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

AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$\dfrac{1}{3}$
›Reveal solutionSolution
Capacitive reactance triples when frequency drops to a third; matching this against the given halving of current pins down $X_C/R=\sqrt{3/5}$ at the original frequency. Answer: (A).

Concept and Intuition

In a series R-C circuit, $X_C=\dfrac{1}{2\pi fC}$ is inversely proportional to frequency, while $R$ stays fixed. The impedance $Z=\sqrt{R^2+X_C^2}$ sets the current $I=V/Z$ at constant voltage. Lowering the frequency raises $X_C$ (capacitor becomes more 'blocking'), raising $Z$ and lowering $I$ — exactly the behaviour described.

Step-by-Step Solution

Let the original frequency be $f$ , with reactance $X_C$ and impedance $Z_1=\sqrt{R^2+X_C^2}$ , current $I=V/Z_1$ .

At frequency $f/3$ : $X_C'=\dfrac{1}{2\pi(f/3)C}=3X_C$ . New impedance $Z_2=\sqrt{R^2+9X_C^2}$ .

Given new current is $I/2$ , and $V$ unchanged: $Z_2=2Z_1$ , so $Z_2^2=4Z_1^2$ .

$R^2+9X_C^2=4(R^2+X_C^2)=4R^2+4X_C^2$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$V_2$
›Reveal solutionSolution
This is a series $RC$ circuit: the ammeter $A$ reads the common current, $V_2$ reads the resistor voltage (in phase with the current) and $V_1$ reads the capacitor voltage (lagging the current by $90^\circ$ ). Statements I and III are true, so the answer is (C).

Concept

In a single series loop the current is the same everywhere, so the ammeter $A$ measures the current common to $C$ and $R$ . The voltmeter $V_1$ is across the capacitor and $V_2$ is across the resistor. The two fixed facts you need:

Across a resistor, voltage and current are in phase.

Across a capacitor, the current leads the voltage by $90^\circ$ (ELI the ICE-man: in a Capacitor, I leads E).

Statement I - $A$ and $V_2$ are in phase

$V_2$ is the resistor voltage, which is always in phase with the current read by $A$ . True.

Statement II - $V_1$ leads $V_2$ …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$\dfrac{10^3}{\pi}\ \Omega$
›Reveal solutionSolution
Direct substitution into $X_C=1/(2\pi f C)$ gives $X_C = 10^3/\pi\ \Omega$ .

Concept and Intuition

A capacitor's opposition to alternating current, the capacitive reactance, is $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ . It decreases with increasing frequency because a capacitor charges and discharges more easily (offers less opposition) when the applied voltage reverses direction faster.

Step-by-Step Solution

Given: $C = 500\ \text{nF} = 500\times10^{-9}\ \text{F} = 5\times10^{-7}\ \text{F}$ , $f = 1\ \text{kHz} = 10^3\ \text{Hz}$ .

$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 10^3 \times 5\times10^{-7}}$

Compute the denominator: $2\pi\times10^3\times5\times10^{-7} = 2\pi \times 5\times10^{-4} = 10\pi\times10^{-4} = \pi\times10^{-3}$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$Q. A bulb is connected in series with a capacitor to an ac supply. If the capacitance of the capacitor increases, then power of light emitted by the bulb (A) decreases (B) increases (C) does not change (D) becomes zero$
›Reveal solutionSolution
This tests how a series R–C AC circuit's impedance changes with capacitance; increasing $C$ lowers the capacitive reactance, raising the current and hence the bulb's power. Answer: power increases.

Concept and Intuition

A capacitor in an AC circuit offers an opposition to current flow called capacitive reactance, $X_C = \dfrac{1}{\omega C}$ . Unlike a resistor, this reactance depends inversely on capacitance: a bigger capacitor 'blocks' AC less. Since the bulb (a resistor) is in series with the capacitor, the same current flows through both. The total impedance of the series combination is
$Z=\sqrt{R^2+X_C^2}.$
Step-by-Step Solution

Capacitive reactance: $X_C=\dfrac{1}{\omega C}$ . As $C$ increases, $X_C$ decreases.

Impedance of the series R–C circuit: $Z=\sqrt{R^2+X_C^2}$ . Since $X_C$ decreases, $Z$ decreases.

Current in the circuit: $I=\dfrac{V_{rms}}{Z}$ . Since $Z$ decreases while $V_{rms}$ (supply) is fixed, $I$ increases. …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$\dfrac{1}{3}$
›Reveal solutionSolution
Using the current-halving condition to relate the reactance at two frequencies gives $X_0^2=\tfrac35R^2$ , and the power factor $R/Z_0$ works out to $\sqrt{5/8}$ .

Concept and Intuition

In a series RC AC circuit, impedance is $Z=\sqrt{R^2+X_C^2}$ where $X_C=\dfrac{1}{\omega C}$ is the capacitive reactance. Lowering the frequency raises $X_C$ (harder for the capacitor to "pass" current), which raises $Z$ and lowers the current for the same applied voltage — matching the given fact that current drops when frequency drops. The power factor $\cos\phi=R/Z$ measures how resistive (vs. reactive) the circuit's response is at a given frequency; it needs both $R$ and $X_C$ at that specific frequency.

Step-by-Step Solution

Let the initial angular frequency be $\omega_0$ , reactance $X_0=1/(\omega_0 C)$ , impedance $Z_0=\sqrt{R^2+X_0^2}$ , current $I_0=V/Z_0$ .

At $\omega_0/3$ : reactance becomes $X'=1/((\omega_0/3)C)=3X_0$ (reactance is inversely proportional to frequency).

Current halves: $I'=I_0/2=V/Z' \Rightarrow Z'=2Z_0$ .

Square both impedances: $Z_0^2=R^2+X_0^2$ and $Z'^2=R^2+(3X_0)^2=R^2+9X_0^2$ .

Substitute $Z'^2=4Z_0^2$ : $R^2+9X_0^2=4(R^2+X_0^2)=4R^2+4X_0^2$ .

Rearranging: $5X_0^2=3R^2 \Rightarrow X_0^2=\dfrac{3}{5}R^2$ . …
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AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
$3\,\mathrm{k}\Omega$
›Reveal solutionSolution
$X_C\propto1/f$ , so doubling the source frequency halves the capacitive reactance from 3 k $\Omega$ to 1.5 k $\Omega$ .

Concept and Intuition

A capacitor's opposition to AC current, $X_C=\dfrac{1}{2\pi fC}$ , decreases as frequency increases — at higher frequency the capacitor charges and discharges faster, letting more current through for the same voltage.

Step-by-Step Solution

$X_{C,1}=\dfrac{1}{2\pi f C}=3\,\mathrm{k}\Omega$ at frequency $f$ .

At frequency $2f$ : $X_{C,2}=\dfrac{1}{2\pi(2f)C}=\dfrac{X_{C,1}}{2}$ .

$X_{C,2}=\dfrac{3}{2}=1.5\,\mathrm{k}\Omega$ . …
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AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
$\frac{50}{\pi}\ \mu F$
›Reveal solutionSolution
Compute the capacitive reactance from $f$ and $C$ , then apply Ohm's law for AC: $I_{rms}=V_{rms}/X_C = 1.25$ A.

Concept and Intuition

A capacitor's opposition to AC is the reactance $X_C = 1/(\omega C)$ , which plays the role of "resistance" in $V=IX_C$ for rms quantities in a purely capacitive AC circuit.

Step-by-Step Solution

$\omega = 2\pi f = 2\pi(50) = 100\pi$ rad/s.

$X_C = \dfrac{1}{\omega C} = \dfrac{1}{100\pi\times\left(\dfrac{50}{\pi}\times10^{-6}\right)}$ .

The $\pi$ 's cancel: $100\pi\times\dfrac{50}{\pi}\times10^{-6} = 100\times50\times10^{-6} = 5000\times10^{-6} = 5\times10^{-3}$ .

$X_C = \dfrac{1}{5\times10^{-3}} = 200\ \Omega$ . …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$\mu$
›Reveal solutionSolution
Standard AC-capacitor problem: compute the reactance, then use Ohm's law for AC to get the rms current, 3.14 A.

Concept and Intuition

In an AC circuit, a capacitor opposes current with a frequency-dependent reactance $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ (unlike a resistor, this depends on frequency). Once $X_C$ is known, rms current follows exactly as in a resistive circuit: $I_{rms}=V_{rms}/X_C$ .

Step-by-Step Solution

$C = 100\ \mu\text{F} = 10^{-4}\ \text{F}$ , $f=50$ Hz, $V_{rms}=100$ V.

$X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi(50)(10^{-4})} = \dfrac{1}{0.0314159} \approx 31.83\ \Omega$ . …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$6\ \mathrm{k}\Omega$
›Reveal solutionSolution
Capacitive reactance is inversely proportional to frequency, so doubling frequency halves it: $6\,\mathrm{k}\Omega\to3\,\mathrm{k}\Omega$ .

Concept and Intuition

Capacitive reactance $X_C = \dfrac{1}{2\pi f C}$ decreases as frequency increases — a capacitor "conducts" AC more easily at higher frequencies. Since $C$ is fixed here, $X_C \propto 1/f$ directly.

Step-by-Step Solution

Original reactance: $X_C = \dfrac{1}{2\pi f C} = 6\,\mathrm{k}\Omega$ .

New frequency: $f' = 2f$ . New reactance: $X_C' = \dfrac{1}{2\pi(2f)C} = \dfrac{X_C}{2}$ . …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$180^\circ$
›Reveal solutionSolution
For a capacitor in an AC circuit, the current is maximum when the voltage is changing fastest (zero crossing), which places current $90^\circ$ ahead of voltage in phase.

Concept and Intuition

The current through a capacitor is $I=C\dfrac{dV}{dt}$ . If $V=V_0\sin\omega t$ , then $I=C V_0\omega\cos\omega t = I_0\sin(\omega t+90^\circ)$ — the current waveform is a cosine relative to the voltage sine, i.e., it peaks a quarter cycle before the voltage.

Step-by-Step Solution

Let applied voltage $V=V_0\sin\omega t$ .

Current through capacitor: $I=C\dfrac{dV}{dt}=C V_0\omega\cos\omega t$ .

Rewrite $\cos\omega t=\sin(\omega t+90^\circ)$ , so $I=I_0\sin(\omega t+90^\circ)$ where $I_0=\omega C V_0$ .

Comparing phases, current is ahead of (leads) voltage by $90^\circ$ . …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$\omega$
›Reveal solutionSolution
Using $I=V/Z$ at two different frequencies and the fact that capacitive reactance scales inversely with frequency lets us solve for the ratio $X_C/R$ at the original frequency: $\sqrt{3/5}$ .

Concept and Intuition

For a series R-C circuit, impedance is $Z=\sqrt{R^2+X_C^2}$ with $X_C = \dfrac{1}{\omega C}$ . Lowering the frequency increases $X_C$ (reactance rises as frequency falls), which increases impedance and hence decreases current for the same applied voltage. Given how much the current drops when frequency is reduced to $\omega/3$ , we can back out the reactance-to-resistance ratio at the original frequency.

Step-by-Step Solution

At frequency $\omega$ : current $I = V/Z_1$ , where $Z_1=\sqrt{R^2+X_C^2}$ and $X_C=1/(\omega C)$ .

At frequency $\omega/3$ : reactance becomes $X_C' = \dfrac{1}{(\omega/3)C} = 3X_C$ . Current becomes $I/2 = V/Z_2$ , where $Z_2=\sqrt{R^2+9X_C^2}$ .

Since voltage $V$ is the same in both cases:

$I = \frac{V}{Z_1}, \qquad \frac{I}{2}=\frac{V}{Z_2} \Rightarrow Z_2 = 2Z_1$

Square both sides: $R^2+9X_C^2 = 4(R^2+X_C^2) = 4R^2+4X_C^2$ …
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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

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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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