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

Q: Why do capacitors act as short circuits at high frequencies?

Capacitive reactance is inversely proportional to frequency ($X_C = 1/\omega C$). At very high frequencies, the reactance approaches zero, meaning the capacitor offers almost no opposition to the AC current, acting like a perfect short circuit.

Q: Why does the inductor act as an open circuit?

Inductive reactance is directly proportional to frequency ($X_L = \omega L$). As frequency approaches infinity, the reactance becomes infinitely large. This massive opposition completely blocks the current, acting like a broken wire or open circuit.

Q: Why did we ignore the $1\,\Omega$ resistor in the bottom branch?

Because it is in series with the inductor. When the inductor becomes an open circuit, that entire path is broken. No current can flow through a broken path, so the resistor in that path has no effect on the circuit.

Q: How did the two $2\,\Omega$ resistors end up in parallel?

The vertical capacitor became a short circuit, directly connecting the node before the top $2\,\Omega$ resistor to the node before the bottom $2\,\Omega$ resistor. Since their other ends are also connected together at the right main wire, they share the exact same potential difference, placing them perfectly in parallel.

Q: Does the $220\,\text{V}$ source value affect the impedance?

No. Impedance is an intrinsic property of the circuit components and the frequency of the source. The voltage value only determines the total current that will flow, not the impedance itself.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine an alternating current as a rhythmic vibration of electrons. When we look at the given circuit, we see a complex web of resistors, capacitors, and an inductor connected across an AC voltage source. At first glance, calculating the total impedance of this network seems like a daunting task requiring heavy complex algebra. However, the key to unlocking this problem lies in the phrase **"at very high frequencies."**

### The High-Frequency Limit

To understand how the circuit behaves, we must look at the mathematical definitions of reactance. The opposition offered by a capacitor is called capacitive reactance, given by $X_C = \frac{1}{\omega C}$. Notice that the angular frequency $\omega$ is in the denominator. As the frequency approaches infinity, the reactance $X_C$ approaches zero. Physically, the AC signal is vibrating so fast that charge doesn't have time to build up on the capacitor plates to oppose the flow. Thus, at very high frequencies, **capacitors act as perfect short circuits**.

Conversely, the opposition offered by an inductor is inductive reactance, given by $X_L = \omega L$. Here, $\omega$ is in the numerator. As the frequency approaches infinity, the reactance $X_L$ becomes infinitely large. The inductor, which inherently opposes changes in current, acts like a massive brick wall against the ultra-fast vibrations. Therefore, **inductors act as open circuits**.

### Redrawing the Circuit

Armed with these limiting behaviors, we can dramatically simplify our circuit. We replace every $0.5\,	ext{F}$ capacitor with a solid wire (short circuit) and remove the $20\,	ext{H}$ inductor entirely, leaving a gap (open circuit).

Let's trace the consequences of this transformation:
1.  **The Dead Branch:** In the bottom branch, the inductor has become an open circuit. Because this path is broken, absolutely no current can flow through the left side of the bottom branch. Consequently, the $1\,\Omega$ resistor located there becomes completely redundant and can be ignored.
2.  **The Shorted Nodes:** Look at the vertical branch that originally contained a capacitor. It is now a solid wire connecting the top branch to the bottom branch. Let's call the top connection point Node A and the bottom connection point Node B. Because they are connected by a short circuit, Node A and Node B are at the exact same electrical potential.

### Calculating the Impedance

From this common potential (Nodes A and B), the current has two parallel paths to return to the right side of the AC source:
-   Through the $2\,\Omega$ resistor in the top branch.
-   Through the $2\,\Omega$ resistor in the bottom branch.

Since these two $2\,\Omega$ resistors are connected between the exact same two nodes, they are in parallel. We can find their equivalent resistance $R_p$ using the product-over-sum rule:

$$R_p = \frac{2 	imes 2}{2 + 2} = \frac{4}{4} = 1\,\Omega$$

Finally, all the current from the source must first pass through the $1\,\Omega$ resistor in the top branch before reaching Node A. Therefore, this $1\,\Omega$ resistor is in series with our parallel combination. The total effective impedance $Z_{eq}$ is simply the sum of these resistances:

$$Z_{eq} = 1\,\Omega + 1\,\Omega = 2\,\Omega$$

By understanding the extreme limits of AC components, a complex network collapses into a trivial combination of basic resistors. The final effective impedance is exactly $2\,\Omega$.

Animated Solution for Physics - Alternating Current: At very high frequencies, the effective impedance of the given circuit will be ......... $Ω$ .

Enter Numerical Value:

Visualized Solution

Analyzing the AC Circuit

Capacitive Reactance at High Frequency

Inductive Reactance at High Frequency

High Frequency Equivalent Circuit

Simplifying the Open Circuit

Identifying Parallel Resistors

Calculating Parallel Resistance

Total Effective Impedance

What if Frequency was Zero?

The Sigma Insight: AC Circuits and Power in AC Circuits

Analyzing the Setup

The High-Frequency Limit

Redrawing the Circuit

Calculating the Impedance

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Animated Solution for Physics - Alternating Current: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } At very high frequencies, the effective impedance of the given circuit will be ......... Ω.