(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its

Question

Q. (a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source.

(b) What is the phase difference between the voltages across inductor and the capacitor at resonance in the LCR circuit?

(c) When an inductor is connected to a 200 V dc voltage, a current of 1 A flows through it. When the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Explain, why? Also, calculate the self inductance of the inductor.

(OR)

(a) Draw the diagram of a device which is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.

(b) A small town with a demand of 1200 kW of electric power at 220 V is situated 20 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is

0.5\,\Omega

per km. The town gets the power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.

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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 — Transformer Principle

Transformer Principle: From Intuition to Precision

Imagine you have a water pipe with a narrow section and a wide section. Water flows through the narrow part fast but with low pressure; through the wide part it flows slow but with high pressure. The total amount of water (flow × pressure) stays the same. A transformer does something similar — but for electricity.

A transformer takes AC power at one voltage and current, and delivers nearly the same power at a different voltage and current. If voltage goes up, current must come down, and vice versa. The total power (voltage × current) is almost unchanged — minus a tiny loss.

The Core Idea: Mutual Induction

Two coils of wire are placed near each other, usually wound around a common iron core. When AC flows through the first coil (the primary), it creates a changing magnetic field. That changing field passes through the second coil (the secondary) and induces a voltage across it. This is mutual induction — a changing current in one coil induces a voltage in a neighbouring coil.

The iron core is crucial: it guides the magnetic field from one coil to the other with very little leakage, making the transfer efficient.

Note

A transformer works only with AC. A steady DC current produces a constant magnetic field, which induces nothing in the secondary coil. Change is essential.

The Precise Statement

For an ideal transformer (no energy losses), the relationship between primary and secondary voltages and currents is:

\frac{V_s}{V_p} = \frac{N_s}{N_p} \quad \text{and} \quad \frac{I_s}{I_p} = \frac{N_p}{N_s}

where:

$V_p$ , $V_s$ = primary and secondary voltages
$I_p$ , $I_s$ = primary and secondary currents
$N_p$ , $N_s$ = number of turns in primary and secondary coils

V_p I_p = V_s I_s

What This Means

If the secondary has more turns than the primary ( $N_s > N_p$ ), the secondary voltage is higher — this is a step-up transformer. Current in the secondary is correspondingly lower.

If the secondary has fewer turns ( $N_s < N_p$ ), the secondary voltage is lower — a step-down transformer. Current in the secondary is higher.

Watch out

A step-up transformer raises voltage but lowers current. It does not create energy. The product $V \times I$ stays constant (ignoring losses). Many beginners think a step-up transformer "amplifies" power — it does not.

Why the Turns Ratio Works

The voltage induced in each turn of a coil is the same (because the same changing magnetic flux links every turn). So the total induced voltage is proportional to the number of turns:

V_p \propto N_p, \quad V_s \propto N_s

Dividing gives the ratio. For current, conservation of power forces the inverse relationship.

A Real Transformer: Small Losses …

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(a) Series-LCR impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ is minimum ( $=R$ ) at resonance; at resonance $V_L$ and $V_C$ are $180^\circ$ out of phase; the inductor's $L=\frac{2\sqrt3}{\pi}\approx1.10$ H.

(b) A step-down transformer lowers ac voltage; for the transmission line the current is 300 A, line resistance 20 $\Omega$ , so the power lost as heat is 1800 kW.

Graph of the impedance Z of a series LCR circuit plotted against the angular frequency omega of the ac source, showing the characteristic U-shaped curve whose minimum value Z = R occurs exactly at the resonant frequency omega-naught, where the inductive and capacitive reactances are equal and cancel.

Part (a)

Impedance of a series LCR circuit

The resistor opposes current by $R$ ; the inductor by $X_L=\omega L$ (rising with frequency); the capacitor by $X_C=1/\omega C$ (falling with frequency). Because $V_L$ leads and $V_C$ lags the current by $90^\circ$ , they are $180^\circ$ apart and combine as $|X_L-X_C|$ . Phasor addition of $V_R$ (along the current) with $(V_L-V_C)$ gives

V_0=I_0\sqrt{R^2+(X_L-X_C)^2}\ \Rightarrow\ Z=\sqrt{R^2+\Big(\omega L-\frac{1}{\omega C}\Big)^2}.

Variation with frequency. At $\omega\to0$ , $X_C\to\infty$ so $Z\to\infty$ ; as $\omega$ rises $X_C$ falls and $X_L$ rises until, at $\omega_0=1/\sqrt{LC}$ , $X_L=X_C$ and $Z=R$ (minimum); beyond resonance $X_L>X_C$ and $Z$ rises again. The plot of $Z$ versus $f$ is a U-shaped curve with its minimum $R$ at $f_0=\frac{1}{2\pi\sqrt{LC}}$ .

(b) Phase difference at resonance …

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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 47 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 ANNUAL1 markMCQ
$Q. An ideal transformer has 500 turns in the primary and 5000 turns in the secondary. If the primary be connected to a 6 V battery, then the secondary voltage is$
(a) 0
(b) 0.6 V
(c) 60 V
(d) 6 V
›Reveal solutionSolution

A transformer needs a changing current/flux to work. A DC battery gives a constant current, so once steady state is reached the secondary voltage is 0.

A transformer works on the principle of mutual induction: the emf induced in the secondary is

$e_s = -M\dfrac{di_p}{dt}$

…

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CBSE 2026Set ANNUAL1 mark
$Q. Why cannot a transformer be used to step up direct current (D.C.)?$
›Reveal solutionSolution

No changing flux, no induced EMF — a transformer needs AC to work at all.

A transformer operates on the principle of mutual induction: a time-varying current in the primary coil produces a time-varying magnetic flux in the core, which links the secondary coil and induces an EMF in it, given by $\varepsilon_2 = -N_2\dfrac{d\Phi}{dt}$ . With a constant DC current in the primary, the flux in the core, once established, remains steady (constant) — its rate of change $d\Phi/dt$ is zero in the steady state. Since the induced EMF depends entirely on this rate of change, no EMF (and hence no stepped-up voltage) is induced in the secondary for steady DC, so a t …

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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 X11 markMCQ
$Q. Transformer cores are usually laminated. This is to reduce energy loss due to$
(a) flux leakage
(b) winding resistance
(c) eddy currents
(d) hysteresis
›Reveal solutionSolution

(c) eddy currents. The changing flux in the core induces circulating (eddy) currents in the solid metal, which dissipate energy as heat ( $\propto$ resistance path). Laminating the core with thin i …

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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
$Q. Which quantity is increased in a step-up transformer ?$
(a) current
(b) voltage
(c) power
(d) frequency
›Reveal solutionSolution

A step-up transformer increases voltage (and correspondingly decreases current), since power and frequency stay the same.

For an ideal transformer,

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$ …

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CBSE 2025Set ANNUAL1 mark
$Q. On which principle does transformer work?$
›Reveal solutionSolution

A transformer transfers energy from primary to secondary coil through mutual induction of a changing magnetic flux.

A transformer consists of two coils (primary and secondary) wound on a common laminated soft-iron core. When an alternating current is passed through the primary coil, it produces a continuously changing magnetic flux in the core. Since the secondary coil is linked to the same core, this changing flux also links the secondary coil.

By Faraday's law of electromagnetic induction, a changing flux linked with the secondary coil induces an alternating emf in it — this is mutual induction, i.e., induction of emf in one coil due to a changing current in a nearby (magnetically coupled) coil.

…

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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 mark
$Q. Why is electric power transmission from power stations to sub-stations near consumers done at high voltages ?$
›Reveal solutionSolution

For a fixed power to be delivered, $P=VI$ , so raising the transmission voltage lowers the current; since resistive line loss goes as $I^2R$ , a lower current means far less energy is wasted as heat.

Electrical power transmitted is $P = VI$ . For a given power $P$ to be delivered by the transmission line, increasing the transmission voltage $V$ proportionally decreases the current $I = P/V$ .

The power dissipated as heat in the transmission line's resistance $R$ is

$P_{loss} = I^2 R$

…

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Q.(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with frequency of the ac source.

Part (a)

Impedance of a series LCR circuit

(b) Phase difference at resonance …

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