Physics 9 Friday, April 18, 2014 Turn in HW12. I ll put HW13 online tomorrow. For Monday: read all of Ch33 (optics) For Wednesday: skim Ch34 (wave optics) I ll hand out your take-home practice final exam (due just before Reading Days start, 10% of grade) by Wednesday. If you want to learn more about diodes, LEDs, and p-n junctions, here are two YouTube videos that are ( extremely ) informative, but maybe a little slow: https://www.youtube.com/watch?v=jbteckh3l9q https://www.youtube.com/watch?v=godkgafzsh4 Today: mainly high-pass & low-pass filters. Also, we can try analyzing a multi-battery circuit together for practice.
Quick demos/hand-arounds: Three-way stairway light switch. Household light switch with bulb & battery. Take apart Wednesday s transformer.
What are the two solutions to the equation z 2 + 1 = 0
What are the two solutions to the equation z 2 + 1 = 0 Define i = 1. Then solution to above equation is z = ±i. Complex numbers: z = x + iy, where x, y R. real part of z: Re(z) = x. imaginary part of z: Im(z) = y. (Note: it s y, not iy.) addition: (a + ib) + (c + id) = (a + c) + i(b + d) multiplication: (a + ib) (c + id) = (ac bd) + i(bc + ad) division: a + ib c + id = (a + ib)(c id) (c + id)(c id) = (ac + bd) + i(bc ad) c 2 + d 2
Represent complex numbers as vectors on complex plane. Then adding (4 + 2i) + ( 1 + 3i) = (3 + 5i) is just like adding vectors. (4, 2) + ( 1, 3) = (3, 5)
Sometimes a vector in a plane is most conveniently represented (x,y). Sometimes it s more convenient to use length and angle: x = R cos θ y = R sin θ R = x 2 + y 2 tan θ = y/x Sometimes a complex number is most conveniently written R cos θ + ir sin θ instead of x + iy.
That shocking t-shirt depends on DeMoivre s Theorem: e iθ = cos θ + i sin θ If you take that equation as true, and plug in θ = π, then e iπ = cos π + i sin π = ( 1) + i(0) = 1 To prove this, use taylor series: e θ = 1 + θ + θ2 2 + θ3 6 + θ4 24 + θ5 120 + θ6 720 + θ7 5040 + e iθ = 1 + (iθ) + (iθ)2 2 + (iθ)3 6 then using i 2 = 1, i 4 = 1, etc., you get + (iθ)4 24 + (iθ)5 120 + (iθ)6 720 + (iθ)7 5040 + e iθ = 1 + iθ θ2 2 iθ3 6 + θ4 24 + i θ5 120 θ6 720 i θ7 5040 +
e iθ = 1 + iθ θ2 2 iθ3 6 + θ4 24 + i θ5 120 θ6 720 i θ7 5040 + Meanwhile, we can expand cos θ and sin θ: cos θ = 1 θ2 2 + θ4 24 θ6 720 + sin θ = θ θ3 6 + θ5 120 θ7 5040 + So then cos θ + i sin θ has the same talyor series as e iθ. e iθ = cos θ + i sin θ That means I can write a complex number as x + iy = R cos θ + i sin θ = Re iθ where R = x 2 + y 2 and tan θ = x/y.
How does this relate to AC circuits? Textbook is drawing oscillating quantities as little arrows that turn ccw at frequency f = ω/(2π). At time t, the vertical position of the tip of each arrow represents voltage V (t), current I(t), etc. In this example (inductor), V (t) and I(t) are 90 out of phase. Notice that if you took the I(t) arrow and multiplied it by i = 1, it would line up with the V (t) arrow if you interpret the arrows as lying in the complex plane. The part we are graphing at each instant to get V (t) or I(t) is the vertical (imaginary) component of the arrow.
The part we are graphing at each instant to get V (t) or I(t) is the vertical (imaginary) component of the arrow. The current phasor arrow in the graph is the complex number I(t) = I max e iωt = I max (cos(ωt) + i sin(ωt)) whose vertical (imaginary) component is I(t) = Im(I(t)) = I max sin(ωt) (The imaginary part of a complex number is a real number!)
The part we are graphing at each instant to get V (t) or I(t) is the vertical (imaginary) component of the arrow. The voltage phasor arrow in the graph is the complex number V(t) = V max e iωt+π/2 = iv max e iωt = V max ( sin(ωt) + i cos(ωt)) whose vertical (imaginary) component is V (t) = Im(V(t)) = V max cos(ωt) We got the 90 phase shift between V and I just by multiplying by i = 1. Using complex numbers, V(t) is proportional to I(t), even if there is a phase shift between them.
For a resistor, V = IR. There is no phase shift between voltage and current. The complex phasor arrows are: V(t) = V max e iωt I(t) = (V max /R)e iωt The vertical components (the actual voltage and current) are: V (t) = Im(V(t)) = V max sin(ωt) I(t) = Im(I(t)) = (V max /R) sin(ωt)
For a capacitor, Q = CV, which means I = C dv. The phase of dt the current is 90 ahead of the phase of the voltage. The complex voltage phasor arrow is: V(t) = V max e iωt Taking the derivative to get the current gives: I(t) = C dv(t) = C iω V max e iωt dt So I(t) and V(t) are proportional, in spite of the phase shift: I(t) = (iωc) V(t)
For an inductor, V = L di. The phase of the current is 90 dt behind the phase of the voltage. The complex current phasor arrow is: I(t) = I max e iωt Taking the derivative to get the voltage gives: V(t) = L di(t) = L iω I max e iωt dt So I(t) and V(t) are proportional, in spite of the phase shift: V(t) = (iωl) I(t)
The derivative of e iωt is simple: just multiply by iω. The derivative is proportional to the original. de iωt dt But with sines and cosines: d sin(ωt) dt = iωe iωt = ω cos(ωt) d cos(ωt) = ω sin(ωt) dt That s a pain, because derivative is not proportional to original. So we pretend that voltage and current are complex, writing: V(t) = Ve iωt I(t) = Ie iωt This lets us write V = IZ where Z is called the impedance. This lets us treat resistors, capacitors, and inductors in a unified way.
We pretend that voltage and current are complex, writing: V(t) = Ve iωt I(t) = Ie iωt To get instantaneous V (t) or I(t), take the vertical (imaginary) part of the complex phasor: V (t) = Im(V(t)) I(t) = Im(I(t)) Usually, you care more about the amplitude. To get the amplitude, just take the magnitude of the complex phasor: V max = V(t) I max = I(t) We ll see how this works in a minute. It s easier than you think. (Most people use the real (horizontal) component to get the instantaneous V (t) or I(t), but I m using the vertical (imaginary) component for consistency with the book s phasor diagrams. Also, engineers use j = 1, because they use lowercase i for current.)
Generalized Ohm s law: V = IZ For a resistor, Z = R. V(t) = I(t) R = Z R = R For an inductor, Z = iωl (where i = 1 ) V(t) = L di(t) dt = (iωl) I(t) = Z L = iωl For a capacitor, Z = 1 (iωc) = i/(ωc). I(t) = C dv(t) dt = (iωc) V(t) = Z C = 1/(iωC) Z = R + ix. Real part of impedance is called resistance. Imaginary part is called reactance: X L = ωl. X C = 1/(ωC). Whew! That was the hard part. Now the fun part.
Let s take AC signal source E(t) as our input signal, and V cb (t) (across R 2 ) as our output signal. How does the output amplitude compare with the input amplitude? What happens to V out when R 2 R 1? algebra = V out V in = V cb(t) E(t) = R 2 R 1 + R 2
Now replace one resistor with a capactor. Just by replacing resistance with impedance we can analyze the circuit in exactly the same way as the previous one. Which impedance is smaller at high frequency? What happens to the amplitude of V out (V across the capacitor) as the frequency changes?
Now swap resistor and capactor. Which impedance is smaller at high frequency? What happens to the amplitude of V out (voltage across the resistor) as the frequency changes?
The graphs show V out V in vs. frequency for low-pass filter (left) and high-pass filter (right).
( ) The graphs show log Vout 10 V in vs. log 10 (frequency) for low-pass filter (left) and high-pass filter (right).
Quick way to get total current drawn from the batteries. R 1 + (R 5 (R 2 + (R 3 R 4 )))) = 160 Ω I batt = 3 V 160 Ω
Predict the relative brightness for the three bulbs (assuming the bulbs are identical). (A) A < B < C (B) A < B = C (C) A = B = C (D) A > B = C (E) A > B > C
You just predicted A > B = C when all 3 bulbs are present. Now predict what will happen to the brightness of bulbs A and B if bulb C is unscrewed. (A) A and B will both become brighter. (B) A and B will both become dimmer. (C) A will become brighter, and B will become dimmer. (D) A will become dimmer, and B will become brighter. (E) The brightness of A and B will not change.
If you were to build this circuit, when would bulb A be brighter? (A) A is brighter when the switch is open (B) A is brighter when the switch is closed (C) A is the same brightness in both cases
How does the resistance of bulb B compare with the resistance of a closed switch? (A circuit diagram usually shows a switch in its open position, as this one does.) (A) a closed switch has much smaller resistance than bulb B (B) a closed switch has much larger resistance than bulb B (C) the resistance of a closed switch is similar to the resistance of bulb B
By the way, what is the resistance of an open switch? (Is it very easy or is it very difficult for current to flow through an open switch?) (A) an open switch has a very small resistance, effectively zero (B) an open switch has a very large resistance, effectively infinite
Physics 9 Friday, April 18, 2014 Turn in HW12. I ll put HW13 online tomorrow. For Monday: read all of Ch33 (optics) For Wednesday: skim Ch34 (wave optics) I ll hand out your take-home practice final exam (due just before Reading Days start, 10% of grade) by Wednesday. If you want to learn more about diodes, LEDs, and p-n junctions, here are two YouTube videos that are ( extremely ) informative, but maybe a little slow: https://www.youtube.com/watch?v=jbteckh3l9q https://www.youtube.com/watch?v=godkgafzsh4 If you found today s discussion of filters interesting and mostly understandable (and you get an A or A+ in Phys 009), then you re qualified to take my electronics course (Phys 364), which is entirely lab-based (like a studio). Tu/Th 2 5pm. Filters, amplifiers, transistors, Arduinos, etc.