Alternating Current — NEET Notes

AC voltage applied to a resistor

Begin with the simplest case. A resistor of resistance R is connected to a source that delivers a sinusoidally varying potential difference v = v_m sin ωt, where v_m is the peak voltage (amplitude) and ω is the angular frequency (= 2πν, with ν the frequency in hertz). Apply Kirchhoff's loop rule and the current at every instant must satisfy iR = v_m sin ωt, giving

i = i_m sin ωt,   where   i_m = v_m / R

Ohm's law works for AC across a pure resistor

The current and the voltage cross zero, reach their maxima, and reach their minima at exactly the same instants. They are in phase. Plotting both sinusoids against time, the two curves rise and fall together — there is no lag or lead. The instantaneous power dissipated is p = i²R = i_m² R sin²ωt. The current itself averages to zero over a complete cycle (positive and negative halves cancel), but its square does not — i² is always positive, and so is the Joule heating it produces.

RMS values — the equivalent DC current

If the average current is zero, how do we describe how "strong" an AC current is? The answer is the root mean square (rms) value. Take i², average it over one full cycle, and take the square root. Using the identity sin²ωt = ½(1 − cos 2ωt) and noting that <cos 2ωt> = 0 over a full cycle, <sin²ωt> = ½. So the average power dissipated in a resistor is

P = ½ i_m² R = I² R,   I = i_m/√2 ≈ 0.707 i_m

RMS current — the AC value that behaves like DC

Similarly V = v_m/√2. The rms (or "effective") current is the DC current that would produce the same Joule heating in the same resistor. Indian household mains is quoted as 220 V at 50 Hz — these are rms values; the actual peak voltage is v_m = √2 × 220 ≈ 311 V. AC ammeters and voltmeters always read rms, never peak. NEET 2022 asked exactly this relationship — peak = √2 × rms — phrased as a one-line statement.

Phasor representation of AC

Voltage and current in a resistor stay in phase. In an inductor or capacitor — and certainly in a series combination — they do not. Adding two sinusoids of the same frequency but different phases by simple algebra is painful; adding them as vectors is easy. That is the whole purpose of phasors.

A phasor is a vector that rotates anticlockwise about the origin at angular speed ω. Its magnitude equals the amplitude (peak value) of the quantity it represents; its vertical projection at any instant gives the instantaneous value. Because all currents and voltages in a single-frequency circuit rotate at the same ω, the angles between phasors stay fixed — the diagram is frozen in time and we can do vector addition on it. For a pure resistor, the V and I phasors point along the same line: phase angle zero.

AC voltage applied to an inductor

Now replace the resistor with a pure inductor of inductance L (assume negligible winding resistance). With v = v_m sin ωt driving the loop and no resistor in the loop, Kirchhoff's rule reads v − L(di/dt) = 0, so

di/dt = (v_m/L) sin ωt   ⇒   i = − (v_m/ωL) cos ωt = i_m sin(ωt − π/2)

Current in a pure inductor

The current amplitude is i_m = v_m / (ωL). The quantity ωL plays the role of resistance — it has units of ohms, and it limits the current. We call it the inductive reactance, X_L = ωL. The current lags the voltage by π/2, or one-quarter cycle. Physically: the back-emf in the coil opposes any change in current, so the current can never catch up with the driving voltage — it always reaches its peak a quarter cycle later. Reactance grows linearly with frequency, so inductors block high frequencies and pass low ones (and pass DC, where ω = 0, freely).

The instantaneous power supplied is p = v · i = − i_mv_m sin ωt · cos ωt = −(i_mv_m/2) sin 2ωt. Over a full cycle this averages to zero. A pure inductor consumes no average power. Energy flows into the magnetic field for one quarter cycle, then flows back out into the source during the next — a perfect oscillation.

AC voltage applied to a capacitor

Replace the inductor with a capacitor C. The instantaneous voltage across the capacitor is q/C, equal to the source voltage v_m sin ωt. So q = C v_m sin ωt, and differentiating:

i = dq/dt = ωC v_m cos ωt = i_m sin(ωt + π/2)

Current in a pure capacitor

The current amplitude is i_m = v_m/(1/ωC) = v_m/X_C, where capacitive reactance X_C = 1/ωC. The current leads the voltage by π/2: charge has to be deposited onto the plates before the voltage across them can build up, so the current flow peaks before the voltage does. Capacitive reactance is inversely proportional to frequency: a capacitor blocks DC entirely (X_C → ∞ at ω = 0) but passes high frequencies easily. NEET 2023 (Q.27) tested this directly: decrease the AC frequency, and the capacitive reactance increases.

As with an inductor, the average power supplied to a pure capacitor over one full cycle is zero. Energy oscillates between the source and the electric field stored on the plates. Only the resistor — out of R, L, and C — dissipates energy as heat. NEET 2016 (Q.173) asked this in plain language: over a full cycle, an ideal capacitor consumes no energy from the source.

R, L, C — the three AC elements

Before tackling the LCR series combination, freeze the three single-element cases side by side. Each one has its own "ohms-equivalent" — resistance, inductive reactance, or capacitive reactance — and its own phase relationship between current and voltage.

AC voltage applied to a series LCR circuit

Now connect R, L, and C in series with the AC source. The same current flows through all three at every instant — let it be i = i_m sin(ωt + φ), where φ is the (unknown) phase difference between source voltage and current. The voltage across the resistor (V_R) is in phase with the current. The voltage across the inductor (V_L) is π/2 ahead of the current. The voltage across the capacitor (V_C) is π/2 behind the current. So in the phasor diagram, V_L points up, V_C points down, and V_R points along the current axis. Their vector sum equals the source voltage V_m.

Because V_L and V_C are exactly opposite, they combine into a single phasor of magnitude |V_L − V_C|. The source voltage is then the hypotenuse of a right triangle with sides V_R and (V_L − V_C):

v_m² = (i_mR)² + (i_mX_L − i_mX_C)²   ⇒   Z = √( R² + (X_L − X_C)² )

Impedance of a series LCR circuit

The impedance Z plays the role of resistance — units of ohms, and it relates rms voltage to rms current by V = IZ. The phase angle between source voltage and current comes from the same triangle:

tan φ = (X_L − X_C) / R

Phase angle between source voltage and current

Three cases follow directly. If X_L > X_C, tan φ > 0, so φ is positive — the circuit is predominantly inductive and the current lags the voltage. If X_L < X_C, tan φ < 0, φ is negative — the circuit is predominantly capacitive and the current leads. If X_L = X_C, φ = 0, the circuit behaves as purely resistive at that frequency — and this is the resonant state.

Resonance and the Q-factor

Vary the frequency of the source. X_L = ωL grows linearly; X_C = 1/ωC falls hyperbolically. At one special frequency the two cross — and at that frequency the inductive and capacitive contributions to impedance cancel exactly. Z drops to its minimum value R, and the current amplitude i_m = v_m/R reaches its maximum. Setting X_L = X_C:

ω₀ = 1/√(LC)   ⇒   f₀ = 1/(2π√(LC))

Resonant frequency of a series LCR circuit

NEET 2023 Q.26 tested precisely this formula — L = 10 mH, C = 1 μF gives f₀ = 1/(2π√(10⁻⁸)) ≈ 1.59 kHz. NEET 2022 Q.45 set ω = 100 rad/s for L = 10 H, C = 10 μF, asking the candidate to recognise that the source frequency equals the resonant frequency.

How sharp is the resonance? That depends on how quickly the current falls away as you move off ω₀, which in turn depends on R. The quality factor or Q-factor measures this sharpness:

Q = ω₀L / R = 1 / (ω₀CR)

Quality factor — sharpness of resonance

A high Q means a tall, narrow current-vs-frequency peak. Small R → high Q → sharply selective circuit. Operationally, the half-power bandwidth Δω of the resonance curve obeys Q = ω₀/Δω. NEET 2021 Q.47 used exactly this — for ω₀ = 50 rad/s and R/(2L) = 4, the half-power frequencies are 50 ± 4 = 46 and 54 rad/s.

Resonance is what makes a radio work. The antenna picks up signals from every station simultaneously; the tuning knob adjusts a variable capacitor so that f₀ of the LC tank circuit matches the carrier frequency of the station you want. At that frequency the current is huge; at every other frequency the impedance is large and the response is suppressed.

Power in an AC circuit & power factor

With v = v_m sin ωt and i = i_m sin(ωt + φ), the instantaneous power is

p = v · i = v_m i_m sin ωt · sin(ωt + φ) = ½ v_m i_m [cos φ − cos(2ωt + φ)].

The second term has average zero. The first is steady. So the average power over one cycle is

P = ½ v_m i_m cos φ = V_rms I_rms cos φ

Average power in any AC circuit

The factor cos φ is called the power factor. The product V_rms I_rms is called the apparent power; only the fraction cos φ of it is the real (dissipated) power. Four cases tell you everything:

• Pure resistive (φ = 0): cos φ = 1, maximum power dissipation. All apparent power is real.
• Pure inductive or capacitive (φ = ±π/2): cos φ = 0. No power dissipated. The current that still flows is called the wattless current.
• General LCR (0 < |φ| < π/2): 0 < cos φ < 1. Power dissipated only in R.
• At resonance (X_L = X_C, φ = 0): cos φ = 1, P = I²R is maximum.

NEET 2020 Q.108 used a sharp logical trick. With L removed, the phase angle equals some value; with C removed, the phase angle equals the same value. The only way that can happen in a series LCR circuit is if X_L = X_C originally — i.e. the full circuit was at resonance. So Z = R, power factor cos φ = 1. NEET 2018 Q.39 and NEET 2016 Q.175 both ask the standard "power dissipated = V_rms²R/Z²" computation; both have resonance-or-near-resonance flavour.

Transformers — why AC won

The single biggest reason power grids run on AC and not DC is the transformer. A transformer changes voltage and current by a fixed ratio, with very little energy loss, using nothing but two coils and a piece of iron. Build it with a primary winding of N_p turns and a secondary winding of N_s turns on a common laminated soft-iron core. Apply alternating voltage to the primary. The changing magnetic flux through the core links both windings equally (in the ideal case), so by Faraday's law:

V_s / V_p = N_s / N_p

Ideal transformer voltage ratio

For an ideal transformer (no losses), power in equals power out: V_p I_p = V_s I_s. So the current ratio is the inverse of the turn ratio:

I_s / I_p = N_p / N_s. NEET 2023 Q.29 used this directly: a 220 V mains feeds a step-down transformer running a 12 V, 60 W lamp. Output power = 60 W = input power = 220 × I_p, so I_p = 60/220 = 0.27 A. NEET 2021 Q.45 is the same template: 220 V → 11 V, 44 W lamp gives I_p = 44/220 = 0.2 A.

A step-up transformer has N_s > N_p, increases voltage, and decreases current. A step-down transformer has N_s < N_p, decreases voltage, and increases current. The big practical use is long-distance transmission: a generator at the power plant feeds a step-up transformer that pushes the line voltage to hundreds of kilovolts, so the line current — and therefore the I²R loss in the transmission cables — is tiny. At the consumer end, step-down transformers reduce the voltage back to the safe 240 V that reaches your wall socket.

Real transformers are not 100% efficient. Four loss mechanisms eat into the ideal P_in = P_out:

A well-designed power transformer achieves efficiency above 95%. The fact that AC can be transformed up and down almost losslessly is exactly why Tesla and Westinghouse — and not Edison's DC — won the late-19th-century War of Currents.