A series LCR ac circuit has $L = 2\cdot0$ H, $C = 32\ \mu$F and $R = 10\ \Omega$. (a) At what angular frequency of ac will it resonate ? (b) Calculate the Q value of the circuit. OR An ideal inductor of $\dfrac{5}{\pi}$ H inductance is connected to a 200 V, 50 Hz ac supply. (a) Calculate the rms and peak value of current in the inductor. (b) What is the phase difference between current through the inductor and the applied voltage ? How will it change if a small resistance is connected in series with this inductor in the circuit ?

> [!TLDR] > (a) Series LCR resonates at $\omega_0=1/\sqrt{LC}=125\ \text{rad/s}$ with $Q=\omega_0L/R=25$. > (b) A pure inductor of $5/\pi$ H on 200 V, 50 Hz has $X_L=500\ \Omega$, giving $I_{\text{rms}}=0.4$ A and $I_0\approx0.566$ A; the current lags the voltage by $90^\circ$, reduced below $90^\circ$ when a resistance is added. ## Part (a) At resonance the inductive and capacitive reactances cancel, $X_L=X_C$, i.e. $\omega L=\dfrac{1}{\omega C}$, so $$\omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2.0\times32\times10^{-6}}}=\frac{1}{\sqrt{64\times10^{-6}}}=125\ \text{rad/s}.$$ The quality factor measures the sharpness of resonance: $$Q=\frac{\omega_0 L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{125\times2.0}{10}=25.$$ …

A series LCR ac circuit has L = 2·0 H, C = 32μF and R = 10Ω. (a) At what angular frequency of ac will it reson

Question

L = 2\cdot0

(a) At what angular frequency of ac will it resonate ?

(b) Calculate the Q value of the circuit.

(OR)

\dfrac{5}{\pi}

(a) Calculate the rms and peak value of current in the inductor.

(b) What is the phase difference between current through the inductor and the applied voltage ? How will it change if a small resistance is connected in series with this inductor in the circuit ?

CBSECBSE Class XII Board 2020Subjective· 3mImportance★★★★★

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Part (a)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.

--- …

Part (b)Concept understanding — AC Through an Inductor

AC Through an Inductor: From Intuition to the 90° Lag

The Core Intuition — Why an Inductor "Fights" Change

Imagine you are pushing a heavy box across a floor. If you push steadily, the box moves at a constant speed. But if you try to suddenly jerk it forward, the box resists — its inertia makes it want to stay where it is. The harder you push, the more it pushes back, but only while you are changing the speed.

An inductor does the same thing, but with electric current. It does not resist steady current (DC) — a perfect inductor has zero resistance. What it resists is change in current. The faster you try to change the current, the harder the inductor pushes back with a voltage that opposes that change.

This opposition is called self-induction, and the property that causes it is inductance $L$ , measured in henries (H).

The Physics: Faraday's Law in Action

When current flows through a coil, it creates a magnetic field. If the current changes, the magnetic field changes, and that changing field induces a voltage in the same coil — a back emf. Faraday's law gives the magnitude:

v_L = L \frac{di}{dt}

The sign matters: the induced voltage always acts to oppose the change that produced it (Lenz's law). So if current is increasing ( $di/dt > 0$ ), the inductor generates a voltage that tries to push current the other way — like a spring that pushes back harder the faster you compress it.

Watch out

A common mistake is to think the inductor "resists" current like a resistor. It does not. It resists change in current. For steady DC, $di/dt = 0$ , so $v_L = 0$ — the inductor acts like a short circuit.

AC Through an Inductor: The Mathematics

Now feed the inductor with an AC voltage source:

v(t) = V_m \sin(\omega t)

where $\omega = 2\pi f$ is the angular frequency. The circuit equation (Kirchhoff's voltage law) gives:

v(t) = L \frac{di}{dt}

So:

\frac{di}{dt} = \frac{V_m}{L} \sin(\omega t)

Integrate to find the current:

i(t) = \int \frac{V_m}{L} \sin(\omega t) \, dt = -\frac{V_m}{\omega L} \cos(\omega t) + C

The constant $C$ is zero for steady-state AC (no DC offset). Using $\cos(\omega t) = \sin(\omega t + 90^\circ)$ :

i(t) = \frac{V_m}{\omega L} \sin(\omega t - 90^\circ)

i(t) = I_m \sin(\omega t - 90^\circ) \quad \text{where} \quad I_m = \frac{V_m}{\omega L}

The 90° Lag — What It Means Physically

Compare the voltage and current:

Voltage: $v(t) = V_m \sin(\omega t)$
Current: $i(t) = I_m \sin(\omega t - 90^\circ)$

The current reaches its peak exactly one-quarter cycle after the voltage does. We say current lags voltage by 90° (or $\pi/2$ radians).

Why? Look at the derivative relationship. When voltage is at its peak, $di/dt$ is maximum — the current is changing fastest. But the current itself is passing through zero at that instant (think of a sine wave: its steepest slope is at the zero crossing). When voltage passes through zero, $di/dt = 0$ , and the current is at its peak (the sine wave is flat at the top).

Tip

Visualise it: voltage drives the rate of change of current, not the current itself. So the current waveform is the integral of the voltage waveform — and integrating a sine gives a negative cosine, which is a sine shifted by -90°.

Inductive Reactance — The "AC Resistance"

The amplitude of the current is:

I_m = \frac{V_m}{\omega L}

Define inductive reactance $X_L$ :

X_L = \omega L = 2\pi f L

Then $I_m = V_m / X_L$ , which looks like Ohm's law — but $X_L$ is not a resistance. It depends on frequency: …

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Series LCR resonates at $\omega_0=1/\sqrt{LC}=125\ \text{rad/s}$ with $Q=\omega_0L/R=25$ .
A pure inductor of $5/\pi$ H on 200 V, 50 Hz has $X_L=500\ \Omega$ , giving $I_{\text{rms}}=0.4$ A and $I_0\approx0.566$ A; the current lags the voltage by $90^\circ$ , reduced below $90^\circ$ when a resistance is added.

Part (a)

At resonance the inductive and capacitive reactances cancel, $X_L=X_C$ , i.e. $\omega L=\dfrac{1}{\omega C}$ , so

\omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{2.0\times32\times10^{-6}}}=\frac{1}{\sqrt{64\times10^{-6}}}=125\ \text{rad/s}.

The quality factor measures the sharpness of resonance:

$Q=\frac{\omega_0 L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{125\times2.0}{10}=25.$ …

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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 45 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 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 A1 markMCQ
$Q. Choke coil works on the principle of (A) self induction (B) transient current (C) mutual induction (D) wattless current$
›Reveal solutionSolution

A choke coil works on self-induction: its inductance opposes changes in AC current, limiting the current with negligible power loss.

A choke coil is an inductor of large inductance $L$ and small resistance. In an AC circuit it presents an inductive reactance $X_L = \omega L$ , which arises from the back-emf produced by the coil's own changing flux — i.e. self-induction. This reactance limits the current. Because an ideal inductor's current lags the voltage by $90^\circ$ , the average power …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In an a.c. circuit having pure inductor, current$
(a) leads the voltage by an angle of π/2
(b) leads the voltage by an angle of π
(c) lags the voltage by an angle of π/2
(d) lags the voltage by an angle of π
›Reveal solutionSolution

In a pure inductor, the back-emf opposes the change in current, forcing the current to lag 90° behind the applied voltage.

For a pure inductor with applied voltage $v = v_0\sin(\omega t)$ , the induced back-emf balances the applied voltage: $v = L\dfrac{di}{dt}$ . Solving,

$i = \frac{v_0}{\omega L}\int \sin(\omega t)\,dt = -\frac{v_0}{\omega L}\cos(\omega t) = \frac{v_0}{\omega L}\sin\!\left(\omega t - \frac{\pi}{2}\right)$

…

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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 55/6/11 markMCQ
$Q. An ac source is connected to a resistor and an inductor in series. The voltage across the resistor and inductor are 8 V and 6 V respectively. The voltage of the source is: (A) 10 V (B) 12 V (C) 14 V (D) 16 V$
›Reveal solutionSolution
In a series RL circuit, the resistor and inductor voltages are 90° out of phase, so the source voltage is the phasor sum (Pythagorean sum) of the two: $\sqrt{8^2 + 6^2} = 10\ \text{V}$ . The correct option is (A).

Concept and Intuition

When an AC source drives a resistor and an inductor in series, the two components respond differently to the alternating current. The resistor’s voltage is in phase with the current — it rises and falls exactly when the current does. The inductor’s voltage, however, leads the current by 90° (a quarter-cycle ahead). This phase difference means you cannot simply add the two voltages as ordinary numbers (8 V + 6 V = 14 V). That would be a common mistake — it ignores the fact that the peaks of these voltages occur at different times.

Instead, we treat them as phasors — rotating arrows whose lengths represent the voltage magnitudes and whose angles represent the phase. The resistor voltage points along the reference direction (say, the x-axis), and the inductor voltage points 90° ahead (the y-axis). The source voltage is the vector sum of these two perpendicular phasors. That’s why the Pythagorean theorem gives the answer.

Watch out

Never add RMS voltages of different components in an AC circuit as if they were DC voltages. The phase difference (here 90° for a pure inductor) makes the total less than the arithmetic sum. 8 + 6 = 14 V is a trap — the correct answer is 10 V.

Step-by-Step Solution

Identify the given data.

Voltage across resistor: $V_R = 8\ \text{V}$ (RMS value).

Voltage across inductor: $V_L = 6\ \text{V}$ (RMS value).

The circuit is a series RL combination connected to an AC source.

Recall the phase relationship.

In a series RL circuit, the current $I$ is the same through both components.

For the resistor: $V_R = I R$ , and $V_R$ is in phase with $I$ .

For the inductor: $V_L = I X_L$ , and $V_L$ leads $I$ by $90^\circ$ . …
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CBSE 2025Set D1 markMCQ
$Q. In a purely inductive circuit, the power factor is (A) 0 (B) 1 (C) 0.5 (D) infinity$
›Reveal solutionSolution

In a pure inductor current lags voltage by 90°, so power factor cos φ = cos 90° = 0 (no average power consumed).

In a purely inductive a.c. circuit, the current lags the applied voltage by a phase angle of 90° (φ = 90°).

The power factor is

$\cos\phi = \cos 90^\circ = 0$

…

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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 mark
$Q. In a purely inductive a.c. circuit the alternating current lags behind the alternating voltage by ____________ phase angle.$
›Reveal solutionSolution

In a purely inductive circuit, the back-EMF from the inductor makes the current reach its peak a quarter-cycle after the voltage — a 90 degree phase lag.

For a purely inductive AC circuit with applied voltage $V = V_0\sin\omega t$ , the current is:

$I = \dfrac{V_0}{\omega L}\sin\left(\omega t - \dfrac{\pi}{2}\right)$

…

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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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Q.A series LCR ac circuit has L=2⋅0L = 2\cdot0L=2⋅0 H, C=32 μC = 32\ \muC=32 μF and R=10 ΩR = 10\ \OmegaR=10 Ω.

Part (a)

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Q.A series LCR ac circuit has $L = 2\cdot0$ H, $C = 32\ \mu$ F and $R = 10\ \Omega$ .

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