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)/R = V₀e^−t/RC/ R

Only plot B is right.

Answer: B

Electromagnetism - Capacitor

Two real capacitors of equal capacitance (C₁ = C₂) are shown in the figure. Initially, while the switch S is open, one of the capacitors is uncharged and the other carries charge Q₀. The energy stored in the charged capacitor is U₀. Sometime after the switch is closed, the capacitors C₁ and C₂ carry charges Q₁ and Q₂, respectively; the voltages across the capacitors are V₁ and V₂; and the energies stored in the capacitors are U₁ and U₂.Which of the following statements is INCORRECT?

A. Q₀ = ½ (Q₁ + Q₂)
B. Q₁ = Q₂
C. V₁ = V₂
D. U₁ = U₂
E. U₀ = U₁ + U₂

(GR9677 #25)

Solution:

Initial charge: Q₀
Charge is conserved and distribute evenly
Q₀ = ½ (Q₁ + Q₂)
(A) CORRECT

C = Q/V
For parallel capacitor V = V₁ = V₂
Given C₁ = C₂ → Q₁ = Q₂
(B) and (C) CORRECT

U = ½QV
V₁ = V₂ and Q₁ = Q₂ 
→ U₁ = U₂
(D) CORRECT

U = ½QV → Q = 2U/V
From (A) Q₀ = ½ (Q₁ + Q₂),
2U₀/V₀ = ½ (2U₁/V₁ + 2U₂/V₂)
V₀ = V₁ = V₂ → 2U₀ = U₁ + U₂

(E) INCORRECT

Answer: E

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 - Capacitor

Two capacitors of capacitance 1.0 microfarad and 2.0 microfarad are each charged by being connected across a 5.0-Volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0 microfarad capacitor?

A. 0 V
B. 0.6 V
C. 1.7 V
D. 3.3 V
E. 5.0 V

(GR9677 #62)

Solution:

The charges stored at each capacitor:

Q₁ = C₁ × V_t = 1 × 5 = 5 μF
Q₂ = C₂ × V_t = 2 × 5 = μF

The plates of opposite charge are connected together (series connection):

Q₁'  =  Q₂' = Q_t
1/C_t  = 1/C₁  +  1/C₂ = 1  +  1/2 = 3/2
C_t = 2/3
Since  V_t = 5
Q₂' = Q_t  =  C_t × V_t = 2/3 × 5 = 10/3

→ V₂ = Q₂' / C₂ = 10/3 × 1/2 =  10/6 = 1.66 V

Answer: C

Electromagnetism - Capacitor

The capacitor in the circuit shown above is initially charged. After closing the switch, how much time elapses until one-half of the capacitor’s initial stored energy is dissipated?

A. RC
B. RC/2
C. RC/4
D. RC ln 2
E. ½ RC ln 2

(GR9277 #11)

Solution:

U_t  = ½U₀ , t = ?
Energy of capacitor:  U = ½CV² 

U_t  = ½U₀ → ½CV_t ²  = ½ × ½CV₀2
V_t ²  = ½ V₀2

Potential difference of capacitor: V_t = V₀e^−t/RC
V_t ² = V₀2e^−2t/RC
 ½ V₀2 = V₀2e^−2t/RC → ½  = e^−2t/RC
ln ½ = − 2t/RC →  ln 2 = 2t/RC 
t =  ½ RC ln 2

Answer: E

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.

V₀ is 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(t) is 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 - Capacitor

A parallel-plate capacitor is connected to a battery. V₀ is the potential difference between the plates, Q₀ is the charge on the positive plate, E₀ the magnitude of the electric field, and D₀ the magnitude of the displacement vector. The original vacuum between the plates is filled with a dielectric and then the battery is disconnected. If the corresponding electrical parameters for the final state of the capacitor are denoted by a subscript f, which of the following is true?

A. V_f  > V₀ 
B. V_f  < V₀
C. Q_f  = Q₀
D. E_f  > E₀
E. D_f  > D₀

(GR9277 #88)

Solution:

(A), (B) FALSE
The battery is disconnected after the plates is filled with a dielectric
V_f  = V₀ = constant

(C) FALSE
Charge: Q = CV

V_f  = V₀ and when dielectric is inserted: C_f  = κC₀
Q_f  = C_f V_f  = κC₀V₀ = κQ₀

(D) FALSE
Potential: V = Ed
V_f  = V₀ and d is not changed
E_f  = E₀

(E) TRUE
Displacement vector in empty space: D₀ = ε₀E
with dielectric constant: D_f  = κε₀E
D_f  > D₀

Answer: E

Electromagnetism - Capacitor

A 3-microfarad capacitor is connected in series with 6-microfarad capacitor. When a 300-volt potential difference is applied across this combination, the total energy stored in the two capacitor is

A. 0.09 J
B. 0.18 J
C. 0.27 J
D. 0.41 J
E. 0.81 J

(GR0177 #10)

Solution:

C in Series:

¹/_Cnet  = ¹/_C1  + ¹/_C2 = ( ¹/₃  + ¹/₆ )  ( ¹/₁₀-6 ) = ¹/_2 ∙ 10-6 

C_net = 2 ∙ 10^-6 F 

U = ½ QV = ½ C_netV² = ½ (2 ∙ 10^-6) (300)² = 9 (10^-6) (10⁴) = 0.09 J

Answer: A

Electromagnetism - RLC Circuit

An AC circuit consists of the elements shown above, with R = 10,000 ohms, L = 25 millihenrie and C an adjustable capacitance. The AC voltage generator supplies a signal with amplitude of 40 volts and angular frequency of 1,000 radians per second. For what value of C is the amplitude of the current maximized?

A. 4 nF
B. 40 nF
C. 4 μF
D. 40 μF
E. 400 μF

(GR0177 #38)

Solution:

I_max when X_L = X_C

→ ωL = 1 / ωC

C = 1 / ω²L 
= 1 / [(10³)² × 25 × 10⁻³]
= 1 / (25 × 10³)
= 40 × 10⁻⁶ 
= 40 μF

Answer: D

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

Electromagnetism - LC circuit

For an inductor and capacitor connected in series, the equation describing the motion of charge is

where L is the inductance, C is capacitance, and Q is the charge. An analogous equation can be written for a simple harmonic oscillator with position x, mass m, and spring constant k. Which of the following correctly lists the mechanical analogs of L, C, and Q? 

	L	C	Q
A	m	k	x
B.	m	1/k	x
C.	k	x	m
D.	1/k	1/m	x
E.	x	1/k	1/m

(GR0177 #59)

Solution:

The form of SHO:

LC circuit equation: 

L = m
C = 1/k
Q = x

Answer: B