Alternating Current: Full Notes

7.1 AC Voltage, Current & RMS Values

An AC source drives a current that oscillates sinusoidally rather than flowing one direction at a constant rate:

AC voltage

V = V_0\sin(\omega t)

Resulting current

I = I_0\sin(\omega t - \phi)

where $\phi$ is the phase by which current lags (or leads) the voltage, depending on what the circuit contains. Since the instantaneous current averages to zero over a full cycle, a more useful measure is the root-mean-square (RMS) value, the equivalent steady current that would dissipate the same average power in a resistor:

RMS current and voltage

I_{rms} = \dfrac{I_0}{\sqrt{2}}, \qquad V_{rms} = \dfrac{V_0}{\sqrt{2}}

Household “230 V AC” and similar ratings are always quoted as RMS values, not peak values.

Worked example

Peak from RMS

An AC supply is rated 220 V (RMS). Find its peak voltage.

V_0 = V_{rms}\sqrt{2} = 220\sqrt{2} = \textcolor{#e08a1e}{311.1\text{ V}}

7.2 Reactance of L and C

7.2.1 Pure inductor

For an inductor alone, the source voltage at every instant equals the back-EMF needed to drive the changing current, $V = L\,dI/dt$ . Driving $I = I_0\sin(\omega t)$ through it requires:

V = L\dfrac{d}{dt}\left[I_0\sin(\omega t)\right] = \omega L I_0\cos(\omega t)

This is a cosine while the current is a sine, so voltage leads current by 90° in a pure inductor. Writing the voltage amplitude as (effective resistance) × (current amplitude) identifies the inductive reactance:

Inductive reactance

X_L = \omega L

XL grows with frequency: an inductor opposes rapidly changing current more than slowly changing current, and offers zero opposition to true DC (ω = 0).

7.2.2 Pure capacitor

For a capacitor alone, charge follows $Q = CV$ , and current is the rate charge flows onto the plates, $I = dQ/dt = C\,dV/dt$ . Driving $V = V_0\sin(\omega t)$ across it gives:

I = C\dfrac{d}{dt}\left[V_0\sin(\omega t)\right] = \omega C V_0\cos(\omega t)

Here current is a cosine while voltage is a sine, so current leads voltage by 90° in a pure capacitor (the exact opposite of the inductor). The capacitive reactance is read off the same way:

Capacitive reactance

X_C = \dfrac{1}{\omega C}

XC falls with frequency: a capacitor charges and discharges easily under a rapidly alternating voltage, but blocks true DC entirely ( $X_C \to \infty$ as $\omega \to 0$ ).

Worked example

Reactance at a given frequency

Find XL and XC for L = 0.1 H and C = 10 µF at ω = 1000 rad/s.

X_L = \omega L = (1000)(0.1) = \textcolor{#e08a1e}{100\ \Omega}

X_C = \dfrac{1}{\omega C} = \dfrac{1}{(1000)(10\times10^{-6})} = \textcolor{#e08a1e}{100\ \Omega}

XL = XC here, meaning ω = 1000 rad/s is exactly this LC combination’s resonant frequency (check: 1/√LC = 1/√(0.1 × 10⁻⁵) = 1/√10⁻⁶ = 1000 rad/s ✓).

7.3 Series LCR Circuit: Impedance & Phasors

With R, L and C all in series, the same current I flows through all three, but each component’s voltage is phase-shifted differently relative to that current: VR is in phase with I, VL leads I by 90°, VC lags I by 90°. Representing each voltage as a phasor (a rotating arrow whose length is the amplitude and whose angle tracks the phase), VL and VC point in exactly opposite directions, since they’re 180° apart from each other, while VR points perpendicular to both. Adding the three phasors (VR along one axis, VL − VC along the perpendicular axis) by the Pythagorean theorem gives the net voltage amplitude, and dividing by it defines the circuit’s total opposition to current, the impedance:

Impedance

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

Current amplitude

I = \dfrac{V}{Z}

The phase angle by which current lags the source voltage follows from the same phasor triangle:

Phase angle

\tan\phi = \dfrac{X_L - X_C}{R}

When $X_L > X_C$ , the circuit is net inductive (current lags voltage); when $X_C > X_L$ , it’s net capacitive (current leads voltage).

Worked example

Impedance of a series LCR circuit

A series LCR circuit has R = 30 Ω, XL = 70 Ω, XC = 30 Ω. Find the impedance and the current amplitude for a 100 V source.

Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{30^2+(70-30)^2} = \sqrt{900+1600} = \sqrt{2500} = \textcolor{#e08a1e}{50\ \Omega}

I = \dfrac{V}{Z} = \dfrac{100}{50} = \textcolor{#e08a1e}{2\text{ A}}

7.4 Resonance & Voltage Magnification

Since only the $(X_L-X_C)^2$ term in Z depends on frequency, and it’s squared (so never negative), Z is smallest exactly when that term is zero, i.e. when $X_L = X_C$ . Setting $\omega L = 1/(\omega C)$ and solving gives the resonant angular frequency:

Resonance condition

\omega_0 = \dfrac{1}{\sqrt{LC}}

At $\omega = \omega_0$ , $Z = R$ , its absolute minimum, so current $I = V/R$ is at its absolute maximum. This is the sharpest, most exam-tested result in the chapter, and it comes with a genuinely surprising consequence.

Take L = 0.5 H, C = 2 × 10⁻⁶ F, R = 10 Ω, driven by V = 10 V. Resonance is at $\omega_0 = 1/\sqrt{(0.5)(2\times10^{-6})} = 1000$ rad/s. There, $X_L = \omega_0 L = 500\ \Omega$ and $X_C = 1/(\omega_0 C) = 500\ \Omega$ , and $I = V/R = 1.0$ A. Now look at the voltage across just the inductor: $V_L = IX_L = (1.0)(500) = 500$ V. And across just the capacitor: $V_C = IX_C = (1.0)(500) = 500$ V. The source supplies only 10 V, yet VL and VC are each 500 V, fifty times the source.

This doesn’t violate Kirchhoff’s voltage law. VL and VC are 180° out of phase with each other (one leads the current by 90°, the other lags by 90°), so as phasors they point in opposite directions and very nearly cancel, $V_L - V_C = 0$ at resonance exactly, leaving the loop’s phasor sum equal to VR = 10 V, consistent with the 10 V source. VL and VC are fighting each other, not adding onto VR, which is exactly why each is free to individually dwarf the source.

The ratio of this resonant internal voltage to the source voltage is called the quality factor:

Quality factor

Q = \dfrac{X_L}{R}\bigg|_{\omega_0} = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}

Here $Q = 500/10 = 50$ , matching the 50× magnification directly. A high Q means a sharp, narrow resonance peak and large internal voltages, exactly what lets a radio tuning circuit isolate one station from its crowded neighbours, and exactly why real resonant LC circuits need components rated well above the nominal supply voltage. See the interactive simulation to watch VL and VC cross the source voltage as ω sweeps through resonance.

Worked example

Resonance and Q factor

A series LCR circuit has L = 0.2 H, C = 50 µF, R = 20 Ω, driven by a 20 V source. Find the resonant frequency, the maximum current, and the Q factor.

\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(0.2)(50\times10^{-6})}} = \dfrac{1}{\sqrt{10^{-5}}} = \textcolor{#e08a1e}{316.23\text{ rad/s}}

I_{\max} = \dfrac{V}{R} = \dfrac{20}{20} = \textcolor{#e08a1e}{1.0\text{ A}}

Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}} = \dfrac{1}{20}\sqrt{\dfrac{0.2}{50\times10^{-6}}} = \dfrac{1}{20}\sqrt{4000} = \dfrac{63.25}{20} = \textcolor{#e08a1e}{3.16}

A much lower Q than 50, so the voltage magnification here is real but far milder: VL at resonance is only $I_{\max}X_L = (1.0)(63.25) = 63.25$ V, about 3.16 times the 20 V source, not 50 times.

7.4.1 LC oscillations: the same ω₀, from energy conservation alone

Strip the resistor out entirely and just connect a charged capacitor directly to an inductor. With nothing to dissipate energy, charge sloshes back and forth between them indefinitely, an electrical echo of a mass oscillating on a spring. The total energy stored, in the capacitor plus the inductor, never changes:

Total stored energy (constant)

U = \dfrac{Q^2}{2C} + \dfrac{1}{2}LI^2

Since $U$ is constant, $dU/dt=0$ . Using $I=dQ/dt$ and $dI/dt=d^2Q/dt^2$ :

\dfrac{dU}{dt} = \dfrac{Q}{C}\dfrac{dQ}{dt} + LI\dfrac{dI}{dt} = \dfrac{Q}{C}\dfrac{dQ}{dt} + L\dfrac{dQ}{dt}\dfrac{d^2Q}{dt^2} = 0

Dividing through by $dQ/dt$ (nonzero whenever the circuit is actually oscillating):

LC oscillator equation

\dfrac{d^2Q}{dt^2} + \dfrac{Q}{LC} = 0

Compare this directly to the mechanical SHM equation, $d^2x/dt^2 + (k/m)x = 0$ , which has solution $x=x_0\cos(\omega t+\phi)$ with $\omega=\sqrt{k/m}$ . The two equations have exactly the same form, $Q$ playing the role of $x$ , with $1/LC$ playing the role of $k/m$ , so the charge oscillates the same way, $Q(t)=Q_0\cos(\omega t+\phi)$ , with:

LC oscillation frequency

\textcolor{#e08a1e}{\omega} = \sqrt{\dfrac{1}{LC}} = \dfrac{1}{\sqrt{LC}}

the identical $\omega_0$ this section already reached by a completely different route, impedance matching in a driven LCR circuit. There, $\omega_0$ was the frequency an external AC source had to be tuned to, to make $X_L$ and $X_C$ cancel; here, with no external source and no resistance at all, it is simply the frequency the LC circuit oscillates at on its own, left to itself. The two derivations are really the same physics seen from opposite directions: a driven circuit resonates exactly at its own natural, undriven oscillation frequency.

Worked example

Checking the same L and C give the same frequency, two ways

Using the same inductor and capacitor from this section’s opening example, $L=0.5\text{ H}$ , $C=2\times10^{-6}\text{ F}$ , confirm the LC oscillator’s own natural frequency matches the resonant frequency found there.

\omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(0.5)(2\times10^{-6})}} = \dfrac{1}{\sqrt{10^{-6}}} = \textcolor{#e08a1e}{1000\text{ rad/s}}

Exactly the 1000 rad/s found earlier by setting $X_L=X_C$ , as it must be: disconnect the resistor and the source from that same circuit, and left alone, it would ring at precisely this frequency.

7.5 Power in an AC Circuit

Instantaneous power oscillates, but the quantity that matters for heating and energy billing is the time-averaged power, which depends on the phase angle φ between current and voltage:

Average power

P = V_{rms}I_{rms}\cos\phi

$\cos\phi$ is called the power factor. For a pure resistor, φ = 0 and the power factor is 1 (maximum power transfer for given Vrms, Irms). For a pure inductor or capacitor, φ = 90° and $\cos\phi = 0$ : despite current and voltage both being nonzero, no net power is dissipated over a full cycle, the energy sloshes into the field and back out again every half-cycle instead of being lost as heat.

Worked example

Average power with a phase lag

A circuit draws Vrms = 200 V and Irms = 5 A, with current lagging voltage by φ = 60°. Find the average power.

P = V_{rms}I_{rms}\cos\phi = (200)(5)\cos60^\circ = (1000)(0.5) = \textcolor{#e08a1e}{500\text{ W}}

7.6 Transformers

A transformer transfers AC power between two coils linked by a common iron core, stepping voltage up or down with the turns ratio (ideally, with no power loss):

Transformer turns ratio

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

where the subscripts p and s denote the primary (input) and secondary (output) coils. More secondary turns than primary steps voltage up; fewer steps it down. This is only a brief mention here, full transformer efficiency and losses are beyond this chapter’s scope.

— From the NCERT exercises

Two of the chapter’s own exercise questions, with original worked solutions.

NCERT Exercise 7.6

A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

Solution

With no source and no resistor at all, this is exactly the bare LC circuit of §7.4.1: charge sloshes between capacitor and inductor at the natural frequency derived there straight from energy conservation,

\omega_0 = \dfrac{1}{\sqrt{LC}}

with $L=27\times10^{-3}\text{ H}$ and $C=30\times10^{-6}\text{ F}$ :

\omega_0 = \dfrac{1}{\sqrt{(27\times10^{-3})(30\times10^{-6})}} = \dfrac{1}{\sqrt{8.1\times10^{-7}}} \approx \textcolor{#e08a1e}{1.1\times10^{3}\text{ s}^{-1}}

No driving source is needed for this number to mean something: disconnect this exact capacitor and inductor from everything else and, idealised as resistance-free, they will ring at 1100 rad/s on their own, indefinitely.

NCERT Exercise 7.7

A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 µF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Solution

“Frequency of the supply equals the natural frequency” is just resonance (§7.4): at $\omega=\omega_0$ , $X_L=X_C$ , so they cancel entirely inside the impedance and $Z=R$ , its minimum possible value. The 200 V rating is, as always, an rms value (§7.1), so the rms current is simply:

I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{V_{rms}}{R} = \dfrac{200}{20} = 10\text{ A}

At resonance $X_L-X_C=0$ , so the phase angle $\phi=0$ and the power factor $\cos\phi=1$ (§7.5), its maximum possible value too. The average power delivered is then just:

P = V_{rms}I_{rms}\cos\phi = (200)(10)(1) = \textcolor{#e08a1e}{2000\text{ W}}

Equivalently, $P=I_{rms}^2R = (10)^2(20) = 2000$ W: at resonance, with the reactances cancelled, the circuit behaves exactly like a plain 20 Ω resistor as far as power is concerned, and L = 1.5 H, C = 35 µF never enter the answer directly (they only fixed the frequency at which this happens).