(a) How does resistance differ from impedance? With the help of a suitable phasor diagram, obtain an expressio

Question

v = v_m \sin\omega t

(OR)

(b) Find the condition for resonance in a series LCR circuit connected to a source

v = v_m \sin\omega t

, where

\omega

can be varied. Give the factors on which the resonant frequency of a series LCR circuit depends. Plot a graph showing the variation of electric current with frequency in a series LCR circuit.

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

🔒 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 →

Part (a)Concept understanding — Parallel AC Circuit Impedance and Current Phasors

Parallel AC Circuit Impedance and Current Phasors

When branches are in parallel across an ac source $v = v_m \sin \omega t$ , every branch shares the same voltage, so the natural quantities to add are the branch currents, not the branch voltages. The total current is the phasor sum $i = i_1 + i_2$ .

Branch Currents

A pure resistor draws $i_1$ in phase with $v$ , amplitude $v_m/R$ . A series $L$ – $C$ branch has net reactance $X = X_L - X_C = \omega L - \dfrac{1}{\omega C}$ ; it draws $i_2$ of amplitude $v_m/|X|$ , lagging $v$ by $90^\circ$ when $X_L > X_C$ (net inductive) or leading by $90^\circ$ when $X_C > X_L$ (net capacitive).

Combining the Branch Currents

Because $i_1$ (along $v$ ) and $i_2$ (perpendicular to $v$ ) are $90^\circ$ apart, the resultant amplitude is

i_m = \sqrt{i_1^2 + i_2^2} = v_m \sqrt{\frac{1}{R^2} + \left(\frac{1}{X_L - X_C}\right)^2},

and the current makes a phase angle $\phi$ with the voltage given by $\tan\phi = i_2/i_1$ .

Impedance

Writing $i_m = v_m/Z$ , the admittances add in quadrature:

\frac{1}{Z} = \sqrt{\frac{1}{R^2} + \left(\frac{1}{X_L - X_C}\right)^2}.

What Happens Near Resonance …

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.

--- …

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

Figure — Part (a) explicitly requires a phasor diagram to derive the series-LCR impedance; the catalog figure shows exa

Part (a): $R$ is frequency-independent, $Z=\sqrt{R^2+(X_L-X_C)^2}$ depends on frequency and causes a phase shift.

Part (b): resonance at $X_L=X_C\Rightarrow f_0=\dfrac{1}{2\pi\sqrt{LC}}$ (depends on $L,C$ ); current is maximum there.

Part (a)

Resistance vs impedance. Resistance $R$ is the opposition to current offered by a resistor; it is the same for dc and ac and does not depend on frequency, and the voltage and current stay in phase. Impedance $Z$ is the effective opposition of an ac circuit that also contains inductance and/or capacitance; it depends on the source frequency and produces a phase difference $\phi$ between voltage and current.

Series LCR impedance (phasor method). The same current $i=i_m\sin\omega t$ flows through $R$ , $L$ , $C$ . On a phasor diagram, taking $I$ as reference:

$V_R=i_mR$ is in phase with $I$ ,
$V_L=i_mX_L$ leads $I$ by $90^\circ$ ,
$V_C=i_mX_C$ lags $I$ by $90^\circ$ .

$V_L$ and $V_C$ are opposite, so their resultant is $(V_L-V_C)$ . Combining with $V_R$ at right angles:

v_m=\sqrt{V_R^2+(V_L-V_C)^2}=i_m\sqrt{R^2+(X_L-X_C)^2}.

Hence the impedance …

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.

Showing the 12 most recent of 31 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. …
Start your 14-day free trial to see the full solution →
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) $X_L = X_C$ …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Write the formula of impedance of LCR circuit.$
›Reveal solutionSolution

Z = √(R² + (X_L − X_C)²), with X_L = ωL, X_C = 1/ωC.

In a series LCR circuit, the resistance R is in phase with the current, the inductive reactance X_L = ωL leads by 90°, and the capacitive reactance X_C = 1/ωC lags by 90°. The resistance and the net reactance (X_L − X_C) are perpendicular in the phasor diagram, so they add in quad …

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

…

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

…

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

…

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

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

Start your 14-day free trial to see the full solution →
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 $X_C = \dfrac{1}{\omega C}$ .

…

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

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

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

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. For a series L-C-R circuit at resonance, the relation among inductance (L), capacitance (C) and frequency (ω) is$
(a) ω = LC
(b) ω = 1/LC
(c) ω = √(L/C)
(d) ω = 1/√(LC)
›Reveal solutionSolution

At resonance the inductive and capacitive reactances of a series LCR circuit are equal, giving ω = 1/√(LC).

In a series L-C-R circuit driven by an AC source of angular frequency ω, the total reactance is X = X_L − X_C = ωL − 1/(ωC). The circuit is said to be at resonance when this net reactance is zero, i.e. the impedance is purely resistive (Z = R) and is minimum, so the current is maximum.

Setting X_L = X_C:

ωL = 1/(ωC)

ω² = 1/(LC)

ω = 1/√(LC)

…

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

Q.(a) How does resistance differ from impedance? With the help of a suitable phasor diagram, obtain an expression for the impedance of a series LCR circuit connected to a source v=vmsin⁡ωtv = v_m \sin\omega tv=vm​sinωt.

Part (a)

🎓Unlock everything free for 14 days

Q.(a) How does resistance differ from impedance? With the help of a suitable phasor diagram, obtain an expression for the impedance of a series LCR circuit connected to a source $v = v_m \sin\omega t$ .

Unlock everything free for 14 days