Chapter 7

Alternating Current

CBSE · 6 subtopics

Formula SheetEvery formula this chapter can be examined on, in one table.

The ac source: basic quantities

Formula	Symbols, units and conditions
$v = v_m \sin \omega t$	$v$ = instantaneous voltage (V), $v_m$ = peak (amplitude) voltage (V), $\omega$ = angular frequency (rad s $^{-1}$ ), $t$ = time (s). This is the standard source used in every question of this chapter.
$\omega = 2\pi\nu = \dfrac{2\pi}{T}$	$\nu$ = frequency (Hz), $T$ = period (s). Use it to read $\nu$ off a given source equation: $v = v_m\sin(314\,t)$ means $\omega = 314$ , so $\nu = 50$ Hz.
$i = i_m \sin(\omega t + \phi)$	$i_m$ = peak current (A), $\phi$ = phase of the current relative to the source voltage. $\phi > 0$ means the current leads; $\phi < 0$ means it lags. Indian mains: $\nu = 50$ Hz, $V = 220$ V (rms), so $T = 0.02$ s and $v_m = \sqrt2 \times 220 = 311$ V.

rms (effective) values

Formula	Symbols, units and conditions
$I = I_{rms} = \dfrac{i_m}{\sqrt2} = 0.707\, i_m \qquad V = V_{rms} = \dfrac{v_m}{\sqrt2} = 0.707\, v_m$	Definition of rms: the value of steady dc that would produce the same average Joule heating in the same resistor in the same time. Every ac ammeter/voltmeter reads rms, and every rating (220 V, 100 W) is an rms/average rating.
$\langle \sin^2\omega t\rangle = \dfrac12 \qquad \langle \sin 2\omega t\rangle = \langle\cos 2\omega t\rangle = 0$	Averages over one complete cycle. These are the only averages needed to derive rms values and average power.

AC voltage applied to a resistor

Formula	Symbols, units and conditions
$i = i_m\sin\omega t \qquad i_m = \dfrac{v_m}{R} \qquad I = \dfrac{V}{R} \qquad \phi = 0$	$R$ in ohm ( $\Omega$ ). Voltage and current are in phase in a pure resistor; Ohm's law holds for peak values and for rms values separately.
$P = \dfrac12 i_m^2 R = I^2R = VI = \dfrac{V^2}{R}$	$P$ = average power over a cycle, in watt (W). Used for bulb-rating numericals: a bulb rated $P$ at $V$ has $R = V^2/P$ .

Phasors

Formula	Symbols, units and conditions
$v = v_m\sin\omega t \;\longleftrightarrow\; \text{phasor } \mathbf{V} \text{ of length } v_m \text{ at angle } \omega t$	A phasor is a rotating arrow whose length is the peak value and whose vertical projection is the instantaneous value. Angle between two phasors = phase difference between the quantities. Phasors are not vectors; they only add like vectors.

AC voltage applied to an inductor

Formula	Symbols, units and conditions
$X_L = \omega L = 2\pi\nu L$	$X_L$ = inductive reactance (ohm, $\Omega$ ), $L$ = self-inductance (henry, H). $X_L \propto \nu$ : a straight line through the origin on an $X_L$ – $\nu$ graph. For dc ( $\nu = 0$ ), $X_L = 0$ .
$i = i_m\sin\left(\omega t - \dfrac{\pi}{2}\right) \qquad i_m = \dfrac{v_m}{X_L} = \dfrac{v_m}{\omega L} \qquad I = \dfrac{V}{X_L}$	Current lags the voltage by $\pi/2$ (a quarter cycle) in a pure inductor.
$P_{av} = V I \cos\dfrac{\pi}{2} = 0$	An ideal inductor dissipates no average power over a complete cycle.

AC voltage applied to a capacitor

Formula	Symbols, units and conditions
$q = Cv = Cv_m\sin\omega t \qquad i = \dfrac{dq}{dt} = \omega C v_m\cos\omega t$	$q$ = instantaneous charge (C), $C$ = capacitance (farad, F). This is the starting line of the capacitor derivation.
$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi\nu C}$	$X_C$ = capacitive reactance (ohm, $\Omega$ ). $X_C \propto 1/\nu$ : a rectangular hyperbola on an $X_C$ – $\nu$ graph. For dc ( $\nu = 0$ ), $X_C \to \infty$ , so a capacitor blocks dc.
$i = i_m\sin\left(\omega t + \dfrac{\pi}{2}\right) \qquad i_m = \dfrac{v_m}{X_C} = \omega C v_m \qquad I = \dfrac{V}{X_C}$	Current leads the voltage by $\pi/2$ in a pure capacitor. In a pure capacitor $I = 2\pi\nu C V \propto \nu$ , a straight line through the origin on an $I$ – $\nu$ graph.
$P_{av} = V I\cos\dfrac{\pi}{2} = 0$	An ideal capacitor dissipates no average power over a complete cycle.

Series LCR circuit

Formula	Symbols, units and conditions
$L\dfrac{di}{dt} + iR + \dfrac{q}{C} = v_m\sin\omega t$	Kirchhoff's loop equation for the series LCR circuit; the phasor method solves it for the steady state.
$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2}$	$Z$ = impedance in ohm ( $\Omega$ ); the ac analogue of resistance. Minimum value $Z = R$ , at resonance.
$i_m = \dfrac{v_m}{Z} \qquad I = \dfrac{V}{Z}$	—
$\tan\phi = \dfrac{X_L - X_C}{R} = \dfrac{\omega L - 1/\omega C}{R}$	$\phi$ = angle by which the applied voltage leads the current. If $X_L > X_C$ the circuit is inductive (current lags); if $X_C > X_L$ it is capacitive (current leads); if $X_L = X_C$ , $\phi = 0$ .
$V = \sqrt{V_R^2 + (V_L - V_C)^2} \qquad V_R = IR,\; V_L = IX_L,\; V_C = IX_C$	Element voltages add as phasors, never algebraically — $V_L$ and $V_C$ are $180^\circ$ apart, and each is $90^\circ$ from $V_R$ .

Two-element special cases

Formula	Symbols, units and conditions
$\text{Series RL:}\quad Z = \sqrt{R^2 + X_L^2},\qquad \tan\phi = \dfrac{X_L}{R} = \dfrac{\omega L}{R}$	Voltage leads current; $V = \sqrt{V_R^2 + V_L^2}$ .
$\text{Series RC:}\quad Z = \sqrt{R^2 + X_C^2},\qquad \tan\phi = \dfrac{X_C}{R} = \dfrac{1}{\omega RC}$	Current leads voltage; $V = \sqrt{V_R^2 + V_C^2}$ .
$\text{Series LC only } (R \approx 0):\quad Z = \lvert X_L - X_C\rvert,\qquad I = \dfrac{V}{\lvert X_L - X_C\rvert},\qquad P = 0$	—

Resonance and sharpness

Formula	Symbols, units and conditions
$X_L = X_C \;\Rightarrow\; \omega_0 L = \dfrac{1}{\omega_0 C} \;\Rightarrow\; \omega_0 = \dfrac{1}{\sqrt{LC}} \qquad \nu_0 = \dfrac{1}{2\pi\sqrt{LC}}$	$\omega_0$ = resonant angular frequency (rad s $^{-1}$ ), $\nu_0$ = resonant frequency (Hz). Resonance needs both $L$ and $C$ present — an RL or RC circuit cannot resonate.
$\text{At resonance: } Z = R \text{ (minimum)},\quad i_m = \dfrac{v_m}{R} \text{ (maximum)},\quad \phi = 0,\quad \cos\phi = 1$	—
$C = \dfrac{1}{4\pi^2\nu_0^2 L} \qquad L = \dfrac{1}{4\pi^2\nu_0^2 C}$	Rearranged resonance condition; used whenever a question fixes the resonant frequency and asks for the missing element.
$Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 CR} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} = \dfrac{\omega_0}{2\Delta\omega}$	$Q$ = quality factor (sharpness of resonance), dimensionless, no SI unit. $2\Delta\omega = R/L$ is the bandwidth. Large $Q$ (large $L$ , small $R$ ) means a sharp, selective circuit.

Power in an ac circuit

Formula	Symbols, units and conditions
$p = v i = \dfrac{v_m i_m}{2}\left[\cos\phi - \cos(2\omega t + \phi)\right]$	Instantaneous power; the second term averages to zero over a cycle.
$P_{av} = \dfrac{v_m i_m}{2}\cos\phi = V I\cos\phi = I^2 Z\cos\phi = I^2 R$	$P_{av}$ in watt. $V$ , $I$ are rms values. Power is dissipated only in $R$ .
$\cos\phi = \dfrac{R}{Z} = \dfrac{R}{\sqrt{R^2 + (X_L - X_C)^2}}$	$\cos\phi$ = power factor, dimensionless, between 0 and 1. Pure $R$ : $\cos\phi = 1$ . Pure $L$ or pure $C$ : $\cos\phi = 0$ . Series LCR at resonance: $\cos\phi = 1$ and $P = V I = V^2/R$ .
$I_{\text{wattless}} = I\sin\phi \qquad I_{\text{power component}} = I\cos\phi$	The component $I\sin\phi$ (perpendicular to $V$ in the phasor diagram) carries no power and is called the wattless current.

Transformers

Formula	Symbols, units and conditions
$\varepsilon_p = -N_p\dfrac{d\phi_B}{dt} \qquad \varepsilon_s = -N_s\dfrac{d\phi_B}{dt}$	$N_p$ , $N_s$ = number of turns in primary and secondary; $\phi_B$ = flux per turn (Wb). Same flux links both coils in an ideal transformer.
$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s} = k$	$k$ = turns ratio (transformation ratio). $k > 1$ step-up ( $V_s > V_p$ , $I_s < I_p$ ); $k < 1$ step-down. Valid for peak or rms values alike, since both oscillate at the source frequency.
$I_p V_p = I_s V_s \qquad (\text{ideal, } 100\% \text{ efficient})$	—
$\eta = \dfrac{P_{out}}{P_{in}} = \dfrac{V_s I_s}{V_p I_p} \times 100\%$	$\eta$ = efficiency; a well-designed transformer exceeds 95%.
$P_{loss} = I^2 R_{line} \qquad R_{line} = 2 \times (\text{length in km}) \times (\text{resistance per km})$	$I = P/V_{line}$ . The factor 2 is for the two-wire line (go and return). This is the standard long-distance transmission numerical.

Numerical short-cuts CBSE sets

Formula	Symbols, units and conditions
$R = \dfrac{V^2}{P}\;\text{(rated bulb)} \qquad P_{\text{actual}} = I^2R \;\text{(bulbs in series)}$	In a series combination the same $I$ flows, so the bulb with the largest $R$ (the lowest wattage rating) dissipates the most power.
$\text{Coil on dc: } R = \dfrac{V_{dc}}{I_{dc}} \qquad \text{same coil on ac: } Z = \dfrac{V_{ac}}{I_{ac}},\quad X_L = \sqrt{Z^2 - R^2},\quad L = \dfrac{X_L}{2\pi\nu}$	—
$\text{Capacitors in parallel: } C_{eq} = C + C_1 \quad\text{(used to bring the power factor to unity)}$	—

Constants and values CBSE expects

$\sqrt2 = 1.414$ , $\dfrac{1}{\sqrt2} = 0.707$ , $\pi = 3.14$ (unless the paper prints $\sqrt2 = 1.4$ , in which case use the printed value).
SI units: $R$ , $X_L$ , $X_C$ , $Z$ in ohm ( $\Omega$ ); $L$ in henry (H); $C$ in farad (F); $\omega$ in rad s $^{-1}$ ; $\nu$ in hertz (Hz); $P$ in watt (W). $Q$ and $\cos\phi$ are dimensionless.
Prefixes: $1\ \text{mH} = 10^{-3}$ H, $1\ \mu\text{F} = 10^{-6}$ F, $1\ \text{mF} = 10^{-3}$ F, $1\ \text{nF} = 10^{-9}$ F.

Short NotesThe chapter subtopic by subtopic, in the textbook's own order.6 subtopics

The whole chapter in brief

What alternating current means

An alternating voltage varies sinusoidally with time and periodically reverses direction: $v = v_m\sin\omega t$ . The current it drives, $i = i_m\sin(\omega t+\phi)$ , also reverses direction every half cycle.
On an $i$ – $t$ graph, ac is the curve that is symmetric about the time axis — equal positive and negative excursions, repeatedly crossing zero. A curve that stays entirely positive (however it wiggles) is a varying dc, not ac.
Average of $i$ over a full cycle is zero, but the average of $i^2$ is not — so Joule heating $i^2R$ still occurs and energy is still dissipated.
Why ac is preferred over dc: (i) ac voltage can be stepped up or stepped down easily and efficiently by a transformer; (ii) that makes economical long-distance transmission possible, because stepping up the voltage cuts the current and therefore the $I^2R$ line loss.

rms (effective) values

$I = \dfrac{i_m}{\sqrt2} = 0.707\,i_m$ and $V = \dfrac{v_m}{\sqrt2} = 0.707\,v_m$ .
Definition to write: the rms value of an alternating current is the value of that steady direct current which produces the same average heating effect in a given resistor in a given time.
Every quoted ac value is rms unless the word "peak" or "amplitude" appears. Mains "220 V" is rms; its peak is $311$ V. Mains "240 V" has a peak of $340$ V.
Key cycle averages: $\langle\sin^2\omega t\rangle = 1/2$ ; $\langle\sin 2\omega t\rangle = \langle\cos2\omega t\rangle = 0$ .

Phasors

A phasor is a rotating arrow: its length is the peak value of the quantity, it rotates anticlockwise with angular speed $\omega$ , and its vertical projection at any instant gives the instantaneous value.
The angle between two phasors is the constant phase difference between the two quantities.
Phasors are drawn like vectors but voltage and current are scalars; the arrows are a device for adding sinusoids of the same frequency.

AC voltage applied to a resistor

$v = v_m\sin\omega t \Rightarrow i = \dfrac{v_m}{R}\sin\omega t = i_m\sin\omega t$ . Current and voltage are in phase: $\phi = 0$ .
Phasor diagram: $\mathbf V$ and $\mathbf I$ are along the same line, angle zero, at all times.
Graph: the $v$ and $i$ curves rise, peak and cross zero at the same instants.
Average power $P = \dfrac12 i_m^2R = I^2R = VI = V^2/R$ , and it is a maximum for a given $V$ and $I$ (power factor unity).

AC voltage applied to an inductor

Loop rule with no resistor: $v - L\dfrac{di}{dt} = 0$ , so $\dfrac{di}{dt} = \dfrac{v_m}{L}\sin\omega t$ . Integrating (the constant of integration is zero, since the current oscillates symmetrically about zero):

$i = -\dfrac{v_m}{\omega L}\cos\omega t = \dfrac{v_m}{\omega L}\sin\left(\omega t - \dfrac{\pi}{2}\right) = i_m\sin\left(\omega t - \dfrac{\pi}{2}\right)$

Current lags the voltage by $\pi/2$ (one quarter cycle). Inductive reactance $X_L = \omega L = 2\pi\nu L$ in ohm, with $i_m = v_m/X_L$ .
$X_L \propto \nu$ and $X_L \propto L$ : the $X_L$ – $\nu$ graph is a straight line through the origin. An inductor offers no reactance to dc.
Average power supplied to an ideal inductor over a complete cycle is zero: $p = -\dfrac{v_m i_m}{2}\sin2\omega t$ , whose average vanishes. The inductor stores energy in its magnetic field for a quarter cycle and returns all of it in the next.
Standard board application: inserting an iron rod into the coil raises $L$ (high permeability), so $X_L$ and $Z$ rise, the current falls, and a series bulb dims. Removing it makes the bulb brighter. Reducing the number of turns lowers $L$ and also brightens the bulb.

AC voltage applied to a capacitor

$q = Cv = Cv_m\sin\omega t$ , so $i = \dfrac{dq}{dt} = \omega Cv_m\cos\omega t = i_m\sin\left(\omega t + \dfrac{\pi}{2}\right)$ .
Current leads the voltage by $\pi/2$ . Capacitive reactance $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi\nu C}$ in ohm, with $i_m = v_m/X_C$ .
$X_C \propto 1/\nu$ : the $X_C$ – $\nu$ graph is a rectangular hyperbola in the first quadrant, very large as $\nu\to0$ and tending to zero as $\nu\to\infty$ . Hence a capacitor blocks dc but passes ac.
Current in a pure capacitor, $I = 2\pi\nu C V$ , increases linearly with frequency — the $I$ – $\nu$ graph is a straight line through the origin.
Average power supplied to an ideal capacitor over a complete cycle is zero.
Standard board application: a lamp in series with a capacitor does not glow on dc but glows on ac; reducing $C$ raises $X_C$ and dims it, while inserting a dielectric raises $C$ , lowers $X_C$ and brightens it.

The three elements compared

Element	Reactance / opposition	Phase of current w.r.t. voltage	Effect of raising frequency	Average power
Resistor $R$	$R$ , independent of $\nu$	in phase, $\phi = 0$	no change	$I^2R$ , maximum
Inductor $L$	$X_L = 2\pi\nu L$	lags by $\pi/2$	opposition rises	zero
Capacitor $C$	$X_C = \dfrac{1}{2\pi\nu C}$	leads by $\pi/2$	opposition falls	zero

Series LCR circuit

Same current $i = i_m\sin(\omega t+\phi)$ flows through all three; the voltages differ in phase.
Phasor construction: draw $\mathbf I$ along a reference direction; $\mathbf V_R = IR$ along $\mathbf I$ ; $\mathbf V_L = IX_L$ at $90^\circ$ ahead of $\mathbf I$ ; $\mathbf V_C = IX_C$ at $90^\circ$ behind $\mathbf I$ . $\mathbf V_L$ and $\mathbf V_C$ are anti-parallel, so they combine to a single phasor of magnitude $\lvert V_L - V_C\rvert$ perpendicular to $\mathbf V_R$ . By Pythagoras,

$V = \sqrt{V_R^2 + (V_L - V_C)^2} = I\sqrt{R^2 + (X_L - X_C)^2} = IZ$

Impedance $Z = \sqrt{R^2 + (X_L-X_C)^2}$ ; impedance triangle: base $R$ , height $(X_L - X_C)$ , hypotenuse $Z$ , angle $\phi$ at the base.
$\tan\phi = \dfrac{X_L - X_C}{R}$ , the angle by which the applied voltage leads the current.
$X_L > X_C$ : inductive, current lags. $X_C > X_L$ : capacitive, current leads. $X_L = X_C$ : in phase.
Series RL: $Z = \sqrt{R^2+X_L^2}$ , $\tan\phi = X_L/R$ , voltage leads. Series RC: $Z = \sqrt{R^2+X_C^2}$ , $\tan\phi = X_C/R = 1/\omega RC$ , current leads.

Resonance

Resonance is the condition $X_L = X_C$ , i.e. $\omega_0 = \dfrac{1}{\sqrt{LC}}$ , $\nu_0 = \dfrac{1}{2\pi\sqrt{LC}}$ .
At resonance: $Z = R$ (minimum), $I = V/R$ (maximum), $\phi = 0$ , power factor $= 1$ , power $= V I = V^2/R$ (maximum).
$V_L$ and $V_C$ are equal in magnitude and $180^\circ$ out of phase at resonance, so they cancel exactly; the whole source voltage appears across $R$ . Each of $V_L$ and $V_C$ individually can be far larger than the source voltage.
Resonance requires both $L$ and $C$ . An RL or RC circuit never resonates.
$Z$ – $\nu$ graph: large at low $\nu$ (capacitive), falling to a minimum $R$ at $\nu_0$ , rising again at high $\nu$ (inductive) — a V-shaped curve.
$I$ – $\nu$ graph: a peak at $\nu_0$ of height $V/R$ . A smaller $R$ gives a taller, narrower peak.
Applications: tuning a radio or TV (the capacitance is varied until $\nu_0$ matches the station), and metal detectors (a metal object changes the circuit impedance away from resonance and the change in current triggers the alarm).

Sharpness of resonance and quality factor

Sharpness of resonance is measured by the quality factor $Q$ : the ratio of the voltage across $L$ (or across $C$ ) at resonance to the voltage across $R$ .

$Q = \dfrac{\omega_0 L}{R} = \dfrac{1}{\omega_0 CR} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} = \dfrac{\omega_0}{2\Delta\omega}$

$2\Delta\omega = R/L$ is the bandwidth: the width of the resonance curve where the current falls to $1/\sqrt2$ of its peak.
$Q$ is a pure number with no SI unit. Large $L$ and small $R$ give large $Q$ , a sharp peak, and high selectivity — the circuit responds strongly only to a narrow band about $\omega_0$ , which is what makes tuning precise.

Power in an ac circuit and the power factor

$P_{av} = V I\cos\phi = I^2Z\cos\phi = I^2R$

$\cos\phi = R/Z$ is the power factor: it measures how close a circuit is to expending the maximum power. It is dimensionless and lies between 0 and 1.
Case by case: pure $R$ gives $\cos\phi = 1$ , maximum dissipation; pure $L$ or pure $C$ gives $\cos\phi = 0$ , no dissipation; a general LCR circuit has $0 < \cos\phi < 1$ and dissipates power only in $R$ ; at resonance $\cos\phi = 1$ again.
Wattless current: when $\cos\phi = 0$ a current flows but no power is consumed. In general, resolving $\mathbf I$ into $I\cos\phi$ along $\mathbf V$ (the power component) and $I\sin\phi$ perpendicular to it (the wattless component) shows only the first carries energy.
A low power factor in a transmission circuit means a large current for the same delivered power, hence a large $I^2R$ loss. The remedy is to connect a capacitor of suitable value in parallel so that the leading wattless current cancels the lagging one, pushing $\cos\phi$ towards 1.

Transformers

Function: to change (step up or step down) an alternating voltage. Principle: mutual induction.
Construction: a primary coil of $N_p$ turns and a secondary coil of $N_s$ turns wound on a common laminated soft-iron core, either one on top of the other or on separate limbs of the core.
Working: alternating $V_p$ on the primary drives an alternating current, producing an alternating flux $\phi_B$ in the core. The same flux links the secondary and induces $\varepsilon_s = -N_s\,d\phi_B/dt$ ; the back emf in the primary is $\varepsilon_p = -N_p\,d\phi_B/dt \approx V_p$ .

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} \qquad\text{and, for an ideal transformer,}\qquad I_pV_p = I_sV_s \Rightarrow \dfrac{I_p}{I_s} = \dfrac{N_s}{N_p}$

$N_s > N_p$ : step-up, voltage raised and current lowered. $N_s < N_p$ : step-down, voltage lowered and current raised. A step-up transformer does not create energy — the current falls in the same proportion as the voltage rises.
Assumptions behind the relations: primary resistance and primary current are small, all the flux links both coils (no leakage), and the secondary current is small.
A transformer cannot work on dc: a steady current gives a constant flux, so no emf is induced in the secondary.

Energy losses in a transformer

Loss	Cause	How it is reduced
Copper loss	$I^2R$ heating in the resistance of the windings	use thick copper wire, especially in the low-voltage high-current winding
Eddy-current (iron) loss	circulating currents induced in the core by the changing flux	laminate the core into thin, mutually insulated sheets
Hysteresis loss	repeated magnetisation and demagnetisation of the core each cycle	use a core material of low hysteresis loss, such as soft iron
Flux leakage	not all the primary flux links the secondary	wind the two coils one over the other on a well-designed core

Magnetostriction (the humming of a transformer) is a further small loss. A thick primary wire is not a cause of loss — it is the cure for copper loss.

Long-distance transmission of power

The generator output is stepped up to a high voltage, so that for the same power the line current, and hence the $I^2R$ line loss, is small.
The high-voltage power is carried to an area sub-station, stepped down there, stepped down again at distributing sub-stations and utility poles, and finally reaches homes at about 240 V.
Line resistance for a two-wire line: $R_{line} = 2\times(\text{length in km})\times(\text{resistance per km})$ ; line current $I = P/V_{line}$ ; loss $= I^2R_{line}$ .

Standard graphs you may be asked to draw

$v$ and $i$ against $\omega t$ for a resistor: two sinusoids in step.
$v$ and $i$ against $\omega t$ for an inductor: the current curve shifted a quarter cycle to the right of the voltage curve.
$v$ and $i$ against $\omega t$ for a capacitor: the current curve shifted a quarter cycle to the left of the voltage curve.
Power curve for a pure $L$ or pure $C$ : a sinusoid of twice the frequency with equal areas above and below the axis.
$X_L$ against $\nu$ : straight line through the origin. $X_C$ against $\nu$ : rectangular hyperbola.
$Z$ against $\nu$ for series LCR: V-shaped with minimum $R$ at $\nu_0$ .
$I$ against $\nu$ for series LCR, two resistances $R_1 > R_2$ : both peak at $\nu_0$ ; the $R_2$ curve is taller and sharper.

Traps that cost marks

Do not add $V_R$ , $V_L$ and $V_C$ algebraically. Use $V = \sqrt{V_R^2 + (V_L-V_C)^2}$ . The arithmetic sum can legitimately exceed the source voltage — that is the "paradox" CBSE asks you to resolve, and the resolution is the $90^\circ$ phase differences.
Sign convention for $\phi$ : NCERT writes $i = i_m\sin(\omega t+\phi)$ with $\tan\phi = (X_C-X_L)/R$ ; most board answers write $\tan\phi = (X_L-X_C)/R$ for the angle by which the voltage leads the current. Either is accepted, but state in words which quantity leads.
"Current leads voltage" means capacitive; "current lags voltage" means inductive. Getting these the wrong way round loses the identification mark.
A phase difference of exactly $\pi/2$ is impossible if any resistance is present: $\phi = \tan^{-1}(X/R) < \pi/2$ always.
rms versus peak: if the source is given as $V = V_0\sin\omega t$ , then $V_0$ is a peak value — divide by $\sqrt2$ before computing average power.
At resonance the impedance is $R$ , not zero, and the power factor is 1, not 0.
$Q$ has no unit; do not write ohm or hertz for it.
A transformer steps voltage up or down only for ac. On dc it does nothing (and may burn the primary).
Laminating the core reduces eddy currents, not hysteresis; low-hysteresis material reduces hysteresis, and thick wire reduces copper loss. Do not swap the remedies.
Average power in a pure $L$ or pure $C$ is exactly zero — say "zero", not "small".
In a series combination of bulbs, the lowest-wattage bulb has the largest resistance and therefore glows brightest, because $P = I^2R$ with the same $I$ .

Solved Numerical ExamplesWorked numericals to attempt first, each with a hint before the full solution.Coming soon

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

Important DerivationsEvery derive / obtain / prove question in this chapter.11

1
Derivation 1
An ac source of emf $V=V_0\sin\omega t$ is connected to a capacitor of capacitance $C$ . Deduce the expression for the current ( $I$ ) flowing in it. Plot the graphs of (i) $V$ vs $\omega t$ , and (ii) $I$ vs $\omega t$ . (2020) OR Show that the current leads the voltage in phase by $\pi/2$ in an ac circuit containing an ideal capacitor. (Foreign 2014)
Asked in: CBSE Class 12 Physics — SA · Section B, 2 marks
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Charge on the capacitor at time $t$ : $q=CV=CV_0\sin\omega t$ Current, $I=\dfrac{dq}{dt}=\dfrac{d}{dt}(CV_0\sin\omega t)=\omega CV_0\cos\omega t$ $I=\dfrac{V_0}{1/\omega C}\cos\omega t=I_0\sin\left(\omega t+\dfrac{\pi}{2}\right)$ , where $I_0=\omega CV_0=\dfrac{V_0}{X_C}$ Thus the current leads the applied voltage in phase by $\pi/2$ . Graphs: both $V$ and $I$ are sinusoids of the same angular frequency $\omega$ ; the $I$ – $\omega t$ curve is shifted a quarter cycle ( $\pi/2$ ) to the left of the $V$ – $\omega t$ curve, i.e. $I$ attains its peak a quarter period before $V$ does, though both share the same period $2\pi/\omega$ .
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Derivation 2
A device 'X' is connected to an ac source $V=V_0\sin\omega t$ . The variation of voltage, current and power in one cycle is shown in the graph: three curves $A$ , $B$ , $C$ are plotted against $\omega t$ from 0 to $2\pi$ — $B$ is a plain sine wave (zero at $\omega t=0$ ); $C$ is a sine wave shifted a quarter-cycle ahead of $B$ ; $A$ oscillates at twice the frequency of $B$ and $C$ , with equal areas above and below the axis in each cycle. (a) Identify the device 'X'. (b) Which of the curves $A$ , $B$ and $C$ represent the voltage, current and the power consumed in the circuit? Justify your answer. (c) How does its impedance vary with frequency of the ac source? Show graphically. (d) Obtain an expression for the current in the circuit and its phase relation with the ac voltage.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2017, Ev · Section E, 5 marks
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(a) Since curve $C$ (current) leads curve $B$ (voltage) by a quarter cycle, and the power curve averages to zero over a cycle, device $X$ is a capacitor. (b) $B\to$ voltage; $C\to$ current (leads voltage by $\pi/2$ , as in a pure capacitor); $A\to$ power (its net area over a full cycle is zero — positive for one quarter, negative for the next — consistent with zero average power in a pure capacitor). (c) $Z=X_C=\dfrac{1}{\omega C}=\dfrac{1}{2\pi f C}$ , so impedance decreases as frequency increases — a downward hyperbola, high at low $f$ , approaching zero as $f\to\infty$ . (d) With $q=CV_0\sin\omega t$ , $I=\dfrac{dq}{dt}=\omega CV_0\cos\omega t=I_0\sin\left(\omega t+\dfrac{\pi}{2}\right)$ , where $I_0=\omega CV_0$ . The current leads the voltage in phase by $\pi/2$ .
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Derivation 3
A source of ac voltage $V=V_0\sin\omega t$ is connected to a series combination of a resistor 'R' and a capacitor 'C'. Draw the phasor diagram and use it to obtain the expression for (i) the impedance of the circuit and (ii) the phase angle.
Asked in: CBSE Class 12 Physics — SA · AI 2015C, Ev · Section C, 3 marks
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Let $I=I_0\sin(\omega t+\phi)$ be the current. $V_R=IR$ is in phase with $I$ ; $V_C=IX_C$ lags $I$ by $90°$ . Phasor diagram: draw $V_R$ along the current direction and $V_C$ perpendicular to it (lagging by $90°$ ); their resultant (right-triangle sum) gives the applied voltage: $V=\sqrt{V_R^2+V_C^2}$ . (i) $V=I\sqrt{R^2+X_C^2}=I\sqrt{R^2+\dfrac{1}{\omega^2C^2}}$ , so $I=\dfrac{V}{Z}$ with impedance $Z=\sqrt{R^2+\dfrac{1}{\omega^2C^2}}$ (ii) $\tan\phi=\dfrac{V_C}{V_R}=\dfrac{X_C}{R}=\dfrac{1}{\omega RC} \Rightarrow \phi=\tan^{-1}\left(\dfrac{1}{\omega RC}\right)$ , with the current leading the voltage by this angle $\phi$ .
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Derivation 4
(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 the 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.
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2019 · Section E, 5 marks
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(a) With $I$ the circuit current, $V_L=IX_L$ leads $I$ by $\pi/2$ , $V_C=IX_C$ lags $I$ by $\pi/2$ , $V_R=IR$ is in phase with $I$ . The net voltage is $E=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}=IZ$ , where $Z=\sqrt{R^2+\left(\omega L-\dfrac1{\omega C}\right)^2}$ The $Z$ -versus-frequency graph is large at both very low and very high frequency, dipping to a minimum $Z=R$ at the resonant frequency $f_0=\dfrac1{2\pi\sqrt{LC}}$ (a V-shaped curve). (b) At resonance $X_L=X_C$ , so $V_L$ and $V_C$ are equal in magnitude; but $V_L$ leads the current by $90°$ while $V_C$ lags the current by $90°$ , so $V_L$ and $V_C$ are exactly opposite in phase — the phase difference between them is $180°$ ( $\pi$ rad), which is why they cancel and the net (applied) voltage equals $V_R$ alone at resonance. (c) With dc, only the resistance $R$ of the coil limits the current: $R=\dfrac{V}{I}=\dfrac{200}{1}=200\ \Omega$ . With ac, the inductive reactance $X_L$ also limits the current, raising the impedance to $Z=\sqrt{R^2+X_L^2}>R$ , so a smaller current (0.5 A) flows. $Z=\dfrac{V}{I}=\dfrac{200}{0.5}=400\ \Omega$ $X_L=\sqrt{Z^2-R^2}=\sqrt{400^2-200^2}=\sqrt{120000}\approx346.4\ \Omega$ $L=\dfrac{X_L}{2\pi f}=\dfrac{346.4}{2\pi\times50}\approx1.1\text{ H}$
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Derivation 5
(a) A series LCR circuit is connected to an ac source of variable frequency. Draw a suitable phasor diagram to deduce the expressions for the amplitude of the current and the phase angle. (b) Obtain the condition of resonance. Draw a plot showing the variation of current with the frequency of the ac source for two resistances $R_1$ and $R_2$ ( $R_1>R_2$ ). Hence define the quality factor, $Q$ , and write its role in the tuning of the circuit.
Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2014C, Ap · Section E, 5 marks
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(a) With $V_R=IR$ in phase with $I$ , $V_L=IX_L$ leading $I$ by $\pi/2$ , and $V_C=IX_C$ lagging $I$ by $\pi/2$ , the phasor sum gives $V=\sqrt{V_R^2+(V_L-V_C)^2}=I\sqrt{R^2+(X_L-X_C)^2}$ . So the current amplitude is $I_0=\dfrac{V_0}{\sqrt{R^2+\left(\omega L-\frac1{\omega C}\right)^2}}$ , and the phase angle is $\phi=\tan^{-1}\left(\dfrac{X_L-X_C}{R}\right)$ . (b) Resonance occurs when the current amplitude is maximum, i.e. when $\omega L=\dfrac1{\omega C} \Rightarrow \omega_0=\dfrac1{\sqrt{LC}}$ ; at resonance $Z=R$ (minimum impedance). Graph: current amplitude versus frequency peaks at $\omega=\omega_0$ for both resistances; the peak for the smaller resistance $R_2$ is taller and narrower (sharper), while for $R_1$ it is shorter and broader. Quality factor: $Q=\dfrac1R\sqrt{\dfrac LC}=\dfrac{\omega_0L}{R}=\dfrac{\omega_0}{2\Delta\omega}$ . A higher $Q$ gives a sharper resonance curve, so the circuit responds strongly only to a narrow band of frequencies near $\omega_0$ — this is what makes tuning (frequency selection) more precise.
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Derivation 6
(i) When an ac source is connected to an ideal inductor, show that the average power supplied by the source over a complete cycle is zero. (ii) A lamp is connected in series with an inductor and an ac source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor? Explain.
Asked in: CBSE Class 12 Physics — SA · AI 2016 · Section C, 3 marks
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(i) In an ideal inductor, $I_{rms}$ lags $V_{rms}$ by $\phi=\pi/2$ . $P_{av}=V_{rms}I_{rms}\cos\phi=V_{rms}I_{rms}\cos\left(\dfrac\pi2\right)=0$ So the average power supplied over a complete cycle is zero — the inductor only stores and returns energy, dissipating none. (ii) When the key is plugged in, the lamp lights up at a brightness set by $Z=\sqrt{R^2+X_L^2}$ . Inserting an iron rod increases $L$ (high permeability), so $X_L=\omega L$ increases; $Z$ increases and the current decreases — the brightness of the lamp decreases.
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Derivation 7
A voltage $V=V_0\sin\omega t$ is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit?
Asked in: CBSE Class 12 Physics — SA · AI 2014 · Section C, 3 marks
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Let $I=I_0\sin(\omega t-\phi)$ be the current. Instantaneous power, $p=VI=V_0I_0\sin\omega t\,\sin(\omega t-\phi)=V_0I_0\left[\dfrac{1-\cos2\omega t}2\cos\phi-\dfrac{\sin2\omega t}2\sin\phi\right]$ Averaging over a full cycle, $\langle\cos2\omega t\rangle=\langle\sin2\omega t\rangle=0$ , so $P_{av}=\dfrac{V_0I_0}2\cos\phi=V_{rms}I_{rms}\cos\phi$ (i) No power is dissipated (wattless current) when $\cos\phi=0$ , i.e. $\phi=\pi/2$ — a purely inductive or purely capacitive circuit. (ii) Maximum power is dissipated when $\cos\phi=1$ , i.e. $\phi=0$ — a purely resistive circuit, or an LCR circuit at resonance ( $X_L=X_C$ ), giving $P_{av,max}=V_{rms}I_{rms}$ .
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Derivation 8
Show that an ideal inductor does not dissipate power in an ac circuit.
Asked in: CBSE Class 12 Physics — Long Answer · 2/5, 2020, Ap · Section E, 5 marks
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Let $V=V_0\sin\omega t$ be applied across an ideal inductor; the current lags by $\pi/2$ : $I=I_0\sin\left(\omega t-\dfrac\pi2\right)$ . $p=VI=V_0I_0\sin\omega t\,\sin\left(\omega t-\dfrac\pi2\right)=-V_0I_0\sin\omega t\cos\omega t=-\dfrac{V_0I_0}2\sin2\omega t$ $P_{av}=-\dfrac{V_0I_0}2\langle\sin2\omega t\rangle=0$ , since the average of $\sin2\omega t$ over a full cycle is zero. [Equivalently, $P_{av}=V_{rms}I_{rms}\cos\left(\dfrac\pi2\right)=0$ .] The inductor absorbs energy while the magnetic field builds up in one quarter cycle and returns exactly the same energy as the field collapses in the next — so it dissipates no net power.
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Derivation 9
(a) Draw a schematic arrangement for winding of primary and secondary coil in a transformer when the two coils are wound on top of each other. (b) State the underlying principle of a transformer and obtain the expression for the ratio of secondary to primary voltage in terms of (i) the number of secondary and primary windings, and (ii) the primary and secondary currents. (c) Write the main assumption involved in deriving the above relations. (d) Write any two reasons due to which energy losses may occur in actual transformers.
Asked in: CBSE Class 12 Physics — Long Answer · AI 2014C · Section E, 5 marks
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(a) Diagram: primary coil ( $N_p$ turns) wound directly over (concentric with) the secondary coil ( $N_s$ turns), both on a common soft-iron core, minimizing flux leakage. (b) Principle: mutual induction — an alternating primary current produces a continuously changing flux, which, linked with the secondary through the soft-iron core, induces an alternating emf in it. With primary resistance negligible, $V_p\approx\varepsilon_p=-N_p\dfrac{d\phi}{dt}$ ; with secondary open, $V_s=\varepsilon_s=-N_s\dfrac{d\phi}{dt}$ . (i) $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$ (ii) For an ideal transformer, $I_sV_s=I_pV_p \Rightarrow \dfrac{V_s}{V_p}=\dfrac{I_p}{I_s}=\dfrac{N_s}{N_p}$ (c) Main assumptions: the resistance of the primary and secondary windings is negligible, and the same flux links both windings (no flux leakage from the core). (d) Two reasons: (i) copper losses — $I^2R$ heating in the winding resistance; (ii) hysteresis losses — due to repeated magnetization/demagnetization of the core each cycle. [Also acceptable: eddy-current losses, flux-leakage losses.]
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Derivation 10
An a.c. source generating a voltage $\varepsilon=\varepsilon_0\sin\omega t$ is connected to a capacitor of capacitance $C$ . Find the expression for the current $I$ flowing through it. Plot a graph of $\varepsilon$ and $I$ versus $\omega t$ to show that the current is ahead of the voltage by $\pi/2$ .
Asked in: CBSE Class 12 Physics — SA · 2022-23 · Section C, 3 marks · CBSE sample question
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$q=C\varepsilon=C\varepsilon_0\sin\omega t$ $I=\dfrac{dq}{dt}=\omega C\varepsilon_0\cos\omega t=\dfrac{\varepsilon_0}{1/\omega C}\cos\omega t=I_0\sin\left(\omega t+\dfrac\pi2\right)$ , where $I_0=\omega C\varepsilon_0$ Graph: both $\varepsilon$ and $I$ are sinusoids of the same period; $I$ reaches its peak $I_0$ a quarter-cycle ( $\pi/2$ ) before $\varepsilon$ reaches its peak $\varepsilon_0$ — the $I$ – $\omega t$ curve is shifted to the left of the $\varepsilon$ – $\omega t$ curve by $\pi/2$ , showing the current is ahead of the voltage by $\pi/2$ .
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Derivation 11
(a) Derive the expression for the current flowing in an ideal capacitor and its reactance when connected to an ac source of voltage $V=V_0\sin\omega t$ . (b) Draw its phasor diagram. (c) If resistance is added in series to the capacitor, what changes will occur in the current flowing in the circuit and the phase angle between voltage and current?
Asked in: CBSE Class 12 Physics — Long Answer · 2020-21, Ev · Section E, 5 marks · CBSE sample question
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(a) $C=\dfrac qV \Rightarrow q=CV_0\sin\omega t$ $i=\dfrac{dq}{dt}=\omega CV_0\cos\omega t=\dfrac{V_0}{1/\omega C}\sin\left(\omega t+\dfrac\pi2\right)=i_0\sin\left(\omega t+\dfrac\pi2\right)$ where $i_0=\dfrac{V_0}{X_C}$ and $X_C=\dfrac1{\omega C}$ is the capacitive reactance. In a pure capacitive circuit, current leads voltage by $\pi/2$ . (b) Phasor diagram: current phasor $i_0$ drawn $90°$ ahead of (anticlockwise from) the voltage phasor $V_0$ , both rotating anticlockwise with angular speed $\omega$ . (c) With a resistance $R$ added in series, the circuit becomes a series RC circuit: impedance $Z=\sqrt{R^2+X_C^2}$ (larger than $X_C$ alone), so the current $I=V/Z$ decreases; the phase angle decreases from $\pi/2$ to $\phi=\tan^{-1}(X_C/R)$ , somewhere between $0$ and $\pi/2$ depending on the relative sizes of $R$ and $X_C$ .

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