Problem 2 Part f)
Problem 3
Problem 3 Cont'd
Problem 4
Problem 5
Problem 6
Problem 6 Cont'd
Problem 6 Cont'd
Answer to 7:
The answer to this problem is (D). This is explained on page 10 of Prof.
Greenside's file notes-on-AC-circuits.pdf.
A simpler quick answer is observe that capacitors act like infinite-resistance
device at low frequencies omega and like a zero-resistance wires at high
frequencies. So the potential difference V_0 across the capacitor is large at
low frequencies and tiny at high frequencies and so the curve must be (D).
Answer to 8:
The answer is (D), circuits II and III act like high-pass filters. The quick nonmathematical way to see this is that, at low frequencies of a voltage source,
a capacitor acts like a high-resistance wire (open circuit). So circuit IV would
allow a low-frequency signal through to the output while III would not, and
conversely, since a capacitor acts like a low-resistance wire at highfrequencies, III will allow high-frequency signals through but not IV.
From the formula Z_L = i omega L for the impedance of an inductor, an
inductor at low frequencies as zero impedance and has a large impedance
at high frequencies. So circuit I will allow low-frequency but not highfrequency signals to pass through (making I a low-pass filter) while circuit
II will let high-frequency signals through but not low-frequency. So II and III
are the high-pass filters, while I and IV are low-pass filters.
Problem 9