Physics Problems & Solutions: RL Circuit

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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:
$V_{\text{out}}=\left( \frac{R_2}{R_1+R_2}\right ) V_{\text{in}}$

Case I.
$V_{\text{out}}=\left( \frac{R}{R+X_L}\right) V_{\text{in}}=\left( \frac{R}{R+\infty}\right ) V_{\text{in}}=0$
→ Low-pass filter

Case II.
$V_{\text{out}}=\left( \frac{X_L}{R+X_L}\right) V_{\text{in}}=\left( \frac{\infty}{R+\infty}\right ) V_{\text{in}}=V_{\text{in}}$
Note: R ≪ ∞
→ High-pass filter

Case III.
$V_{\text{out}}=\left( \frac{R}{R+X_C}\right) V_{\text{in}}=\left( \frac{R}{R+0}\right ) V_{\text{in}}=V_{\text{in}}$
→ High-pass filter

Case IV.
$V_{\text{out}}=\left( \frac{X_C}{R+X_C}\right) V_{\text{in}}=\left( \frac{0}{R+0}\right ) V_{\text{in}}=0$
→ Low-pass filter

Answer: D

Electromagnetism - RL Circuit

In the circuit shown above, the switch S is closed at t = 0. Which of the following best represents the voltage across the inductor, as seen on an oscilloscope?

(GR0177 #40)

Solution:

The voltage drops across the resistor and the inductor → (A), (B), (C) are FALSE.

The time constant when the voltage decreases slowly is

$\tau=\frac{L}{R}=\frac{10 \text{ mH}}{2 \Omega }=5\text{ msec}$

Answer: D

Electromagnetism - RL Circuit

In the circuit shown above, R₂ = 3R₁ and the battery of emf ξ has negligible internal resistance. The resistance of the diode when it allows current to pass through it is also negligible. At time t = 0, the switch S is closed and the currents and voltage are allowed to reach their asymptotic values. Then at time t₁, the switch is opened. Which of the following curves most nearly represents the potential at point A as a function of time t?

(GR8677 #94)

Solution:

When the switch is closed, at time t = 0, the voltage drops through the circuit. Thus, C, D and E are FALSE.

The time constant in the first process: $\tau_1=\frac{L}{R_1}$

The time constant, τ is the time it takes the voltage across the component to either decrease or increase.

When the switch is opened, at time t₁, the voltage increases with the time constant: $\tau_2=\frac{L}{R_2}=\frac{L}{3R_1}=\frac{1}{3}\tau_1$

Thus, $\tau_2< \tau_1$

It means, it takes less time for the voltage to increase than decrease.

Answer: B