Electromagnetism - High and Low-Pass Filters

The circuits below consist of two-element combinations of capacitors, diodes and resistors. V_in represents an ac-voltage with variable frequency. It is desired to build a circuit for which V_out ≈ V_in at high frequencies and V_out ≈ 0 at low frequencies. Which of the following circuits will perform this task?

(GR9677 #45)

Solution:

High-pass filter: V_out ≈ V_in at high frequencies, (ω → ∞)
Low-pass filter: V_out ≈ 0 at low frequencies (ω → 0)

Diode is a device to allow current flow in one direction only. A combination of Diode and Resistor will not affect the output frequency.
→ (B), (C) are FALSE

(A), (D), (E) circuits are combination of resistor and capacitor.

Capacitive reactance, X_C  = 1 / ωC
X_C → 0 for ω → ∞
X_C → ∞ for ω → 0

(A) FALSE
It is parallel combination of capacitor and resistor, voltage is uniform throughout the circuit.

(D) FALSE
It is Low-pass filter only.
Using Voltage Divider for series circuit: V_out = X_C / (R + X_C) V_in

X_C → 0 for ω → ∞

V_out = 0 / (R + 0) V_in ≈ 0

(E) TRUE
It is High and Low-pass filter.
Using Voltage Divider: V_out = R / (R + X_C) V_in

X_C → 0 for ω → ∞

V_out = R / (R + 0) V_in 
V_out ≈ V_in

X_C → ∞ for ω → 0
V_out = R / (R + ∞) V_in 
V_out ≈ 0

Answer: E

Notes: for similar problem see GR0177 #39

Electromagnetism - High-Pass Filters

Which two of the following circuits are high-pass filters

A. I and II
B. I and III
C. I and IV
D. II and III
E. II and IV

(GR0177 #39)

Solution:

High-pass filter: V_out ≈ V_in  for frequency, ω → ∞

Inductive reactance, X_L = ωL
Capacitive reactance, X_C  = 1 / ωC

For ω → ∞
X_L → ∞
X_C → 0

Voltage divider:


Case I.

→ Low-pass filter


Case II.

Note: R ≪ ∞
→ High-pass filter


Case III.

→ High-pass filter


Case IV.

→ Low-pass filter

Answer: D