(a) A resistor of 30Ω and a capacitor of (250)/(π)muF are connected in series to a 200V, 50Hz ac source. Calcu

Question

30\ \Omega

(i) the current in the circuit, and

(ii) voltage drops across the resistor and the capacitor.

(iii) Is the algebraic sum of these voltages more than the source voltage ? If yes, solve the paradox.

(OR)

(b) A series LCR circuit with

R = 20\ \Omega

L = 2\ \text{H}

and

C = 50\ \mu\text{F}

is connected to a

200\ \text{V}

ac source of variable frequency. What is

(i) the amplitude of the current, and

(ii) the average power transferred to the circuit in one complete cycle, at resonance ?

(iii) Calculate the potential drop across the capacitor.

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★

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Part (a)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. …

Part (b)Concept understanding — Resonance in AC Circuits

Resonance in AC Circuits

A series circuit containing a resistor $R$ , an inductor $L$ and a capacitor $C$ driven by an AC source exhibits resonance — a sharp condition at which the circuit responds most strongly.

The Competing Reactances

In a series RLC circuit the inductor and capacitor oppose the current in opposite senses. Their reactances are

X_L = \omega L, \qquad X_C = \frac{1}{\omega C}

where $\omega = 2\pi f$ is the angular frequency. As frequency rises, $X_L$ grows while $X_C$ shrinks. The total impedance is

Z = \sqrt{R^2 + (X_L - X_C)^2}

The Resonance Condition

At one special frequency the two reactances become exactly equal and cancel:

X_L = X_C \;\Rightarrow\; \omega_0 L = \frac{1}{\omega_0 C} \;\Rightarrow\; \omega_0 = \frac{1}{\sqrt{LC}}

The corresponding resonant frequency is

f_0 = \frac{1}{2\pi\sqrt{LC}}

At this frequency the impedance falls to its minimum, $Z = R$ (purely resistive), so the current reaches its maximum value

I_{\max} = \frac{V_{\text{rms}}}{R}

Because the reactances cancel, the source voltage and current are exactly in phase — the power factor is 1 at resonance.

Physical Picture

At resonance energy sloshes back and forth entirely between the inductor's magnetic field and the capacitor's electric field, cycle after cycle. The source only has to make up the small amount of energy lost as heat in $R$ . This is the electrical analogue of a swing pushed at its natural frequency: a small periodic drive builds a large oscillation.

Sharpness and the Q-factor

How sharply the current peaks around $f_0$ is measured by the quality factor:

Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

A large $Q$ (small $R$ ) gives a tall, narrow resonance curve — the circuit is highly selective, responding to a very narrow band of frequencies. A small $Q$ gives a broad, flat peak.

Why It Matters …

Why this formula?

Resonance in AC Circuits: Why the Key Formulas Hold

Resonance in an AC circuit occurs when the inductive reactance ( $X_L$ ) and capacitive reactance ( $X_C$ ) exactly cancel each other out. Let's build the understanding step-by-step.

1. The Core Condition for Resonance

Consider a series RLC circuit (resistor $R$ , inductor $L$ , capacitor $C$ ) driven by an AC voltage source $V = V_0 \sin(\omega t)$ .

The total impedance $Z$ of the series combination is:

Z = R + j(X_L - X_C)

where:

$X_L = \omega L$ (inductive reactance)
$X_C = \frac{1}{\omega C}$ (capacitive reactance)
$j = \sqrt{-1}$

Why resonance happens:

The circuit "wants" to let maximum current flow. The opposition to current comes from both resistance and reactance. But reactance can be negative (capacitive) or positive (inductive). When they are equal in magnitude but opposite in sign, they cancel:

X_L = X_C

This is the fundamental condition — not a formula to memorize, but a logical consequence of impedance minimization.

2. Deriving the Resonant Frequency

From $X_L = X_C$ :

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

Multiply both sides by $\omega$ :

\omega^2 L C = 1

Thus:

\omega_0 = \frac{1}{\sqrt{LC}}

Since $\omega = 2\pi f$ , the resonant frequency in hertz is:

f_0 = \frac{1}{2\pi \sqrt{LC}}

Why this makes sense:

A larger $L$ or $C$ means the circuit takes longer to "oscillate" — lower frequency.
A smaller $L$ or $C$ means faster oscillations — higher frequency.
The product $LC$ controls the natural time scale of the circuit.

3. What Happens at Resonance — Key Consequences

(a) Impedance is Minimum (Purely Resistive)

At resonance, $X_L - X_C = 0$ , so:

Z = R + j(0) = R

Why: The reactive parts cancel, leaving only the resistance. The circuit behaves like a pure resistor.

(b) Current is Maximum

From Ohm's law for AC:

I = \frac{V}{Z}

At resonance, $Z = R$ (minimum possible), so current is maximum:

I_{\text{max}} = \frac{V}{R}

Why: The opposition to current is smallest when reactance cancels.

(c) Voltage Across L and C Can Be Very Large

The voltage across the inductor:

V_L = I \cdot X_L = \frac{V}{R} \cdot \omega_0 L

The voltage across the capacitor:

V_C = I \cdot X_C = \frac{V}{R} \cdot \frac{1}{\omega_0 C}

Since $X_L = X_C$ at resonance, $V_L = V_C$ in magnitude, but they are 180° out of phase — they cancel each other in the loop.

Why this is important:

If $R$ is small, $V_L$ and $V_C$ can be many times larger than the source voltage $V$ . This is called voltage magnification — a key concept for tuned circuits and filters.

--- …

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Part (a): For the RC circuit $X_C=40\ \Omega$ , $Z=50\ \Omega$ , $I=4\ \text{A}$ , $V_R=120\ \text{V}$ , $V_C=160\ \text{V}$ ; the algebraic sum $=280\ \text{V}$ exceeds the source, but the phasor sum $=200\ \text{V}$ — no paradox.

Part (b): At resonance $I_0=10\sqrt2\approx14.1\ \text{A}$ , $P_{av}=2000\ \text{W}$ , and $V_C=2000\ \text{V}$ .

Part (a): Series RC circuit

$R=30\ \Omega$ , $C=\dfrac{250}{\pi}\ \mu\text{F}$ , $V=200\ \text{V}$ , $f=50\ \text{Hz}$ .

Capacitive reactance:

X_C=\frac{1}{2\pi f C}=\frac{1}{2\pi(50)\left(\frac{250}{\pi}\times10^{-6}\right)}=\frac{1}{2(50)(250\times10^{-6})}=40\ \Omega.

Current. $Z=\sqrt{R^2+X_C^2}=\sqrt{30^2+40^2}=\sqrt{900+1600}=\sqrt{2500}=50\ \Omega,$ $I=\frac{V}{Z}=\frac{200}{50}=4\ \text{A}.$
Voltage drops. The resistor drop is in phase with the current; the capacitor drop lags it by $90^\circ$ . $V_R=IR=4\times30=120\ \text{V},\qquad V_C=IX_C=4\times40=160\ \text{V}.$
Resolving the paradox. $V_R+V_C=120+160=280\ \text{V}>200\ \text{V}.$ This is not a contradiction. Because $V_R$ and $V_C$ are $90^\circ$ out of phase, they must be combined as phasors, not added algebraically: $\sqrt{V_R^2+V_C^2}=\sqrt{120^2+160^2}=200\ \text{V},$ …

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

Showing the 12 most recent of 36 on this concept.

CBSE 2026Set 55/1/11 markMCQ
$10\ \text{V}$
›Reveal solutionSolution

In a series LCR circuit, when each component drops 10 V, the source voltage is 10 V (since $V_C$ and $V_L$ cancel) and the equal drops imply $X_L = X_C = R$ . Shorting the capacitor leaves an RL circuit of impedance $R\sqrt{2}$ , so the current becomes $I' = \dfrac{10}{R\sqrt{2}}$ and the inductor voltage is $V_L' = I' X_L = \dfrac{10}{\sqrt{2}} = 5\sqrt{2}\ \text{V}$ . The answer is (B).

Concept and intuition

The problem gives a series LCR circuit where the voltage across each element — resistor, capacitor, and inductor — is 10 V. That’s a strong clue: in a series circuit, the current is the same through all components, but the voltages are not in phase. The resistor voltage is in phase with current, the inductor voltage leads by 90°, and the capacitor voltage lags by 90°. So the three 10 V readings are phasor magnitudes, not simple arithmetic sums.

The key insight: if the capacitor is shorted, the circuit becomes a simple RL series circuit. The source voltage remains the same (it’s fixed by the supply), but the impedance changes. We need to find the new inductor voltage.

Step-by-step solution

1. Find the source voltage from the initial LCR condition.

In a series LCR circuit, the phasor sum of voltages across R, L, and C equals the source voltage $V_s$ . Since $V_L$ and $V_C$ are opposite in phase (180° apart), they subtract. Given $V_R = V_L = V_C = 10\ \text{V}$ :
$V_s = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{10^2 + (10 - 10)^2} = 10\ \text{V}$
So the source supplies only 10 V. This makes sense: the inductor and capacitor voltages cancel exactly, so the source only “sees” the resistor drop.

Tip

This cancellation is the hallmark of resonance in a series LCR circuit — at resonance, $X_L = X_C$ , and the impedance is purely resistive. Here, $V_L = V_C$ implies $X_L = X_C$ , so the circuit is at resonance.

2. Determine the relationship between $R$ and $X_L$ (or $X_C$ ).

At resonance, the current is $I = V_s / R = 10 / R$ . The voltage across the inductor is $V_L = I X_L = (10/R) X_L = 10\ \text{V}$ . Therefore:
$\frac{10}{R} X_L = 10 \quad \Rightarrow \quad X_L = R$
So the inductive reactance equals the resistance. Similarly, $X_C = R$ as well. …
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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 …
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CBSE 2026Set V11 markMCQ
$Q. Power factor of a series LCR circuit is maximum when :$
(a) $X_L = X_C$
(b) $X_C = 0$
(c) $X_L > X_C$
(d) $X_L < X_C$
›Reveal solutionSolution

(a) $X_L = X_C$ …

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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: The quality factor is ω_r L / R.$
›Reveal solutionSolution

True — for a series resonant circuit, Q = ω_r L / R.

The quality factor (Q-factor) of a series resonant LCR circuit measures the sharpness of resonance. It is defined as the ratio of the inductive reactance at resonance to the resistance:

Q = ω_r L / R = (1/R)√(L/C),

…

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CBSE 2026Set SEM31 markMCQ
$Q. The condition of getting maximum current in an LCR series circuit is$
(a) X_L = 0
(b) X_C = 0
(c) X_L = X_C
(d) R = X_L − X_C
›Reveal solutionSolution

A series LCR circuit carries maximum current at resonance, where the inductive and capacitive reactances are equal (X_L = X_C), leaving impedance Z = R minimum. Option (c).

Step 1 — impedance of a series LCR circuit: Z = √(R² + (X_L − X_C)²), from NCERT/CBSE Class 12 Physics, Alternating Current.

…

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CBSE 2025Set D1 markMCQ
$Q. In resonance condition, the frequency of L-C circuit is (A) (1/2π)\sqrt(1/LC) (B) 2π\sqrt(1/LC) (C) 2π\sqrt(LC) (D) (1/2π)\sqrt(LC)$
›Reveal solutionSolution

At resonance the inductive and capacitive reactances are equal, giving the natural frequency f = 1/(2π√(LC)).

Resonance in an L-C (or series L-C-R) circuit occurs when the inductive reactance equals the capacitive reactance:

$X_L = X_C \ \Rightarrow\ \omega L = \frac{1}{\omega C}$

Solving for the angular frequency,

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A series LCR circuit fed by an ac source with angular frequency ω acts as a purely resistive circuit, when$
(a) ωL > 1/ωC
(b) ωL < 1/ωC
(c) ωL = 1/ωC
(d) ω³L = 1/ωC²
›Reveal solutionSolution

A series LCR circuit behaves as purely resistive at resonance, when the inductive and capacitive reactances cancel.

The impedance of a series LCR circuit is

$Z=\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}$ …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When LCR series circuit is at resonance then the phase angle (phi) between current and voltage is -$
(a) pi/2
(b) pi
(c) 2 pi
(d) 0
›Reveal solutionSolution

At resonance in a series LCR circuit, current and voltage are exactly in phase.

In a series LCR circuit, the phase angle $\phi$ between the applied voltage and current is given by $\tan\phi = \dfrac{X_L - X_C}{R}$ , where $X_L = \omega L$ and $X_C = \dfrac{1}{\omega C}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The power delivered by the AC source of a circuit becomes maximum when$
(i) wL = wC
(ii) wL = 1/(wC)
(iii) wL = -(1/(wC))^2
(iv) wL = sqrt(wC)
›Reveal solutionSolution

Maximum power occurs at resonance, wL = 1/(wC).

In a series LCR circuit the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$ with $X_L = \omega L$ and $X_C = 1/\omega C$ . Power $P = V_{rms} I_{rms}\cos\phi$ is greatest when Z is minimum (Z = R) and the current is in phase with the voltage. …

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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: $X' = \frac{X}{6}$ …
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CBSE 2024Set A11 markMCQ
$Q. The resonance phenomenon is exhibited by a circuit only if following components are present$
(a) L and R
(b) R and C
(c) L and C
(d) None of the above
›Reveal solutionSolution

(c) L and C …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In a series LCR circuit, resonant frequency depends on which of the following -$
(a) $\dfrac{R}{LC}$
(b) $\dfrac{L}{C}$
(c) $\dfrac{1}{\sqrt{LC}}$
(d) $\dfrac{1}{\sqrt{RC}}$
›Reveal solutionSolution

Resonance in a series LCR circuit occurs when inductive and capacitive reactances are equal.

…

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Q.(a) A resistor of 30 Ω30\ \Omega30 Ω and a capacitor of 250π μF\dfrac{250}{\pi}\ \mu\text{F}π250​ μF are connected in series to a 200 V200\ \text{V}200 V, 50 Hz50\ \text{Hz}50 Hz ac source. Calculate

Part (a): Series RC circuit

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Q.(a) A resistor of $30\ \Omega$ and a capacitor of $\dfrac{250}{\pi}\ \mu\text{F}$ are connected in series to a $200\ \text{V}$ , $50\ \text{Hz}$ ac source. Calculate

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