Chapter 7: Alternating Current

Almost all electric power in the world is generated, transmitted and distributed as alternating current (AC). Why? Because the EMF naturally produced by a coil rotating in a magnetic field is sinusoidal (Ch.6), and because AC voltages can be stepped up or down losslessly by a transformer — making efficient long-distance transmission possible. This chapter develops the mathematical tool kit needed to analyse AC circuits: phasors, reactances, impedance, resonance, Q-factor, power factor; and applies it to three essential devices: the series LCR circuit, the LC oscillator, and the transformer.

Concept Map

• AC voltage: $v(t) = v_0\sin\omega t$, frequency $f = \omega/(2\pi)$
• Average over a cycle: zero (for sin/cos)
• RMS values: $V_{\text{rms}} = V_0/\sqrt 2$, $I_{\text{rms}} = I_0/\sqrt 2$ — what an AC ammeter/voltmeter reads
• Phasor: a rotating vector whose projection on the vertical axis is the instantaneous AC value
• Pure R: $V$ and $I$ in phase, $Z = R$
• Pure L: $V$ leads $I$ by $\pi/2$, reactance $X_L = \omega L$
• Pure C: $V$ lags $I$ by $\pi/2$, reactance $X_C = 1/(\omega C)$
• Series LCR: $Z = \sqrt{R^2 + (X_L - X_C)^2}$, $\tan\phi = (X_L-X_C)/R$
• Resonance: $X_L = X_C\Rightarrow\omega_0 = 1/\sqrt{LC}$; sharpness $Q = \omega_0 L/R = (1/R)\sqrt{L/C}$
• Average power: $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$; $\cos\phi$ = power factor
• LC oscillation: $\omega = 1/\sqrt{LC}$, energy sloshes between $L$ and $C$
• Transformer: $V_s/V_p = N_s/N_p$; ideal efficiency 100% but real $\sim 90$–99%

7.1 AC Voltage and Current — Average and RMS

Definition

An alternating voltage is one whose direction reverses periodically; the standard form is sinusoidal:

$v(t) = v_0\sin\omega t,$

where $v_0$ is the peak (amplitude) voltage and $\omega = 2\pi f$ is the angular frequency. The household supply in India has $f = 50\,\text{Hz}$, $V_{\text{rms}} = 230\,\text{V}$, so $V_0 = 230\sqrt 2 \approx 325\,\text{V}$.

Average value

Over one complete cycle (period $T = 2\pi/\omega$):

$\langle v\rangle_T = \frac{1}{T}\int_0^T v_0\sin\omega t\,dt = 0.$

The cycle-average of a pure sinusoid is zero — that's why we use RMS values instead.

Over a half-cycle ($0$ to $T/2$):

$\langle v\rangle_{T/2} = \frac{1}{T/2}\int_0^{T/2} v_0\sin\omega t\,dt = \frac{2v_0}{\pi} \approx 0.637\,v_0.$

Derivation of RMS

The root-mean-square value is defined by

$V_{\text{rms}} = \sqrt{\langle v^2\rangle_T}.$

Compute $\langle v^2\rangle$:

$\langle v^2\rangle = \frac{1}{T}\int_0^T v_0^2\sin^2\omega t\,dt = \frac{v_0^2}{T}\int_0^T\frac{1-\cos 2\omega t}{2}dt = \frac{v_0^2}{2}.$

Therefore

$\boxed{\;V_{\text{rms}} = \frac{V_0}{\sqrt 2},\qquad I_{\text{rms}} = \frac{I_0}{\sqrt 2}\;}$

Why RMS?

The instantaneous power dissipated in a resistor is $p(t) = i^2(t)R$. Average power:

$\langle p\rangle = R\,\langle i^2\rangle = R\,I_{\text{rms}}^2.$

So $I_{\text{rms}}$ is the DC current that would dissipate the same average power. AC ammeters and voltmeters always read RMS values.

Worked Example

The mains supply at "$230\,\text{V}, 50\,\text{Hz}$" means

$V_0 = 230\sqrt 2 \approx 325\,\text{V},\quad f = 50\,\text{Hz},\quad T = 20\,\text{ms},\quad \omega = 100\pi\,\text{rad/s}.$

A 100 W bulb has $R = V_{\text{rms}}^2/P = 529\,\Omega$ and $I_{\text{rms}} = 100/230 = 0.435\,\text{A}$, $I_0 = 0.615\,\text{A}$.

Pitfalls

• $\langle\sin\rangle_T = 0$, but $\langle\sin^2\rangle_T = 1/2$.
• The "$\sqrt 2$" factor is for sinusoidal waveforms; square waves and triangle waves have different $V_{\text{rms}}/V_0$ ratios ($1$ and $1/\sqrt 3$ respectively).
• The peak voltage of a 230 V supply is 325 V — household insulation must withstand this.

7.2 AC Applied to a Pure Resistor

Setup

A pure resistor $R$ across an AC source $v(t) = v_0\sin\omega t$.

Derivation

By Ohm's law applied instantaneously,

$i(t) = \frac{v(t)}{R} = \frac{v_0}{R}\sin\omega t = i_0\sin\omega t,\quad i_0 = v_0/R.$

Voltage and current are in phase — both reach maxima, zeros and minima simultaneously.

Phasor diagram

Represent $v(t)$ and $i(t)$ as vectors $\vec V_0$ and $\vec I_0$ rotating at $\omega$. For a resistor they point in the same direction — phase difference $\phi = 0$.

Power

$p(t) = v(t)i(t) = v_0 i_0 \sin^2\omega t,$

$\langle p\rangle = \tfrac12 v_0 i_0 = V_{\text{rms}}I_{\text{rms}}.$

A pure resistor always dissipates energy.

Worked Example

A $40\,\Omega$ resistor across a $200\,\text{V}$ rms, $50\,\text{Hz}$ source. Peak current $i_0 = 200\sqrt 2/40 = 7.07\,\text{A}$. Power $= V_{\text{rms}}^2/R = 1000\,\text{W}$.

Pitfalls

• $V_{\text{rms}}/I_{\text{rms}} = R$ is not a generalization; in reactive elements it would be $X$, and in general $Z$.
• Even though $\langle i\rangle = 0$, $\langle p\rangle \neq 0$ — power is a quadratic in $i$.

7.3 AC Applied to a Pure Inductor

Setup

A pure inductor $L$ (zero resistance) across $v(t) = v_0\sin\omega t$.

Derivation

Kirchhoff: $v - L\,di/dt = 0\implies di/dt = v/L = (v_0/L)\sin\omega t$. Integrate:

$i(t) = -\frac{v_0}{\omega L}\cos\omega t = \frac{v_0}{\omega L}\sin(\omega t - \pi/2).$

So

$i_0 = \frac{v_0}{\omega L}\equiv\frac{v_0}{X_L},\qquad \boxed{\;X_L = \omega L\;}$

$X_L$ is the inductive reactance (units $\Omega$). It plays the role of resistance for an inductor in an AC circuit.

Phase: the current lags the voltage by $\pi/2$ — equivalently, the voltage leads the current by $\pi/2$. (Mnemonic: "ELI" — Emf $E$ leads current $I$ for $L$.)

Phasor diagram

$\vec V$ leads $\vec I$ by $90^\circ$ — draw $\vec V$ along the vertical, $\vec I$ along the horizontal (or any consistent $90^\circ$ split).

Frequency dependence of $X_L$

• DC ($\omega = 0$): $X_L = 0$ — inductor acts like a short.
• High $\omega$: $X_L\to\infty$ — inductor acts like an open circuit.

Power dissipated

$\langle p\rangle = V_{\text{rms}}I_{\text{rms}}\cos\phi = V_{\text{rms}}I_{\text{rms}}\cos 90^\circ = 0.$

A pure inductor dissipates no average power — energy oscillates between the source and the magnetic field of the inductor.

Worked Example

$L = 0.50\,\text{H}$ across $230\,\text{V}, 50\,\text{Hz}$.

$X_L = 2\pi\times 50\times 0.50 = 157\,\Omega,\quad I_{\text{rms}} = 230/157 = 1.46\,\text{A}.$

Average power = 0.

Pitfalls

• "Reactance" $X_L$ has units of $\Omega$ but is not dissipative.
• A "pure" inductor is an idealisation — real inductors have winding resistance, so real coils do dissipate some power.

7.4 AC Applied to a Pure Capacitor

Setup

A capacitor $C$ across $v(t) = v_0\sin\omega t$. Charge on plate $q(t) = Cv(t)$, current $i = dq/dt$:

$i(t) = C\frac{dv}{dt} = \omega C v_0\cos\omega t = \omega C v_0\sin(\omega t + \pi/2).$

Derivation

$i_0 = \omega C v_0\equiv\frac{v_0}{X_C},\qquad \boxed{\;X_C = \frac{1}{\omega C}\;}$

$X_C$ is the capacitive reactance.

Phase: the current leads the voltage by $\pi/2$. (Mnemonic: "ICE" — Current $I$ leads voltage for $C$.)

Phasor diagram

$\vec I$ leads $\vec V$ by $90^\circ$.

Frequency dependence

• DC ($\omega = 0$): $X_C\to\infty$ — capacitor blocks DC (after the initial transient).
• High $\omega$: $X_C\to 0$ — capacitor acts like a short.

This is opposite to the inductor — hence inductors and capacitors compete in AC circuits.

Power dissipated

$\langle p\rangle = V_{\text{rms}}I_{\text{rms}}\cos(-\pi/2) = 0.$

Capacitors store and return energy via the electric field, but don't dissipate.

Worked Example

$C = 15\,\mu\text{F}$ across $220\,\text{V}, 50\,\text{Hz}$.

$X_C = \frac{1}{2\pi\times 50\times 15\times 10^{-6}} = 212\,\Omega,\quad I_{\text{rms}} = 220/212 = 1.04\,\text{A}.$

Pitfalls

• $X_C$ decreases with $\omega$ (opposite to $X_L$).
• A "DC blocking" capacitor is the same idea: at DC, infinite reactance, no current.

7.5 Series LCR Circuit

Setup

A resistor $R$, inductor $L$ and capacitor $C$ in series across an AC source $v(t) = v_0\sin\omega t$. Same current $i(t)$ flows through all three.

Phasor derivation

Take the current $\vec I$ as the reference phasor (horizontal). Then:

• $\vec V_R$ — in phase with $\vec I$ (horizontal), magnitude $IR$.
• $\vec V_L$ — leads $\vec I$ by $90^\circ$ (vertical up), magnitude $IX_L$.
• $\vec V_C$ — lags $\vec I$ by $90^\circ$ (vertical down), magnitude $IX_C$.

The applied voltage is the phasor sum:

$\vec V = \vec V_R + \vec V_L + \vec V_C.$

$V_L$ and $V_C$ are anti-parallel; their difference combines with $V_R$ at right angle:

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

Define impedance:

$\boxed{\;Z = \sqrt{R^2 + (X_L - X_C)^2}\;}$

The peak current is

$i_0 = \frac{v_0}{Z}.$

The phase angle of $\vec V$ relative to $\vec I$:

$\boxed{\;\tan\phi = \frac{X_L - X_C}{R}\;}$

• $X_L > X_C$: $\phi > 0$ — voltage leads current — circuit is inductive.
• $X_L < X_C$: $\phi < 0$ — voltage lags current — circuit is capacitive.
• $X_L = X_C$: $\phi = 0$ — pure resistance — resonance.

Impedance triangle

Draw a right triangle with legs $R$ (horizontal) and $(X_L - X_C)$ (vertical); hypotenuse is $Z$, angle is $\phi$.

Worked Example

$R = 30\,\Omega$, $L = 25.5\,\text{mH}$, $C = 786\,\mu\text{F}$, $v_0 = 283\,\text{V}$, $f = 50\,\text{Hz}$.

$X_L = 2\pi(50)(0.0255) = 8.0\,\Omega,\quad X_C = \frac{1}{2\pi(50)(786\times 10^{-6})} = 4.05\,\Omega,$

$Z = \sqrt{30^2 + (8 - 4.05)^2} = \sqrt{900 + 15.6} = 30.26\,\Omega,$

$i_0 = 283/30.26 = 9.35\,\text{A},\quad \tan\phi = 3.95/30 = 0.132,\quad \phi = 7.5^\circ.$

Pitfalls

• Phasors don't add by simple algebra — you must add their vector components.
• The sum $V_L + V_C$ is not the magnitude of either: $V_L$ and $V_C$ are anti-parallel, so the difference of magnitudes is what counts.
• Each component voltage can individually be larger than the source voltage at resonance — there is no contradiction because they are out of phase and partially cancel.

7.6 Resonance, Bandwidth and Q-factor

Resonance condition

In a series LCR driven at frequency $\omega$, the impedance $Z(\omega) = \sqrt{R^2 + (\omega L - 1/\omega C)^2}$ is minimised when

$X_L = X_C\implies \omega L = \frac{1}{\omega C}\implies \omega_0 = \frac{1}{\sqrt{LC}}.$

$\boxed{\;\omega_0 = \frac{1}{\sqrt{LC}},\quad f_0 = \frac{1}{2\pi\sqrt{LC}}\;}$

At $\omega = \omega_0$: $Z = R$ (minimum), $i_0 = v_0/R$ (maximum), $\phi = 0$ (current in phase with source).

Resonance curve

A plot of $I_{\text{rms}}$ vs $\omega$ peaks sharply at $\omega_0$ and falls off on either side. The shape of the curve depends on $R$: small $R$ gives a sharp peak (high "quality"), large $R$ gives a broad peak.

Bandwidth and Q-factor

The bandwidth $\Delta\omega$ is the width of the resonance curve between the two half-power points (where $I = I_{\max}/\sqrt 2$). It can be shown that

$\Delta\omega = \frac{R}{L},\qquad \boxed{\;Q = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}\;}$

$Q$ is the quality factor — large $Q$ means sharp resonance (high selectivity). Equivalently, $Q$ is $2\pi\times$ (energy stored / energy lost per cycle).

At resonance, voltage across L and C

At resonance, $V_L = I X_L = (v_0/R)\,X_L = Q\,v_0$; same for $V_C$. So $V_L$ and $V_C$ are each $Q$ times the source voltage — and they cancel exactly because they are anti-phase.

Applications

• Radio tuning: turning the dial varies $C$ (or $L$) to bring $f_0$ in resonance with the desired station's carrier frequency; the high-$Q$ circuit selects that frequency from the antenna's mix.
• Band-pass filters, oscillator circuits, MRI, wireless power transfer.

Worked Example

$L = 30\,\text{mH}$, $C = 27\,\mu\text{F}$, $R = 7.5\,\Omega$. Find resonant frequency, $Q$, and bandwidth.

$\omega_0 = 1/\sqrt{LC} = 1/\sqrt{30\times 10^{-3}\times 27\times 10^{-6}} = 1111\,\text{rad/s},\quad f_0 = 177\,\text{Hz},$

$Q = \omega_0 L/R = 1111\times 0.030/7.5 = 4.44,\quad \Delta\omega = R/L = 250\,\text{rad/s}.$

Pitfalls

• $\omega_0 = 1/\sqrt{LC}$ depends only on $L$ and $C$, not on $R$. $R$ only affects sharpness (Q).
• Bandwidth $\Delta\omega$ has units of rad/s, $\Delta f = \Delta\omega/(2\pi)$ in Hz.
• $Q$ can also be expressed as $Q = (1/\omega_0 RC)$ — equivalent at resonance.

7.7 Power in AC Circuits

Instantaneous and average power

Let $v(t) = v_0\sin\omega t$, $i(t) = i_0\sin(\omega t - \phi)$. Instantaneous power:

$p(t) = v(t)i(t) = v_0 i_0\sin\omega t\sin(\omega t - \phi).$

Using $\sin A\sin B = \tfrac12[\cos(A-B) - \cos(A+B)]$:

$p(t) = \tfrac12 v_0 i_0[\cos\phi - \cos(2\omega t - \phi)].$

The second term averages to zero over a cycle, so

$\boxed{\;\langle P\rangle = \tfrac12 v_0 i_0\cos\phi = V_{\text{rms}}I_{\text{rms}}\cos\phi\;}$

The factor $\cos\phi$ is called the power factor.

Special cases

• Pure resistor: $\phi = 0$, $\cos\phi = 1$, full power dissipated.
• Pure inductor or pure capacitor: $\phi = \pm 90^\circ$, $\cos\phi = 0$, no power dissipated.
• Series LCR at resonance: $\phi = 0$, $\cos\phi = 1$, $\langle P\rangle = V_{\text{rms}}^2/R$.

In general:

$\cos\phi = \frac{R}{Z}.$

Wattless current

The component of the current perpendicular to the voltage phasor (i.e., $I_{\text{rms}}\sin\phi$) does not contribute to average power. This is called the wattless current. It is the part of the current that "sloshes back and forth" without doing work — only loading the wires with resistive losses without doing useful work at the device. Industrial loads with bad power factors (heavily inductive motors) waste capacity in transmission lines — this is corrected by adding capacitor banks.

Worked Example

A circuit with $R = 100\,\Omega$, $X_L = 50\,\Omega$, $X_C = 30\,\Omega$, across $230\,\text{V}$ rms, 50 Hz.

$Z = \sqrt{100^2 + 20^2} = 101.98\,\Omega,\quad I_{\text{rms}} = 2.255\,\text{A},$

$\cos\phi = R/Z = 100/101.98 = 0.981,$

$P = V_{\text{rms}}I_{\text{rms}}\cos\phi = 230\times 2.255\times 0.981 = 509\,\text{W}.$

(Same as $I_{\text{rms}}^2 R = 2.255^2\times 100 = 509\,\text{W}$ — power dissipated only in $R$.)

Pitfalls

• Power factor $\cos\phi$ can be improved by adding the right reactance: leading PF $\to$ add inductor; lagging PF $\to$ add capacitor.
• Power is always dissipated in $R$ and never in pure $L$ or $C$ — but instantaneous power in $L$/$C$ oscillates positive and negative.

7.8 LC Oscillations

Setup

Connect a charged capacitor (charge $q_0$) to an inductor (no resistance). At $t = 0$, all the energy is in the capacitor; the inductor carries no current.

Derivation: equation of motion

Kirchhoff: $q/C - L\,di/dt = 0$, with $i = -dq/dt$ (charge flowing off the capacitor):

$L\frac{d^2 q}{dt^2} + \frac{q}{C} = 0.$

This is simple harmonic motion in $q$:

$\boxed{\;\omega = \frac{1}{\sqrt{LC}}\;}$

Solution: $q(t) = q_0\cos\omega t$, $i(t) = q_0\omega\sin\omega t$.

Energy

$U_E(t) = \frac{q^2}{2C} = \frac{q_0^2}{2C}\cos^2\omega t,\quad U_B(t) = \tfrac12 Li^2 = \frac{q_0^2}{2C}\sin^2\omega t.$

Total $U_E + U_B = q_0^2/(2C) =$ constant. Energy sloshes between electric (in $C$) and magnetic (in $L$). The frequency at which it sloshes is exactly $\omega = 1/\sqrt{LC}$.

Mass-spring analogy

Inductance is "inertia" (resists changes in current), capacitance is "compliance" (a spring-like element).

Damping

A real LC circuit has resistance, leading to a damped LCR oscillation:

$L\ddot q + R\dot q + q/C = 0,$

with quality factor $Q = \omega_0 L/R$. Just like an underdamped mass-spring oscillator. For sufficiently small $R$, the oscillation persists for many cycles before dying out.

Worked Example

$L = 20\,\text{mH}$, $C = 50\,\mu\text{F}$ initially charged to $10\,\text{V}$. Find the angular frequency, peak current.

$\omega = 1/\sqrt{20\times 10^{-3}\times 50\times 10^{-6}} = 1000\,\text{rad/s},$

$i_0 = q_0\omega = CV_0\omega = 50\times 10^{-6}\times 10\times 1000 = 0.5\,\text{A}.$

Pitfalls

• Without resistance, oscillation is undamped and persists forever — only in idealisation.
• In a real LC oscillator (Hartley, Colpitts, etc.), the resistance is compensated by an active element (transistor amplifier) to maintain oscillations.

7.9 Transformer

Principle

A transformer steps voltages up or down using mutual induction. A changing current in the primary coil produces a changing flux in a soft-iron core; the flux passes through the secondary coil, inducing an EMF.

Construction

• Primary coil: $N_p$ turns connected to the input AC source.
• Secondary coil: $N_s$ turns connected to the load.
• Soft-iron core (laminated, to minimise eddy currents) links the two coils, providing a high-permeability path with low leakage.

Derivation: ideal transformer relation

If all the flux $\Phi$ through one turn links every turn of both coils (no leakage), and the primary EMF equals the source voltage,

$V_p = N_p\,\frac{d\Phi}{dt},\quad V_s = N_s\,\frac{d\Phi}{dt}.$

Dividing:

$\boxed{\;\frac{V_s}{V_p} = \frac{N_s}{N_p}\equiv k\;}$

$k$ is the turns ratio. If $k > 1$, step-up (boosts voltage); $k < 1$, step-down.

Current ratio (ideal)

For an ideal lossless transformer, $P_p = P_s$:

$V_p I_p = V_s I_s\implies \frac{I_s}{I_p} = \frac{N_p}{N_s} = \frac{1}{k}.$

A step-up transformer trades current for voltage; a step-down trades voltage for current.

Energy losses (real transformer)

1. Copper losses ($I^2 R$ in the windings): minimised by using thick low-resistance wire.
2. Iron losses (eddy currents in core): minimised by laminating the core.
3. Hysteresis loss in the core: minimised by using soft magnetic materials (low $H_c$, narrow loop) like silicon steel.
4. Flux leakage: not all primary flux links the secondary; minimised by good magnetic coupling (high $\mu_r$, closed-loop core).
5. Magnetostriction (humming noise in transformers): a minor loss.

Typical efficiency: 90–99% in modern power transformers.

Why AC for transmission?

Transformers work only with AC (steady DC produces no $d\Phi/dt$). High voltage transmission ($\sim 400\,\text{kV}$) at low current gives small $I^2R$ losses; step-down transformers convert to safe domestic voltages ($230\,\text{V}$).

Worked Example

A step-down transformer: $V_p = 2200\,\text{V}$, $V_s = 220\,\text{V}$, $I_s = 10\,\text{A}$. Find $N_s/N_p$ and $I_p$ (ideal).

$\frac{N_s}{N_p} = \frac{V_s}{V_p} = 0.10,\quad I_p = \frac{V_s I_s}{V_p} = \frac{220\times 10}{2200} = 1.0\,\text{A}.$

If the efficiency is 90%, $P_s = 0.9\,P_p\implies I_p = (V_s I_s)/(0.9\,V_p) = 1.11\,\text{A}$.

Pitfalls

• A transformer does not "amplify" power — it conserves it (up to losses) while trading $V$ for $I$.
• It works only with AC; using a transformer with DC will burn out the primary because $X_L = 0$ at DC.
• The turn ratio determines $V$, but the load determines the current draw.

Solved Problems

Problem 1 — RMS for a non-sinusoidal wave

A square wave of amplitude $V_0$ has $V_{\text{rms}} = V_0$ (because $V^2 = V_0^2$ always). Verify by integrating $\langle V^2\rangle$ over one period — both halves give $V_0^2$, so $\langle V^2\rangle = V_0^2$.

Problem 2 — Inductive AC circuit

$L = 0.10\,\text{H}$ across $200\,\text{V}, 50\,\text{Hz}$. $X_L = 2\pi(50)(0.10) = 31.42\,\Omega$. $I_{\text{rms}} = 200/31.42 = 6.37\,\text{A}$. Power $= 0$ (pure inductor).

Problem 3 — Series LCR

A series LCR with $R = 8\,\Omega$, $L = 20\,\text{mH}$, $C = 200\,\mu\text{F}$, $f = 50\,\text{Hz}$, source $V_{\text{rms}} = 200\,\text{V}$.

$X_L = 2\pi(50)(0.020) = 6.28\,\Omega,\quad X_C = 1/[2\pi(50)(200\times 10^{-6})] = 15.92\,\Omega,$

$Z = \sqrt{8^2 + (6.28 - 15.92)^2} = \sqrt{64 + 92.93} = 12.53\,\Omega,$

$I_{\text{rms}} = 200/12.53 = 15.96\,\text{A},\quad \tan\phi = -9.64/8 = -1.205,\quad \phi = -50.3^\circ.$

Negative $\phi$: circuit is capacitive (current leads voltage). Power $= V_{\text{rms}}I_{\text{rms}}\cos\phi = 200\times 15.96\times \cos 50.3^\circ = 2040\,\text{W}$.

Problem 4 — Resonance frequency

$L = 5.0\,\text{H}$, $C = 80\,\mu\text{F}$, $R = 40\,\Omega$, source $V_{\text{rms}} = 240\,\text{V}$.

$\omega_0 = 1/\sqrt{5.0\times 80\times 10^{-6}} = 50\,\text{rad/s}\implies f_0 = 7.96\,\text{Hz}.$

At resonance: $Z = R = 40\,\Omega$, $I_{\text{rms}} = 240/40 = 6.0\,\text{A}$.

$Q = \omega_0 L/R = 50\times 5.0/40 = 6.25$, so $V_L = V_C = QV_{\text{rms}} = 1500\,\text{V}$ — far larger than the source voltage.

Problem 5 — Power factor improvement

A factory draws $I_{\text{rms}} = 100\,\text{A}$ at $230\,\text{V}$ with $\cos\phi = 0.6$ (lagging, inductive). The actual power consumed is

$P = 230\times 100\times 0.6 = 13.8\,\text{kW},$

while the apparent power is $S = 230\times 100 = 23\,\text{kVA}$. By adding a capacitor bank to bring $\cos\phi = 1$, the current required for the same 13.8 kW drops to $13800/230 = 60\,\text{A}$ — reducing transmission losses by factor $(100/60)^2 = 2.78$.

Problem 6 — LC oscillation

$L = 25\,\text{mH}$, $C = 100\,\mu\text{F}$ initially charged to $200\,\text{V}$.

$\omega = 1/\sqrt{25\times 10^{-3}\times 100\times 10^{-6}} = 632\,\text{rad/s}, T = 9.93\,\text{ms}.$

$q_0 = CV_0 = 2.0\times 10^{-2}\,\text{C}$, $i_0 = q_0\omega = 12.6\,\text{A}$.

Total energy $= \tfrac12 CV_0^2 = 2.0\,\text{J}$. At quarter period, all energy is in inductor: $\tfrac12 Li_0^2 = \tfrac12\times 0.025\times 12.6^2 = 2.0\,\text{J}$.,\checkmark

Problem 7 — Transformer with losses

A transformer is rated 2 kVA, $220/110\,\text{V}$. Primary current at full load is $I_p$ such that $V_p I_p = 2000\,\text{VA}\implies I_p = 9.09\,\text{A}$. If iron + copper losses are 100 W, output is $1900\,\text{W}$ at $\cos\phi = 1$, efficiency $= 95\%$.

JEE/NEET Edge Cases

• Comparing two AC voltmeters in series with R, L, C: each voltmeter reads RMS of the individual voltage. So $V_L$ and $V_C$ readings can each exceed the source voltage at resonance.
• Choke coil: a high-$L$, low-$R$ coil used in fluorescent lamps to limit current without dissipating energy — uses the fact that pure $L$ has $\cos\phi \approx 0$.
• Power factor for series RLC: $\cos\phi = R/Z$. For purely reactive load, $\cos\phi = 0$.
• Capacitor across DC source: blocks DC; only an initial transient current flows (charging the capacitor). At steady state, $I = 0$.
• Inductor across DC source: passes DC freely once steady. The initial transient builds up current with time constant $L/R$.
• At resonance, the apparent rate of work by the source is real (no wattless current), and the source delivers maximum average power $V^2/R$.
• Q-factor units check: $Q = \omega_0 L/R$ is dimensionless; also $= 1/(\omega_0 RC)$; also $= (1/R)\sqrt{L/C}$. All equivalent.
• Trap: $X_L$ increases with frequency, $X_C$ decreases. A series RC acts like a high-pass; series RL like a low-pass; LCR like a band-pass.
• Trap (transformer): turns ratio gives voltage ratio; current ratio is inverse (with losses, slightly different).

Quick Recap

• $V_{\text{rms}} = V_0/\sqrt 2$, $I_{\text{rms}} = I_0/\sqrt 2$ (sinusoidal).
• Pure R: $V$, $I$ in phase, $Z = R$; pure L: $V$ leads by $90^\circ$, $X_L = \omega L$; pure C: $V$ lags by $90^\circ$, $X_C = 1/\omega C$.
• Series LCR: $Z = \sqrt{R^2 + (X_L - X_C)^2}$, $\tan\phi = (X_L - X_C)/R$.
• Resonance: $\omega_0 = 1/\sqrt{LC}$, $Z = R$ minimum, $I$ maximum.
• Quality factor: $Q = \omega_0 L/R = (1/R)\sqrt{L/C}$, bandwidth $\Delta\omega = R/L$.
• Average power: $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$; $\cos\phi$ = power factor.
• LC oscillation: $\omega = 1/\sqrt{LC}$, undamped without $R$.
• Transformer (ideal): $V_s/V_p = N_s/N_p = I_p/I_s$.

Formula Sheet