(a) (i) A resistor and a capacitor are connected in series to an ac source v = vm sinω t. Derive an expression

Question

Q. (a)

(i) A resistor and a capacitor are connected in series to an ac source

v = v_m \sin\omega t

. Derive an expression for the impedance of the circuit.

(ii) When does an inductor act as a conductor in a circuit ? Give a reason for it.

(iii) An electric lamp is designed to operate at

110\ \text{V}

dc and

11\ \text{A}

current. If the lamp is operated on a

220\ \text{V}

50\ \text{Hz}

ac source with a coil in series, then find the inductance of the coil.

(OR)

(b)

(i) Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes for energy losses in a real transformer.

(ii) A step-up transformer converts a low voltage into high voltage. Does it violate the principle of conservation of energy ? Explain.

(iii) A step-up transformer has

200

and

3000

turns in its primary and secondary coils respectively. The input voltage given to the primary coil is

90\ \text{V}

. Calculate :

(1) the output voltage across the secondary coil,

(2) the current in the primary coil if the current in the secondary coil is

2.0\ \text{A}

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Part (a)Concept understanding — Capacitive Reactance

Capacitive Reactance: The AC Resistance of a Capacitor

When you first meet a capacitor in a DC circuit, it behaves like a break in the wire once it's fully charged — no current flows. But in an AC circuit, something entirely different happens. The voltage keeps reversing, so the capacitor never finishes charging. It's constantly being filled, emptied, refilled, and re-emptied. This continuous back-and-forth means current does flow, but the capacitor resists that flow in a frequency-dependent way. That resistance is called capacitive reactance.

The Intuition: Why Frequency Matters

Imagine a water pipe with a flexible rubber membrane stretched across it (a crude capacitor). If you push water slowly from one side, the membrane bulges and eventually stops the flow — that's DC. But if you push and pull the water rapidly (AC), the membrane just vibrates, and water sloshes back and forth through the pipe. The faster you push-pull (higher frequency), the less the membrane impedes the flow. At very high frequencies, it's almost like the membrane isn't there.

In a capacitor, the "membrane" is the electric field between the plates. Higher frequency means the voltage changes faster, so the capacitor has less time to oppose the current. The result: capacitive reactance decreases as frequency increases.

The Precise Statement

Capacitive reactance $X_C$ is the opposition a capacitor offers to alternating current. It is measured in ohms ( $\Omega$ ), just like resistance. The formula is:

X_C = \frac{1}{2\pi f C}

Where:

$X_C$ = capacitive reactance (ohms)
$f$ = frequency of the AC signal (hertz)
$C$ = capacitance (farads)

What the Formula Tells You

Three key relationships jump out:

Inverse with frequency: Double the frequency, halve the reactance. At DC ( $f = 0$ ), $X_C$ becomes infinite — the capacitor blocks DC completely.
Inverse with capacitance: A larger capacitor (more farads) offers less opposition. It can store more charge per volt, so it "gives way" more easily.
No power dissipation: Unlike a resistor, a pure capacitor doesn't convert electrical energy to heat. Reactance is a reactive opposition — energy is stored and returned, not lost.

Watch out

Do not confuse capacitive reactance with resistance. Resistance dissipates energy as heat; reactance stores and releases it. A capacitor in an AC circuit has zero real power loss (in the ideal case).

Phase: The Hidden Twist

There's a critical detail that separates reactance from resistance. In a purely resistive circuit, voltage and current peak at the same time — they are in phase. In a purely capacitive circuit, current leads voltage by $90^\circ$ (or $\pi/2$ radians).

Why? Because current is the rate of change of charge: $I = C \frac{dV}{dt}$ . When the voltage is at its peak (not changing), the current is zero. When the voltage is crossing zero (changing fastest), the current is maximum. This quarter-cycle shift is baked into the definition of reactance. …

Why this formula?

Capacitive Reactance: Why $X_C = \frac{1}{\omega C}$ ?

Let’s build the intuition from the ground up — starting with what a capacitor does in a circuit.

1. The Fundamental Behavior of a Capacitor

A capacitor stores charge. The defining equation is:

Q = C V

where:

$Q$ = charge on the plates (in coulombs)
$C$ = capacitance (in farads)
$V$ = voltage across the plates

But in an AC circuit, voltage changes continuously. So charge must also change — meaning current flows.

2. Relating Current to Voltage

Current is the rate of flow of charge:

I = \frac{dQ}{dt}

Substitute $Q = C V$ :

I = \frac{d}{dt}(C V)

If $C$ is constant (which it is for a fixed capacitor):

I = C \frac{dV}{dt}

Key insight: The current through a capacitor is proportional to the rate of change of voltage, not the voltage itself.

3. Applying a Sinusoidal Voltage

In AC circuits, voltage is typically sinusoidal:

V(t) = V_0 \sin(\omega t)

where:

$V_0$ = peak voltage
$\omega = 2\pi f$ = angular frequency (rad/s)

Now find the current:

I(t) = C \frac{d}{dt}[V_0 \sin(\omega t)] = C V_0 \cdot \omega \cos(\omega t)

So:

I(t) = \omega C V_0 \cos(\omega t)

4. The Phase Shift — Why It Matters

Notice:

Voltage: $\sin(\omega t)$
Current: $\cos(\omega t) = \sin(\omega t + 90^\circ)$

Current leads voltage by $90^\circ$ in a pure capacitor. This is the opposite of an inductor (where current lags).

5. Extracting the Reactance

Compare the amplitudes:

Voltage amplitude: $V_0$
Current amplitude: $I_0 = \omega C V_0$

By Ohm’s law for AC (magnitude only):

\text{Reactance} = \frac{\text{Voltage amplitude}}{\text{Current amplitude}} = \frac{V_0}{\omega C V_0} = \frac{1}{\omega C}

Thus:

X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}

6. Why "Reactance" and Not "Resistance"?

Resistance ( $R$ ) dissipates energy as heat. …

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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Part (a): $Z_{RC}=\sqrt{R^2+(1/\omega C)^2}$ ; an inductor acts as a conductor at dc ( $X_L=0$ ); the choke needed for the lamp is $L=\sqrt3/10\pi\approx0.055$ H.

Part (b): a step-up transformer works by mutual induction ( $E_s/E_p=N_s/N_p$ ), with copper, eddy-current and hysteresis losses; energy is conserved because current steps down as voltage steps up; $V_s=1350$ V and $I_p=30$ A.

Schematic diagram of a transformer showing the primary and secondary windings wound on a common laminated soft-iron core, with the primary leads on the left and the secondary leads on the right -- the standard NCERT construction diagram used to explain a step-up transformer's working.

Part (a) — RC circuit, inductor at dc, and the choke

Impedance of series RC. With $v=v_m\sin\omega t$ , the resistor voltage is in phase with the current while the capacitor voltage lags by $90^\circ$ . Adding as phasors, $v_m=i_m\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}\;\Rightarrow\; Z=\frac{v_m}{i_m}=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2},\quad X_C=\frac{1}{\omega C}.$ The current leads the voltage, $\tan\phi=\dfrac{1}{\omega CR}$ .
Inductor as a conductor. $X_L=\omega L=2\pi fL$ . At $f=0$ (dc steady state) $X_L=0$ , so the inductor behaves like a plain conductor (short circuit).
Choke for the lamp. Lamp resistance $R=\dfrac{110}{11}=10\,\Omega$ . On the $220$ V, $50$ Hz supply the lamp must still draw $11$ A, so the series RL impedance is $Z=\dfrac{220}{11}=20\,\Omega$ . Then $Z^2=R^2+(\omega L)^2\Rightarrow(\omega L)^2=20^2-10^2=300\Rightarrow\omega L=10\sqrt3.$ $L=\frac{10\sqrt3}{\omega}=\frac{10\sqrt3}{2\pi\times50}=\frac{\sqrt3}{10\pi}\approx0.055\ \text{H}.$ …

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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 23 on this concept.

CBSE 2026Set 55/2/11 markMCQ
$X_C$
›Reveal solutionSolution
Figure — CBSE 2026 55/2/1 Q9

Capacitive reactance $X_C = \frac{1}{\omega C}$ is linear in $\frac{1}{\omega}$ with slope $\frac{1}{C}$ . Reading the slopes from the angles (tan 45° and tan 30°), we find $\frac{C_1}{C_2} = \frac{1}{\sqrt{3}}$ .

The capacitive reactance of an ideal capacitor is given by
$X_C = \frac{1}{\omega C}$
Rearranging this as $X_C = \frac{1}{C} \cdot \frac{1}{\omega}$ , we see that $X_C$ is directly proportional to $\frac{1}{\omega}$ . When we plot $X_C$ versus $\frac{1}{\omega}$ , we get a straight line passing through the origin with slope equal to $\frac{1}{C}$ .

The key insight: a steeper line means a larger slope, which means a larger value of $\frac{1}{C}$ , which in turn means a smaller capacitance. The graph shows two such lines for capacitors $C_1$ and $C_2$ , making angles of 45° and 30° respectively with the horizontal axis.

Find the slope of line $C_1$ : The line makes an angle of 45° with the $\frac{1}{\omega}$ axis. The slope is

$\text{slope}_{C_1} = \tan 45° = 1$
Since slope $= \frac{1}{C_1}$ , we have
$\frac{1}{C_1} = 1 \implies C_1 \propto 1$

Find the slope of line $C_2$ : The line makes an angle of 30° with the $\frac{1}{\omega}$ axis. The slope is

$\text{slope}_{C_2} = \tan 30° = \frac{1}{\sqrt{3}}$
Since slope $= \frac{1}{C_2}$ , we have …
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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 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 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 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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CBSE 2024Set 55/1/11 markMCQ
$C$
›Reveal solutionSolution
Capacitive reactance is $X = \frac{1}{2\pi \nu C}$ . Doubling $C$ and tripling $\nu$ multiplies the denominator by $6$ , so the new reactance becomes $\frac{X}{6}$ . The correct option is (A).

The key to this problem is understanding what capacitive reactance actually means physically. A capacitor in an AC circuit doesn't "resist" current the way a resistor does — instead, it opposes changes in voltage by storing and releasing charge. The faster the voltage changes (higher frequency) or the larger the capacitor (more charge storage per volt), the easier it is for current to flow. That's why reactance $X$ is inversely proportional to both capacitance $C$ and frequency $\nu$ .

Let's work through the change step by step.

Write the standard formula for capacitive reactance. For a capacitor of capacitance $C$ connected to an AC source of frequency $\nu$ , the reactance is:

$X = \frac{1}{2\pi \nu C}$
This is a direct relationship — no tricks, just the definition.

Identify the new values.

The capacitance is doubled: $C' = 2C$

The frequency is tripled: $\nu' = 3\nu$

Substitute these into the formula for the new reactance $X'$ .

$X' = \frac{1}{2\pi \nu' C'} = \frac{1}{2\pi (3\nu)(2C)}$

Simplify the denominator.

$X' = \frac{1}{2\pi \cdot 6 \cdot \nu C} = \frac{1}{6} \cdot \frac{1}{2\pi \nu C}$

Recognize the original reactance in the expression. Since $X = \frac{1}{2\pi \nu C}$ , we have: $X' = \frac{X}{6}$ …
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CBSE 2024Set 55/2/11 markMCQ
$Q. Which of the following quantity/quantities remains same in primary and secondary coils of an ideal transformer ? Current, Voltage, Power, Magnetic flux (A) Current only (B) Voltage only (C) Power only (D) Magnetic flux and Power both$
›Reveal solutionSolution
In an ideal transformer, the magnetic flux linking both coils is the same (by Faraday’s law), and power is conserved (no losses). Current and voltage change with the turns ratio. So the correct choice is (D) Magnetic flux and Power both.

The core idea

An ideal transformer is a perfect magnetic circuit with no energy losses — no resistance in the windings, no hysteresis, no eddy currents, and perfect coupling (all flux from the primary passes through the secondary).

Two fundamental principles govern it:

Faraday’s law of induction — the same changing magnetic flux $\Phi$ links every turn of both coils.

Conservation of energy — in the absence of losses, the power delivered to the primary must equal the power extracted from the secondary.

From these, everything else follows.

Step-by-step reasoning

Magnetic flux is the same in both coils In an ideal transformer, the core is assumed to have zero reluctance and no flux leakage. The alternating current in the primary creates a time-varying magnetic flux $\Phi(t)$ that is entirely confined to the core. Since both coils are wound on the same core, every turn of the secondary is linked by exactly the same flux as every turn of the primary. By Faraday’s law, the induced emf in each coil is proportional to the number of turns:

$\mathcal{E}_p = -N_p \frac{d\Phi}{dt}, \qquad \mathcal{E}_s = -N_s \frac{d\Phi}{dt}$
The flux $\Phi$ itself is identical — only the induced voltages differ because $N_p \neq N_s$ .

Voltage changes with the turns ratio From the above,

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$
So voltage is not the same in primary and secondary unless $N_p = N_s$ (which is not generally true).

Current changes inversely with the turns ratio For an ideal transformer, the magnetising current is negligible, and the primary current adjusts to balance the secondary load. Power conservation gives:

$V_p I_p = V_s I_s \quad \Rightarrow \quad \frac{I_s}{I_p} = \frac{N_p}{N_s}$
So current is also not the same.

Power is conserved …
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CBSE 2024Set IMPROVEMENT1 markMCQ
$Q. In actual transformer, reason of energy losses is —$
(a) Flux Leakage
(b) Eddy Currents
(c) Resistance of the windings
(d) All of above
›Reveal solutionSolution

A real transformer loses energy through flux leakage, eddy currents, AND winding resistance — all three act together.

An ideal transformer has no losses, but an actual transformer loses energy due to: (1) Flux leakage — not all of the flux produced by the primary links the secondary. (2) Eddy currents induced in the iron core, which dissipate energy as heat (reduced, but not eliminated, by lamination). (3) Resistance of the copper windin …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The core in transformers and other electromagnetic devices is laminated, so as to$
(a) increase the magnetic field
(b) increase the magnetic flux
(c) reduce the magnetism in the core
(d) reduce the eddy current losses in the core
›Reveal solutionSolution
A changing magnetic flux through any conducting core induces circulating "eddy" currents in the body of the core itself; these currents dissipate energy as heat ( $I^2R$ ) without doing any useful work. Laminating the core cuts this loss.

Why the core is laminated

In a transformer (or any a.c. electromagnetic device), the iron core carries a time-varying flux $\Phi(t)$ . By Faraday's law this changing flux induces an e.m.f. not only in the windings but also within the bulk of the iron itself, driving circulating eddy currents inside the core.

If the core were a single solid block, these eddy currents could flow in large loops of large cross-sectional area, and the power dissipated by eddy currents scales as $P_{eddy} \propto B_{max}^2\, f^2\, t^2$ where $t$ is the thickness of the conducting slab carrying the loop. …
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CBSE 2023Set 55/3/11 markMCQ
$Q. An inductor, a capacitor and a resistor are connected in series across an ac source of voltage. If the frequency of the source is decreased gradually, the reactance of :$
(a) both the inductor and the capacitor decreases.
(b) inductor decreases and the capacitor increases.
(c) both the inductor and the capacitor increases.
(d) inductor increases and the capacitor decreases.
›Reveal solutionSolution
The inductive reactance $X_L = 2\pi f L$ is directly proportional to frequency, and the capacitive reactance $X_C = \frac{1}{2\pi f C}$ is inversely proportional to frequency. As frequency decreases, $X_L$ decreases and $X_C$ increases — so option (b) is correct.

The core idea here is simple: reactance is not a fixed property — it depends on frequency. An inductor opposes changes in current, and the faster the current changes (higher frequency), the more it opposes. A capacitor, on the other hand, stores and releases charge; at higher frequencies, it has less time to charge up, so it offers less opposition.

Let’s see exactly how each behaves when frequency is lowered.

Inductive reactance is given by

$X_L = 2\pi f L$
Here $f$ is the frequency and $L$ is the inductance (a constant for a given inductor). Since $X_L$ is directly proportional to $f$ , decreasing $f$ makes $X_L$ smaller. So the inductor’s opposition weakens.

Capacitive reactance is given by

$X_C = \frac{1}{2\pi f C}$
$C$ is the capacitance (constant). Here $X_C$ is inversely proportional to $f$ — as $f$ goes down, $X_C$ goes up. So the capacitor’s opposition strengthens. …
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Q.(a)

Part (a) — RC circuit, inductor at dc, and the choke

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