Q.Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.

Concept: Capacitive Reactance — the opposition a capacitor offers to AC, given by $X_C = \frac{1}{\omega C}$.

Reasoning:

1. Capacitive reactance is defined as $X_C = \frac{1}{2\pi f C}$, where $f$ is frequency and $C$ is capacitance.
2. A capacitor charges and discharges in each AC cycle. At higher $f$, the voltage changes faster, so the capacitor has less time to build up opposing charge. …

The Core Idea: Why Capacitors Oppose AC

A capacitor doesn't "resist" current the way a resistor does. Instead, it opposes changes in voltage. When you apply an AC voltage, the capacitor charges and discharges continuously. The key is time: at low frequencies, the voltage changes slowly, so the capacitor has plenty of time to charge up fully, building a large opposing voltage that limits current. At high frequencies, the voltage reverses so quickly that the capacitor barely charges at all — it never builds up much opposing voltage, so current flows almost freely.

This opposition is called capacitive reactance ($X_C$), and it's not a constant — it depends on frequency.

Step-by-Step Explanation

1. Start with the fundamental relationship for a capacitor

The current through a capacitor is proportional to the rate of change of voltage across it:

This is the physical law. It tells us: the faster the voltage changes, the larger the current. For an AC voltage $v(t) = V_0 \sin(\omega t)$, we get:

Notice the current amplitude is $I_0 = \omega C V_0$.

2. Define reactance as the ratio of voltage amplitude to current amplitude

Just as resistance is $R = V/I$ for DC, reactance is defined as:

This is the mathematical origin of the formula. The $\omega$ in the denominator comes directly from the derivative — the rate of change of voltage.

3. Interpret what happens as frequency increases

When $f$ increases, $\omega = 2\pi f$ increases. Since $X_C = 1/(\omega C)$, the reactance decreases. Let's see why physically:

• At low $f$ (say 50 Hz): The voltage changes slowly. The capacitor charges nearly to the peak voltage before the polarity reverses. A large opposing voltage builds up, so little current flows — high reactance.
• At high $f$ (say 10 kHz): The voltage reverses so fast that the capacitor only charges a tiny amount before being forced to discharge. The opposing voltage never gets large, so current flows easily — low reactance.

4. The limiting cases confirm the trend

• DC ($f = 0$): $X_C \to \infty$. A capacitor blocks DC completely — it charges up and then no current flows. …

Method: Explaining WHY a Circuit Element's Opposition to AC Depends on Frequency (Capacitive Case)

This method is the reasoning pattern for any "explain why" question about frequency-dependent reactance — here specifically why capacitive reactance falls as frequency rises.

Steps

Step 1: Start from the defining physical relationship for the element, not the reactance formula

For a capacitor, current is related to the RATE OF CHANGE of the voltage across it, not the voltage itself:

This is the physical law to reason from — everything else follows from it.

Step 2: Substitute a sinusoidal voltage and differentiate

With $v(t) = V_0\sin(\omega t)$, differentiating gives $i(t) = \omega C V_0 \cos(\omega t)$ — notice the current AMPLITUDE is proportional to $\omega$ (and hence to frequency): $I_0 = \omega C V_0$.

Step 3: Define reactance as the voltage-to-current amplitude ratio, and see how it depends on $\omega$

Since $I_0$ grows with $\omega$ (Step 2) while $V_0$ is fixed, the ratio $X_C$ must shrink as $\omega$ grows — this is the mathematical origin of the inverse relationship. …