speed of Electromagne4c wave " 1 = =c k #0µ0 Maxwell unified E&M&O by realizing that his equations allow for waves that propagate at the speed of light! Relationship of E and B " E p = B p = cBp k ! The Electromagnetic wave spectrum Polarization The orientation of E in the y-z plane is arbitrary. Polarized EM waves have well defined orientation. However, once E is chosen, the orientation of B is fixed. Polarized light sources - examples (typically) polarized: radio, TV, radar, ... EM from simple antennas. Lasers. (typically) unpolarized: sun light light from light bulbs, or other hot sources (e.g. the glow of fire). Polarizers Law of Malus 2 I = I0 cos " E.g.: Intensity,I, of unpolarized light drops by factor 2 when going through a polarizer because “unpolarized” = random polarization and the average of the square of a cos is Q to ponder: • You have unpolarized light that traverses a sequence of two polarizers whose polariza4on axes form and angle of 60 degrees. • What frac4on of the intensity of the unpolarized light makes it through both polarizers? Answer: • Unpolarized light drops by factor 2 when going through ﬁrst polarizer. • Cos260o = 0.52= 0.25 • The remaining intensity ager both polarizers is 12.5% of the ini4al intensity of the unpolarized light. Q to ponder: • Is it possible to rotate the polariza4on axis of light by 90 degrees ? • If yes, how would one do it? Answer • Yes. • Imagine subjec4ng polarized light with a sequence of polarizers, all of which make a small angle with each other. • Each polarizer reduces the intensity by cos2 • However, each polarizer also forces the light to be polarized along the polarizers axis. • Imagine 900 polarizers, each with 0.1 degree from the previous: cos18000.1o = 99.7% • In contrast: cos290o = 0. • It is thus in principle possible to rotate light with consecu3ve polarizers at small angles, and with only small losses in intensity. For Next Time (FNT) " Start preparing for ﬁnal exam