Showing posts with label RC Circuit. Show all posts
Showing posts with label RC Circuit. Show all posts

Electromagnetism - Capacitor





The capacitor shown in Figure 1 above is charged by connecting switch S to contact a. If switch S is thrown to contact b at time t = 0, which of the curve in Figure 2 above represents the magnitude of the current through the resistor R as a function of time?

A. A
B. B
C. C
D. D
E. E
(GR9677 #01)
Solution:

The capacitor is charged by connecting switch S to contact a, so the current after connecting switch to contact b (t = 0) must start with I = V/R.

Only plot A and B are right.

The current can not be the same all the time and since the voltage of a capacitor follows an exponential decay:

V(t) = V0e−t/RC 
 I(t) = V(t)/RV0e−t/RC/ R

Only plot B is right.

Answer: B

Electromagnetism - High and Low-Pass Filters

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



(GR9677 #45)
Solution:

High-pass filter: Vout ≈ Vin at high frequencies, (ω → ∞)
Low-pass filter: Vout ≈ 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, XC  = 1 / ωC
XC → 0 for ω → ∞
XC → ∞ 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: Vout X/ (R + XCVin

XC → 0 for ω → ∞
Vout = 0 / (R + 0) Vin ≈ 0

(E) TRUE
It is High and Low-pass filter.
Using Voltage Divider: Vout R / (R + XCVin

XC → 0 for ω → ∞
Vout R / (R + 0) Vin 
Vout ≈ Vin

XC → ∞ for ω → 0
Vout R / (R + ∞) Vin 
Vout ≈ 0

Answer: E

Notes: for similar problem see GR0177 #39

Electromagnetism - Oscilloscope


The circuit shown above is used to measure the size of the capacitance C. The y-coordinate of the spot on the oscilloscope screen is proportional to the potential difference across R, and the x-coordinate of the spot is swept at a constant speed s. The switch is closed and then opened. One can then calculate C from the shape and the size of the curve on the screen plus a knowledge of which of the following?

A. V0 and R
B. s and R
C. s and V0
D. R and R'
E. The sensitivity of the oscilloscope
(GR9277 #86)
Solution:

The voltage of a capacitor follows an exponential decay:

V(t) = V0e−t/RC 

C depends on V(t), V0, t, and R.

Vis given and known from the beginning of measurement.

The y-coordinate of the spot on the oscilloscope screen is proportional to the potential difference across R. Thus, V(tis the shape and the size of the curve on the screen.
 
The x-coordinate of the spot is swept at a constant speed s. From this we can calculate time t.

Therefore, we can calculate C from the shape and the size of the curve plus a knowledge of s and R.

Answer: B

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: VoutVin  for frequency, ω → ∞

Inductive reactance, XL = ωL
Capacitive reactance, XC  = 1 / ωC

For ω → ∞
XL → ∞
XC → 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