In an alternating current circuit consisting of elements in series, the current increases on increasing the fr

NCERT Exemplar · Q27

Q. In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?

(a) Only resistor.

(b) Resistor and an inductor.

(d) Only a capacitor.

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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. …

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Current increases with frequency only when the net reactance falls as $f$ rises, and the sole element whose reactance decreases with frequency is the capacitor ( $X_C=1/2\pi fC$ ). So the circuit contains a capacitor (with a resistor) — an RC combination.

The current in a series AC circuit is set by the total impedance, $I=V/Z$ . For a fixed supply voltage, the current grows whenever $Z$ shrinks. So the question reduces to: which element makes the impedance fall as the frequency is raised?

1. How each element depends on frequency

Resistor: $Z_R=R$ , constant — unaffected by frequency.
Inductor: $X_L=2\pi f L$ — increases in proportion to frequency.
Capacitor: $X_C=\dfrac{1}{2\pi f C}$ — decreases as frequency increases.

2. Test each simple circuit

Pure resistor: $I=V/R$ is independent of $f$ — no change. Ruled out.
Pure / series inductor: $Z=\sqrt{R^2+(2\pi f L)^2}$ grows with $f$ , so current falls. This is the opposite of what is observed. Ruled out.
Pure / series capacitor: $Z=\sqrt{R^2+\left(\dfrac{1}{2\pi f C}\right)^2}$ ; as $f$ rises the capacitive term shrinks, $Z$ falls, and the current rises. This matches the observation. …

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Q.In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?

Concept understanding — Capacitive Reactance

Capacitive Reactance: The AC Resistance of a Capacitor

The Intuition: Why Frequency Matters

The Precise Statement

What the Formula Tells You

Phase: The Hidden Twist

Why this formula?

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

1. The Fundamental Behavior of a Capacitor

2. Relating Current to Voltage

3. Applying a Sinusoidal Voltage

4. The Phase Shift — Why It Matters

5. Extracting the Reactance

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

1. How each element depends on frequency

2. Test each simple circuit

Method: Determining Circuit Composition from a Monotonic Current-vs-Frequency Trend

Steps

Q.In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?

1. How each element depends on frequency

2. Test each simple circuit

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