Question

At very high frequencies, the effective impedance of the given circuit will be ________ \(\Omega\). 

Hint:

High frequency: Inductor = Cut wire, Capacitor = Plain wire.
Low frequency (DC): Inductor = Plain wire, Capacitor = Cut wire.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
The reactance of a capacitor is \( X_C = \frac{1}{\omega C} \) and the reactance of an inductor is \( X_L = \omega L \).
At very high frequencies (\( \omega \to \infty \)):
1. \( X_C \to 0 \Omega \) (Capacitor acts as a short circuit/wire).
2. \( X_L \to \infty \Omega \) (Inductor acts as an open circuit/broken wire).
Step 2: Detailed Explanation:
Let's analyze the branches:
- Top-left branch: \( 1 \Omega \) resistor in series with a capacitor. Capacitor becomes a short. Impedance = \( 1 \Omega \).
- Bottom-left branch: \( 1 \Omega \) resistor in series with a 20 H inductor. Inductor becomes an open circuit. This branch is disconnected.
- Mid horizontal branch: Capacitor becomes a short. This connects the left and right sections.
- Top-right branch: \( 2 \Omega \) resistor with a capacitor in series. Capacitor is a short. Impedance = \( 2 \Omega \).
- Bottom-right branch: \( 2 \Omega \) resistor with a capacitor in series. Capacitor is a short. Impedance = \( 2 \Omega \).
The right section consists of two \( 2 \Omega \) resistors in parallel (due to the shorts). Equivalent right section = \( \frac{2 \times 2}{2 + 2} = 1 \Omega \).
Total impedance = Left section (\( 1 \Omega \)) + Right section (\( 1 \Omega \)) = \( 2 \Omega \).
Step 4: Final Answer:
The effective impedance is 2 \(\Omega\).