The figure shows the variation of capacitive reactance ($X_C$) of two ideal capacitors of capacitances $C_1$ and $C_2$ with the reciprocal of angular frequency $(1/\omega)$ of an ac source. The value of $C_1/C_2$ is (A) $\dfrac{1}{2}$ (B) $2$ (C) $\sqrt{3}$ (D) $\dfrac{1}{\sqrt{3}}$ ![Figure — CBSE 2026 55/2/1 Q9](/figures/pyq/2026-55-2-1/fig-q9.svg)

The figure shows the variation of capacitive reactance (XC) of two ideal capacitors of capacitances C₁ and C₂

Question

X_C

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Capacitive Reactance

Capacitive Reactance: The AC Resistance of a Capacitor

When you first meet a capacitor in a DC circuit, it behaves like a break in the wire once it's fully charged — no current flows. But in an AC circuit, something entirely different happens. The voltage keeps reversing, so the capacitor never finishes charging. It's constantly being filled, emptied, refilled, and re-emptied. This continuous back-and-forth means current does flow, but the capacitor resists that flow in a frequency-dependent way. That resistance is called capacitive reactance.

The Intuition: Why Frequency Matters

Imagine a water pipe with a flexible rubber membrane stretched across it (a crude capacitor). If you push water slowly from one side, the membrane bulges and eventually stops the flow — that's DC. But if you push and pull the water rapidly (AC), the membrane just vibrates, and water sloshes back and forth through the pipe. The faster you push-pull (higher frequency), the less the membrane impedes the flow. At very high frequencies, it's almost like the membrane isn't there.

In a capacitor, the "membrane" is the electric field between the plates. Higher frequency means the voltage changes faster, so the capacitor has less time to oppose the current. The result: capacitive reactance decreases as frequency increases.

The Precise Statement

Capacitive reactance $X_C$ is the opposition a capacitor offers to alternating current. It is measured in ohms ( $\Omega$ ), just like resistance. The formula is:

X_C = \frac{1}{2\pi f C}

Where:

$X_C$ = capacitive reactance (ohms)
$f$ = frequency of the AC signal (hertz)
$C$ = capacitance (farads)

What the Formula Tells You

Three key relationships jump out:

Inverse with frequency: Double the frequency, halve the reactance. At DC ( $f = 0$ ), $X_C$ becomes infinite — the capacitor blocks DC completely.
Inverse with capacitance: A larger capacitor (more farads) offers less opposition. It can store more charge per volt, so it "gives way" more easily.
No power dissipation: Unlike a resistor, a pure capacitor doesn't convert electrical energy to heat. Reactance is a reactive opposition — energy is stored and returned, not lost.

Watch out

Do not confuse capacitive reactance with resistance. Resistance dissipates energy as heat; reactance stores and releases it. A capacitor in an AC circuit has zero real power loss (in the ideal case).

Phase: The Hidden Twist

There's a critical detail that separates reactance from resistance. In a purely resistive circuit, voltage and current peak at the same time — they are in phase. In a purely capacitive circuit, current leads voltage by $90^\circ$ (or $\pi/2$ radians).

Why? Because current is the rate of change of charge: $I = C \frac{dV}{dt}$ . When the voltage is at its peak (not changing), the current is zero. When the voltage is crossing zero (changing fastest), the current is maximum. This quarter-cycle shift is baked into the definition of reactance. …

Why this formula?

Capacitive Reactance: Why $X_C = \frac{1}{\omega C}$ ?

Let’s build the intuition from the ground up — starting with what a capacitor does in a circuit.

1. The Fundamental Behavior of a Capacitor

A capacitor stores charge. The defining equation is:

Q = C V

where:

$Q$ = charge on the plates (in coulombs)
$C$ = capacitance (in farads)
$V$ = voltage across the plates

But in an AC circuit, voltage changes continuously. So charge must also change — meaning current flows.

2. Relating Current to Voltage

Current is the rate of flow of charge:

I = \frac{dQ}{dt}

Substitute $Q = C V$ :

I = \frac{d}{dt}(C V)

If $C$ is constant (which it is for a fixed capacitor):

I = C \frac{dV}{dt}

Key insight: The current through a capacitor is proportional to the rate of change of voltage, not the voltage itself.

3. Applying a Sinusoidal Voltage

In AC circuits, voltage is typically sinusoidal:

V(t) = V_0 \sin(\omega t)

where:

$V_0$ = peak voltage
$\omega = 2\pi f$ = angular frequency (rad/s)

Now find the current:

I(t) = C \frac{d}{dt}[V_0 \sin(\omega t)] = C V_0 \cdot \omega \cos(\omega t)

So:

I(t) = \omega C V_0 \cos(\omega t)

4. The Phase Shift — Why It Matters

Notice:

Voltage: $\sin(\omega t)$
Current: $\cos(\omega t) = \sin(\omega t + 90^\circ)$

Current leads voltage by $90^\circ$ in a pure capacitor. This is the opposite of an inductor (where current lags).

5. Extracting the Reactance

Compare the amplitudes:

Voltage amplitude: $V_0$
Current amplitude: $I_0 = \omega C V_0$

By Ohm’s law for AC (magnitude only):

\text{Reactance} = \frac{\text{Voltage amplitude}}{\text{Current amplitude}} = \frac{V_0}{\omega C V_0} = \frac{1}{\omega C}

Thus:

X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

6. Why "Reactance" and Not "Resistance"?

Resistance ( $R$ ) dissipates energy as heat. …

Start your 14-day free trial to unlock the full solution →

Capacitive reactance $X_C = \frac{1}{\omega C}$ is linear in $\frac{1}{\omega}$ with slope $\frac{1}{C}$ . Reading the slopes from the angles (tan 45° and tan 30°), we find $\frac{C_1}{C_2} = \frac{1}{\sqrt{3}}$ .

The capacitive reactance of an ideal capacitor is given by

X_C = \frac{1}{\omega C}

Rearranging this as $X_C = \frac{1}{C} \cdot \frac{1}{\omega}$ , we see that $X_C$ is directly proportional to $\frac{1}{\omega}$ . When we plot $X_C$ versus $\frac{1}{\omega}$ , we get a straight line passing through the origin with slope equal to $\frac{1}{C}$ .

The key insight: a steeper line means a larger slope, which means a larger value of $\frac{1}{C}$ , which in turn means a smaller capacitance. The graph shows two such lines for capacitors $C_1$ and $C_2$ , making angles of 45° and 30° respectively with the horizontal axis.

Find the slope of line $C_1$ : The line makes an angle of 45° with the $\frac{1}{\omega}$ axis. The slope is

\text{slope}_{C_1} = \tan 45° = 1

Since slope $= \frac{1}{C_1}$ , we have

\frac{1}{C_1} = 1 \implies C_1 \propto 1

Find the slope of line $C_2$ : The line makes an angle of 30° with the $\frac{1}{\omega}$ axis. The slope is

\text{slope}_{C_2} = \tan 30° = \frac{1}{\sqrt{3}}

Since slope $= \frac{1}{C_2}$ , we have …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 55/2/11 markMCQ
$X_C$
›Reveal solutionSolution
Figure — CBSE 2026 55/2/1 Q9

Capacitive reactance $X_C = \frac{1}{\omega C}$ is linear in $\frac{1}{\omega}$ with slope $\frac{1}{C}$ . Reading the slopes from the angles (tan 45° and tan 30°), we find $\frac{C_1}{C_2} = \frac{1}{\sqrt{3}}$ .

The capacitive reactance of an ideal capacitor is given by
$X_C = \frac{1}{\omega C}$
Rearranging this as $X_C = \frac{1}{C} \cdot \frac{1}{\omega}$ , we see that $X_C$ is directly proportional to $\frac{1}{\omega}$ . When we plot $X_C$ versus $\frac{1}{\omega}$ , we get a straight line passing through the origin with slope equal to $\frac{1}{C}$ .

The key insight: a steeper line means a larger slope, which means a larger value of $\frac{1}{C}$ , which in turn means a smaller capacitance. The graph shows two such lines for capacitors $C_1$ and $C_2$ , making angles of 45° and 30° respectively with the horizontal axis.

Find the slope of line $C_1$ : The line makes an angle of 45° with the $\frac{1}{\omega}$ axis. The slope is

$\text{slope}_{C_1} = \tan 45° = 1$
Since slope $= \frac{1}{C_1}$ , we have
$\frac{1}{C_1} = 1 \implies C_1 \propto 1$

Find the slope of line $C_2$ : The line makes an angle of 30° with the $\frac{1}{\omega}$ axis. The slope is

$\text{slope}_{C_2} = \tan 30° = \frac{1}{\sqrt{3}}$
Since slope $= \frac{1}{C_2}$ , we have …
Start your 14-day free trial to see the full solution →
CBSE 2024Set 55/1/11 markMCQ
$C$
›Reveal solutionSolution
Capacitive reactance is $X = \frac{1}{2\pi \nu C}$ . Doubling $C$ and tripling $\nu$ multiplies the denominator by $6$ , so the new reactance becomes $\frac{X}{6}$ . The correct option is (A).

The key to this problem is understanding what capacitive reactance actually means physically. A capacitor in an AC circuit doesn't "resist" current the way a resistor does — instead, it opposes changes in voltage by storing and releasing charge. The faster the voltage changes (higher frequency) or the larger the capacitor (more charge storage per volt), the easier it is for current to flow. That's why reactance $X$ is inversely proportional to both capacitance $C$ and frequency $\nu$ .

Let's work through the change step by step.

Write the standard formula for capacitive reactance. For a capacitor of capacitance $C$ connected to an AC source of frequency $\nu$ , the reactance is:

$X = \frac{1}{2\pi \nu C}$
This is a direct relationship — no tricks, just the definition.

Identify the new values.

The capacitance is doubled: $C' = 2C$

The frequency is tripled: $\nu' = 3\nu$

Substitute these into the formula for the new reactance $X'$ .

$X' = \frac{1}{2\pi \nu' C'} = \frac{1}{2\pi (3\nu)(2C)}$

Simplify the denominator.

$X' = \frac{1}{2\pi \cdot 6 \cdot \nu C} = \frac{1}{6} \cdot \frac{1}{2\pi \nu C}$

Recognize the original reactance in the expression. Since $X = \frac{1}{2\pi \nu C}$ , we have: $X' = \frac{X}{6}$ …
Start your 14-day free trial to see the full solution →
CBSE 2023Set 55/3/11 markMCQ
$Q. An inductor, a capacitor and a resistor are connected in series across an ac source of voltage. If the frequency of the source is decreased gradually, the reactance of :$
(a) both the inductor and the capacitor decreases.
(b) inductor decreases and the capacitor increases.
(c) both the inductor and the capacitor increases.
(d) inductor increases and the capacitor decreases.
›Reveal solutionSolution
The inductive reactance $X_L = 2\pi f L$ is directly proportional to frequency, and the capacitive reactance $X_C = \frac{1}{2\pi f C}$ is inversely proportional to frequency. As frequency decreases, $X_L$ decreases and $X_C$ increases — so option (b) is correct.

The core idea here is simple: reactance is not a fixed property — it depends on frequency. An inductor opposes changes in current, and the faster the current changes (higher frequency), the more it opposes. A capacitor, on the other hand, stores and releases charge; at higher frequencies, it has less time to charge up, so it offers less opposition.

Let’s see exactly how each behaves when frequency is lowered.

Inductive reactance is given by

$X_L = 2\pi f L$
Here $f$ is the frequency and $L$ is the inductance (a constant for a given inductor). Since $X_L$ is directly proportional to $f$ , decreasing $f$ makes $X_L$ smaller. So the inductor’s opposition weakens.

Capacitive reactance is given by

$X_C = \frac{1}{2\pi f C}$
$C$ is the capacitance (constant). Here $X_C$ is inversely proportional to $f$ — as $f$ goes down, $X_C$ goes up. So the capacitor’s opposition strengthens. …
Start your 14-day free trial to see the full solution →
CBSE 2021Set A1 markMCQ
$Q. Capacitive reactance is (A) w/c (B) c/w (C) w . c (D) 1/wc$
›Reveal solutionSolution

Capacitive reactance X_C = 1/(ωC).

In an AC circuit a capacitor opposes the flow of alternating current; this opposition is the capacitive reactance:

$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C},$ …

Start your 14-day free trial to see the full solution →
CBSE 2020Set ANNUAL1 mark
$Q. How does capacitive reactance vary with frequency?$
›Reveal solutionSolution

$X_C = \dfrac{1}{2\pi f C}=\dfrac{1}{\omega C}$ ; capacitive reactance $\propto \dfrac{1}{f}$ — it decreases as the frequency increases.

Concept. A capacitor opposes changes in voltage. In an AC circuit this opposition is the capacitive reactance $X_C$ .

Why this formula. For a capacitor in AC, $X_C = \dfrac{1}{\omega C}=\dfrac{1}{2\pi f C}$ , where $f$ is the supply frequency and $C$ the capacitance. Since $f$ appears in the denominator, higher frequency means the capacitor charges/discharges more rapidly and passes current more easily, so its reactance …

Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 mark
$\left(\dfrac{1}{\pi}\right)\,\mu F$
›Reveal solutionSolution

$X_C = \dfrac{1}{2\pi f C}$ ; substituting $f=50$ Hz and $C=\dfrac{1}{\pi}\,\mu F$ gives $X_C=10^4\,\Omega$ .

The reactance of a capacitor to an AC supply of frequency $f$ is

$X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{\omega C}$

Given $C = \dfrac{1}{\pi}\,\mu F = \dfrac{1}{\pi}\times 10^{-6}\,F$ and $f = 50$ Hz:

$X_C = \dfrac{1}{2\pi (50)\left(\dfrac{1}{\pi}\times10^{-6}\right)}$

The $\pi$ in the numerator (from $2\pi f$ ) cancels the $\pi$ in the denominator of $C$ : …

Start your 14-day free trial to see the full solution →

🎓Unlock everything free for 14 days

Unlock everything free for 14 days