Two ac circuits are driven by identical ac sources of the same rms voltage. In circuit (a), a single resistor $R$ is connected across the source. In circuit (b), the same resistor $R$ is connected in series with a capacitor $C$ and an inductor $L$ across the source (a series LCR circuit). (a) Under which condition would the rms currents in the two circuits be the same? (b) Can the rms current in circuit (b) ever be larger than that in circuit (a)?

> [!TLDR] > The resistor-only circuit carries $I_a=V/R$. The series LCR circuit carries $I_b=V/\sqrt{R^2+(X_L-X_C)^2}$. Because $\sqrt{R^2+(X_L-X_C)^2}\ge R$ always, the two currents are equal only at resonance ($X_L=X_C$) and $I_b$ can never exceed $I_a$. ### Circuit (a): pure resistor The impedance is just $R$, so $$I_a=\frac{V_{rms}}{R}.$$ ### Circuit (b): series LCR The impedance is $$Z=\sqrt{R^2+(X_L-X_C)^2},\qquad X_L=\omega L,\ X_C=\frac{1}{\omega C},$$ so $$I_b=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C)^2}}.$$ ### (a) When are the currents equal? $I_a=I_b$ requires $Z=R$, i.e. $(X_L-X_C)^2=0\Rightarrow X_L=X_C$. This is the resonance condition $\omega L=\dfrac{1}{\omega C}$, i.e. $\omega=\dfrac{1}{\sqrt{LC}}$. At resonance the reactances cancel and (b) behaves exactly like (a). ### (b) Can $I_b>I_a$? The quantity under the root, $R^2+(X_L-X_C)^2$, is never smaller than $R^2$, so $Z\ge R$ for all frequencies. Hence …

Two ac circuits are driven by identical ac sources of the same rms voltage. In circuit (a), a single resistor

NCERT Exemplar · Q10

R

(a) Under which condition would the rms currents in the two circuits be the same?

(b) Can the rms current in circuit

(b) ever be larger than that in circuit (a)?

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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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The resistor-only circuit carries $I_a=V/R$ . The series LCR circuit carries $I_b=V/\sqrt{R^2+(X_L-X_C)^2}$ . Because $\sqrt{R^2+(X_L-X_C)^2}\ge R$ always, the two currents are equal only at resonance ( $X_L=X_C$ ) and $I_b$ can never exceed $I_a$ .

Circuit (a): pure resistor

The impedance is just $R$ , so

I_a=\frac{V_{rms}}{R}.

Circuit (b): series LCR

The impedance is

Z=\sqrt{R^2+(X_L-X_C)^2},\qquad X_L=\omega L,\ X_C=\frac{1}{\omega C},

I_b=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C)^2}}.

(a) When are the currents equal?

$I_a=I_b$ requires $Z=R$ , i.e. $(X_L-X_C)^2=0\Rightarrow X_L=X_C$ . This is the resonance condition $\omega L=\dfrac{1}{\omega C}$ , i.e. $\omega=\dfrac{1}{\sqrt{LC}}$ . At resonance the reactances cancel and (b) behaves exactly like (a).

(b) Can $I_b>I_a$ ?

The quantity under the root, $R^2+(X_L-X_C)^2$ , is never smaller than $R^2$ , so $Z\ge R$ for all frequencies. Hence …

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

KEAM 2025Set pha-2025-0424A4 marksMCQ
$Q. The a.c circuit exhibiting the phenomenon of resonance has/have the circuit element(s) (A) inductor and resistor (B) capacitor and resistor (C) inductor only (D) capacitor only (E) inductor and capacitor$
›Reveal solutionSolution

Electrical resonance needs both energy-storing reactive elements — an inductor and a capacitor.

Resonance in an a.c. circuit occurs when the inductive reactance equals the capacitive reactance:

$X_L = X_C \;\Rightarrow\; \omega L = \frac{1}{\omega C} \;\Rightarrow\; \omega_0 = \frac{1}{\sqrt{LC}}.$ …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\omega_0$
›Reveal solutionSolution

The resonance quality factor is $Q = \dfrac{\omega_0 L}{R}$ (equivalently $\dfrac{1}{R}\sqrt{L/C}$ ).

For a series LCR circuit at resonance $\omega_0 = \dfrac{1}{\sqrt{LC}}$ , the quality factor is …
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KEAM 2024Set eng-2024-06074 marksMCQ
$V = V_0 \sin \omega t$
›Reveal solutionSolution

$\omega_0=\dfrac{1}{\sqrt{LC}}=10^{6}$ rad s $^{-1}$ .

Resonance in a series LCR circuit occurs at $\omega_0=\dfrac{1}{\sqrt{LC}}$ . With $L=1\,\mu\text{H}=10^{-6}$ H and $C=1\,\mu\text{F}=10^{-6}$ F, …
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KEAM 2024Set eng-2024-06084 marksMCQ
$Q. In an LCR circuit, at resonance, the value of the power factor is (A) 1 (B) 0 (C) 0.5 (D) 0.75 (E) infinity$
›Reveal solutionSolution

At resonance the reactances cancel, the circuit is purely resistive, and $\cos\phi=1$ .

In a series LCR circuit at resonance, the inductive and capacitive reactances are equal, $X_L = X_C$ , so the net reactance is zero and the impedance is purely resistive, $Z = R$ . …
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KEAM 2023Set eng-2023-P1-A14 marksMCQ
$100\,\Omega$
›Reveal solutionSolution
Maximum current is at resonance, giving C = 1/(omega^2 L) = 80 uF.

Concept and Intuition

In a series LCR circuit the current is maximum at resonance, where the inductive and capacitive reactances cancel and the impedance equals R. The resonance condition is omega^2 L C = 1.

Step-by-Step Solution

Resonance: omega^2 L C = 1, so C = 1/(omega^2 L).

Substitute omega = 500 rad/s and L = 50 mH = 0.05 H: C = 1/(500^2 x 0.05). …
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Q.Two ac circuits are driven by identical ac sources of the same rms voltage. In circuit (a), a single resistor $R$ is connected across the source. In circuit (b), the same resistor $R$ is connected in series with a capacitor $C$ and an inductor $L$ across the source (a series LCR circuit).

Concept understanding — Resonance in AC Circuits

Resonance in AC Circuits

The Competing Reactances

The Resonance Condition

Physical Picture

Sharpness and the Q-factor

Why It Matters …

Why this formula?

Resonance in AC Circuits: Why the Key Formulas Hold

1. The Core Condition for Resonance

2. Deriving the Resonant Frequency

3. What Happens at Resonance — Key Consequences

(a) Impedance is Minimum (Purely Resistive)

(b) Current is Maximum

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

Circuit (a): pure resistor

Circuit (b): series LCR

(a) When are the currents equal?

(b) Can $I_b>I_a$ ?

Method: Comparing Currents Between a Purely Resistive Circuit and a Series LCR Circuit

Steps

Concept and Intuition

Step-by-Step Solution

Q.Two ac circuits are driven by identical ac sources of the same rms voltage. In circuit (a), a single resistor RRR is connected across the source. In circuit (b), the same resistor RRR is connected in series with a capacitor CCC and an inductor LLL across the source (a series LCR circuit).

Circuit (a): pure resistor

Circuit (b): series LCR

(a) When are the currents equal?

(b) Can Ib>IaI_b>I_aIb​>Ia​?

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Q.Two ac circuits are driven by identical ac sources of the same rms voltage. In circuit (a), a single resistor $R$ is connected across the source. In circuit (b), the same resistor $R$ is connected in series with a capacitor $C$ and an inductor $L$ across the source (a series LCR circuit).

(b) Can $I_b>I_a$ ?

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