**Week 23 Physics, Circuits:**

- Physics reading: Ch 35.
- Focus on 35.1-35.3 (but definitely read the rest, especially 35.5). The main point of 35.4 is that there are things called inductors that have reactance Z
_{L}= iωL (Knight’s “X_{L}” is Horowitz & Hill’s “Z_{L}” – they use different letters for reactance). You can think of inductors as a sort of “inverse capacitor” in terms of their circuit behavior (though not why they work). - Note that Knight doesn’t use imaginary numbers… hopefully you will start to see why they are useful. Otherwise, how can you get results 35.24 and 35.30? (definitely know equation 35.30 – that shows up a lot in circuits).
- Due online Friday, 11:59 pm:
- Ch 31, problems 39, 43, 73, 74
- Ch 35, problems 7, 8, 13, 14, 16, 19, 28, 42, 43
- Notes:
- You may find it easier to do all of these problems using the Horowitz approach (using complex impedances). I would!
- 13, 14: “crossover frequency” is the same as f
_{3db} - 42: watch out; Knight is asking for the delta-V (voltage difference) across the “top” resistor… the V
_{out}you are used to would be ΔV across the bottom two.

- Notes:

- Focus on 35.1-35.3 (but definitely read the rest, especially 35.5). The main point of 35.4 is that there are things called inductors that have reactance Z
- Circuits Reading:
- AOE 2
^{nd}ed, pp. 20-42. - Horowitz and Hayes, Lab Manual, pp. 32 – 50.

- AOE 2
- Due Wednesday, 5pm, box outside my office: AOE Ch 1: 1.13, 1.14 (read the section immediately before the question :), 1.15, 1.16, 1.17, 1.21, 1.22, 1.24
- Notes:
- 1.14: You need to make the assumption that the V
_{in}= 0 until you connect it at t = 0, so at t = 0 V_{out}(V(t) in Fig 1.34) = 0 too. Then You connect V_{in}(t) , which may vary in time… what is V_{out}(t)? Hint: consider the Thevinin equivalent looking into the two resistors, then compare to the ‘Time Constant’ example on p. 34. - 1.21: They are looking for the
*magnitude*of the response here, like they found for the high-pass filter on pp. 35-6. Show this by following that example, adapting it to the low-pass filter geometry (complex voltage divider in Fig 1.58), using the complex impedance for a capacitor. - 1.22 (show by calculating the phase shift (in degrees – you’ll have to convert from radians) at 0.1 f
_{3DB}(for a high-pass filter) and 10 f_{3DB}(for a low-pass filter).

- 1.14: You need to make the assumption that the V

- Notes:

**Calculus**

- Reading: 14.1 – 14.3
- Problems:
**Quiz 3 on Week 3 Material**