(a) (i) You are given three circuit elements X, Y and Z. They are connected one by one across a given ac sourc

Question

Q. (a)

(i) You are given three circuit elements X, Y and Z. They are connected one by one across a given ac source. It is found that V and I are in phase for element X; V leads I by

\frac{\pi}{4}

for element Y; while I leads V by

\frac{\pi}{4}

for element Z. Identify the elements X, Y and Z.

(ii) Establish the expression for the impedance of the circuit when elements X, Y and Z are connected in series to an ac source. Show the variation of current in the circuit with the frequency of the applied ac source.

(iii) In a series LCR circuit, obtain the conditions under which

(1) the impedance is minimum, and

(2) wattless current flows in the circuit.

(OR)

(b)

(i) Describe the construction and working of a transformer and hence obtain the relation for

\frac{V_s}{V_p}

in terms of the number of turns of the primary and secondary.

(ii) Discuss four main causes of energy loss in a real transformer.

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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 Turns Ratio

What the Turns Ratio Really Means

Imagine you have two coils of wire wrapped around a shared iron core. One coil (the primary) has a certain number of loops — say 100 turns. The other (the secondary) has a different number — say 200 turns. When you feed AC voltage into the primary, it creates a changing magnetic field that cuts through both coils. That field induces a voltage in the secondary.

The key insight: each turn of wire gets the same "share" of the magnetic push. If one turn in the primary gets 1 volt induced across it, then every turn in the secondary also gets 1 volt. So if the secondary has twice as many turns, it gets twice the total voltage.

That's the whole idea in a nutshell. The turns ratio directly determines how voltage transforms from primary to secondary.

The Precise Relation

The voltage across each coil is proportional to the number of turns:

\frac{V_s}{V_p} = \frac{N_s}{N_p}

Where $V_p$ and $V_s$ are the primary and secondary voltages, and $N_p$ and $N_s$ are the number of turns on each coil.

\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}

The current relation follows from conservation of energy. An ideal transformer doesn't create or destroy power — it just transforms it. So the power in equals the power out:

V_p I_p = V_s I_s

Rearranging gives $\frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$ .

Watch out

The current ratio is the inverse of the turns ratio. If you step up voltage (more secondary turns), the secondary current drops proportionally. Many students flip this by accident.

What This Tells You

If $N_s > N_p$ , you have a step-up transformer — secondary voltage is higher, secondary current is lower. If $N_s < N_p$ , it's a step-down transformer — secondary voltage is lower, secondary current is higher.

The turns ratio itself is often written as $a = \frac{N_p}{N_s}$ (primary over secondary). Then:

V_s = \frac{V_p}{a}, \quad I_s = a I_p

A Concrete Example

A transformer has 500 turns on the primary and 100 turns on the secondary. The primary is connected to 230 V AC.

Turns ratio $a = \frac{500}{100} = 5$ . This is a step-down transformer.

Secondary voltage: $V_s = \frac{230}{5} = 46$ V. …

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Figure — Part (a)(ii) explicitly asks to 'Show the variation of current in the circuit with the frequency of the applie

Part (a): X = R, Y = L, Z = C; series impedance $Z=\sqrt{R^2+(\omega L-1/\omega C)^2}$ , current peaks at resonance $\omega_0=1/\sqrt{LC}$ (minimum $Z=R$ ); wattless current needs $R=0$ . Part (b): A transformer gives $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$ ; its four main losses are copper, eddy-current, hysteresis and flux-leakage.

Part (a) — Series LCR circuit

(i) Identifying the elements

X: V and I are in phase → a pure resistor (R).
Y: voltage leads current → an inductor (L) (for a pure inductor V leads I by $\pi/2$ ).
Z: current leads voltage → a capacitor (C) (for a pure capacitor I leads V by $\pi/2$ ).

(ii) Impedance in series and frequency response

The three in series carry the same current; their voltage phasors are $V_R$ (in phase), $V_L$ (leading by $90^\circ$ ) and $V_C$ (lagging by $90^\circ$ ). Since $V_L$ and $V_C$ are opposite, the reactances subtract. With $X_L=\omega L$ and $X_C=\dfrac{1}{\omega C}$ ,

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

Variation with frequency:

Low $\omega$ : $X_C=\dfrac{1}{\omega C}$ is huge → $Z$ large → $I$ small.
High $\omega$ : $X_L=\omega L$ is huge → $Z$ large → $I$ small.
At the resonant frequency $\omega_0=\dfrac{1}{\sqrt{LC}}$ , $X_L=X_C$ so $Z=R$ (minimum) and $I=\dfrac{V}{R}$ (maximum).

The $I$ – $\omega$ graph is a resonance curve peaking at $\omega_0$ .

(iii) Special conditions

(1) Minimum impedance. $Z$ is least when $(X_L-X_C)^2=0$ , i.e.

\omega L=\frac{1}{\omega C}\ \Rightarrow\ \omega_0=\frac{1}{\sqrt{LC}},

giving $Z_{\min}=R$ (series 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 42 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 V11 mark
$Q.For a step-down transformer, the ratio of primary current to secondary current is __________. Fill in the blank choosing the appropriate answer from the bracket: (photons, diffraction, polarity, monopoles, greater than unity, less than unity)$
›Reveal solutionSolution

$N_s/N_p$ (equal to $V_s/V_p$ ), i.e. less than 1. …

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CBSE 2026Set A1 markMCQ
$Q. Which quantity decreases in step-down transformer? (A) Current (B) Voltage (C) Frequency (D) Power$
›Reveal solutionSolution

A step-down transformer reduces the secondary voltage; current rises to conserve power.

A transformer works by mutual induction. The turns ratio fixes the voltage ratio:

$\frac{V_s}{V_p}=\frac{N_s}{N_p}$

…

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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 D1 markMCQ
$Q. In a step-up transformer, the value of current in secondary compared to the primary coil is (A) equal (B) less (C) more (D) no relation between the two$
›Reveal solutionSolution

A step-up transformer increases voltage, so to conserve power the secondary current must be smaller than the primary current.

For an ideal transformer, power is conserved:

$V_p I_p = V_s I_s$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A transformer has 400 secondary turns and 500 primary turns. If the secondary voltage is 400 V, the primary voltage is:$
(a) 500V
(b) 400V
(c) 1000V
(d) 800V.
›Reveal solutionSolution

For a transformer, $\dfrac{V_p}{V_s}=\dfrac{N_p}{N_s}$ .

Given: $N_s = 400$ , $N_p = 500$ , $V_s = 400\,V$ .

Using the transformer turns-ratio relation: …

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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
$Q. In a transformer, the number of turns in the primary and the secondary are 410 and 1230 respectively. If the current in primary is 6 A, then that in the secondary coil is :$
(a) 12 A
(b) 2 A
(c) 1 A
(d) 18 A
›Reveal solutionSolution

Using the transformer current relation $I_sN_s=I_pN_p$ with $N_p=410$ , $N_s=1230$ , $I_p=6$ A gives $I_s=2$ A.

Working

For an ideal transformer, the turns and currents in primary/secondary satisfy

$I_pN_p = I_sN_s \ \Rightarrow\ I_s = I_p\dfrac{N_p}{N_s}$

…

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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)

Part (a) — Series LCR circuit

(i) Identifying the elements

(ii) Impedance in series and frequency response

(iii) Special conditions

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