Alternating Current

Every socket in your home carries a current that never settles down: it climbs to a peak, falls through zero, reverses, climbs to a peak the other way, and repeats fifty times a second. This is alternating current, and almost the entire electrical world — generation at a power plant, transmission across hundreds of kilometres, and the humble bulb or charger in your room — runs on it rather than on the steady direct current of a battery. The reason is not an accident of history. Alternating voltage can be stepped up for efficient long-distance transmission and stepped down for safe home use, with no moving parts, simply by exploiting mutual induction between two coils. That single device, the transformer, is the economic reason AC won the "war of the currents" over DC more than a century ago, and it is one of the two big destinations of this chapter. The other destination is a piece of physics that shows up everywhere from radio tuning to metal detectors: resonance. Resistors, inductors and capacitors each respond to an alternating voltage in their own characteristic way — a resistor tracks the voltage exactly, an inductor's current lags behind it, a capacitor's current runs ahead of it. Wire all three together in series and drive them with a variable-frequency source, and at one special frequency the inductor's and capacitor's opposing effects cancel exactly, current surges to its maximum, and the circuit becomes sharply selective about which frequency it accepts. This chapter builds that whole picture from the ground up: how to describe a sinusoidal quantity compactly with rotating vectors called phasors, why "root mean square" is the sensible way to quote an AC voltage or current, how reactance replaces resistance for inductors and capacitors, how impedance and phase angle summarise a full series LCR circuit, and finally how transformers use the same physics to move electrical energy across a country with minimum loss.

Key takeaways

An AC voltage $v = v_m \sin\omega t$ across a resistor drives a current in phase with it, $i = i_m \sin\omega t$ with $i_m = v_m/R$ ; only a resistor dissipates energy in an AC circuit.
The useful "size" of an AC quantity is its rms value: $I = i_m/\sqrt2 \approx 0.707\,i_m$ and $V = v_m/\sqrt2$ , defined so that $P = I^2R$ matches the familiar DC power formula; India's 220 V mains is an rms value with peak $311$ V.
A pure inductor makes the current lag the voltage by a quarter cycle ( $\pi/2$ ); its inductive reactance $X_L = \omega L$ grows with frequency, and it dissipates zero average power over a cycle.
A pure capacitor makes the current lead the voltage by a quarter cycle; its capacitive reactance $X_C = 1/\omega C$ shrinks with frequency, and it too dissipates zero average power.
In a series LCR circuit the current amplitude is $i_m = v_m/Z$ with impedance $Z = \sqrt{R^2 + (X_C - X_L)^2}$ , and the current leads or lags the voltage by phase angle $\varphi = \tan^{-1}\!\big(\frac{X_C - X_L}{R}\big)$ .
Resonance occurs when $X_L = X_C$ , at angular frequency $\omega_0 = 1/\sqrt{LC}$ ; there impedance is minimum ( $Z = R$ ) and current amplitude is maximum ( $i_m = v_m/R$ ) — this needs both $L$ and $C$ present, never just one.
How sharply the current peaks near resonance is measured by the quality factor $Q = \omega_0 L/R = 1/(\omega_0 CR)$ ; a high- $Q$ circuit is narrowly selective, which is exactly what a radio tuner needs.
Average power delivered to an AC circuit is $P = VI\cos\varphi$ , where $\cos\varphi$ is the power factor; purely reactive circuits carry current but deliver zero power (a "wattless" current), while resonance gives maximum power transfer with $\cos\varphi = 1$ .
A transformer uses mutual induction between a primary and secondary coil wound on a common iron core to change AC voltage: $V_s/V_p = N_s/N_p$ and, for an ideal lossless transformer, $I_s/I_p = N_p/N_s$ — it steps voltage up or down while conserving power, which is why power is transmitted at high voltage and low current to cut $I^2R$ losses.
Key idea: an AC voltage of angular frequency ω drives currents whose amplitude and phase are set by reactance (XL = ωL, XC = 1/ωC), which combine in a series LCR circuit into an impedance Z = √(R² + (XC − XL)²) that is minimized at the resonant frequency ω0 = 1/√(LC), while delivered power is governed by the power factor cos φ = R/Z, and transformers exploit the same mutual induction to trade voltage for current in the ratio Ns/Np for efficient long-distance transmission.

The core insight

An alternating voltage $v = v_m\sin\omega t$ produces currents whose amplitude and phase shift are fixed by resistance and by frequency-dependent reactance ( $X_L = \omega L$ , $X_C = 1/\omega C$ ), so a series LCR circuit has impedance $Z = \sqrt{R^2+(X_C-X_L)^2}$ , resonates with maximum current at $\omega_0 = 1/\sqrt{LC}$ , delivers average power $P = VI\cos\varphi$ set by the power factor, and can be voltage-transformed losslessly by a transformer in the ratio $N_s/N_p$ .

What this chapter covers

Why AC, and how to describe it — the sinusoidal voltage $v = v_m\sin\omega t$ , and why rms values, not peak values, are the everyday way to quote AC voltage and current.
Phasors — a sinusoidal voltage or current as a rotating vector, so phase relationships become a simple picture instead of a page of trigonometry.
The three basic elements taken one at a time — how a resistor, an inductor and a capacitor each respond to an AC voltage alone, defining inductive reactance $X_L$ and capacitive reactance $X_C$ .
The series LCR circuit — combining all three, using phasors to find the total impedance $Z$ and the phase angle $\varphi$ .
Resonance — the special frequency where the circuit's opposition to current is minimum, why it needs both $L$ and $C$ , and the quality factor.
Power in AC circuits — average power, the power factor $\cos\varphi$ , wattless current, and why power factor correction matters.
Transformers — mutual induction changing AC voltage up or down, the ideal transformer equations, real energy losses, and why grids transmit at high voltage.

The core ideas

AC voltage applied to a resistor

Picture the simplest possible AC circuit: a source producing a sinusoidal voltage connected straight across a resistor, nothing else. Because a resistor has no memory — its current at any instant depends only on the voltage at that same instant — Kirchhoff's loop rule applied to this single loop gives the current immediately, with no calculus needed.

v = v_m \sin\omega t, \qquad i = \frac{v}{R} = i_m \sin\omega t, \qquad i_m = \frac{v_m}{R}

where: $v_m$ is the peak (amplitude) of the source voltage, $\omega = 2\pi\nu$ is its angular frequency for supply frequency $\nu$ , $R$ is the resistance, and $i_m$ is the peak current. Reading this physically: current and voltage rise, fall, and cross zero at exactly the same instants — they are in phase. This is the one element in this whole chapter for which AC behaves exactly like DC, element for element, at every instant. Because the current is symmetric about zero, its plain time-average over a cycle is zero — yet the resistor clearly gets hot, so it is obviously absorbing energy. The resolution is that instantaneous power $p = i^2R$ depends on $i^2$ , which is always positive, so heating never averages to zero even though current does. This motivates the rms (root mean square) value, derived in full below:

I = \frac{i_m}{\sqrt2} = 0.707\,i_m, \qquad V = \frac{v_m}{\sqrt2} = 0.707\,v_m, \qquad P = I^2R = IV

where: $I$ and $V$ are the rms current and voltage — the values every ammeter, voltmeter, and appliance rating actually quotes. When a supply is described as "220 V, 50 Hz," the 220 V is an rms value; the peak voltage that briefly appears across the wires each cycle is $v_m = \sqrt2\,(220\ \text{V}) \approx 311\ \text{V}$ .

Phasors: a rotating-vector picture of AC quantities

Once an inductor or capacitor enters a circuit, current and voltage stop being in phase, and keeping track of how much out of phase becomes essential. NCERT introduces the phasor: a vector of length equal to the amplitude ( $v_m$ or $i_m$ ), rotating counter-clockwise about the origin at angular speed $\omega$ . Its vertical (projection) component at any instant equals the instantaneous value of the quantity it represents. Phasors are not real vectors in space — voltage and current are scalars — but their amplitudes and phases combine mathematically exactly like the projections of rotating vectors, so the ordinary rule for adding vectors can be borrowed to add sinusoidal quantities correctly. For the resistor above, $\mathbf{I}$ and $\mathbf{V}$ point in the same direction at every instant — "phase difference zero." For an inductor or capacitor, as the next two sections show, the current phasor sits rotated by a fixed $90°$ relative to the voltage phasor — always lagging for an inductor, always leading for a capacitor.

AC voltage applied to an inductor

Now connect the source to a pure inductor of self-inductance $L$ (negligible winding resistance). Kirchhoff's loop rule includes the self-induced back-emf from Faraday's law, $L\,di/dt$ , in place of an $iR$ drop:

v_m \sin\omega t = L\,\frac{di}{dt} \quad\Longrightarrow\quad \frac{di}{dt} = \frac{v_m}{L}\sin\omega t

Integrating with respect to time (the constant of integration must vanish, since the current has no reason to carry a steady, non-oscillating offset) gives

i = i_m \sin\!\left(\omega t - \frac{\pi}{2}\right), \qquad i_m = \frac{v_m}{X_L}, \qquad X_L = \omega L

where: $X_L$ is the inductive reactance, playing the same role as $R$ did for the resistor but depending on frequency. The physical reading of the $-\pi/2$ phase: the current reaches its peak a quarter-cycle after the voltage does — the inductor's back-emf actively opposes any change in current, so current always struggles to "catch up" to the driving voltage. Because $X_L = \omega L$ grows with frequency, an inductor increasingly chokes off high-frequency current while barely affecting DC ( $\omega \to 0$ gives $X_L \to 0$ ). Computing the instantaneous power $p = iv$ and averaging over a cycle gives exactly zero — the inductor alternately stores energy in its magnetic field and gives it straight back, absorbing no net energy.

AC voltage applied to a capacitor

Connect the source instead to a capacitor of capacitance $C$ . Here it is easiest to start from the charge: at any instant the source voltage equals the capacitor voltage, $v_m\sin\omega t = q/C$ , so $q = Cv_m\sin\omega t$ , and the current is the rate of change of charge:

i = \frac{dq}{dt} = \omega C v_m \cos\omega t = i_m \sin\!\left(\omega t + \frac{\pi}{2}\right), \qquad i_m = \frac{v_m}{X_C}, \qquad X_C = \frac{1}{\omega C}

where: $X_C$ is the capacitive reactance. Physically the current now leads the voltage by a quarter cycle: current must flow first to deposit charge before the voltage across the plates can build up, so the current-versus-time curve reaches its peak before the voltage does. Because $X_C = 1/\omega C$ falls as frequency rises, a capacitor passes high-frequency current relatively freely but almost completely blocks DC ( $\omega \to 0$ gives $X_C \to \infty$ ) — the opposite frequency behaviour to an inductor. As with the inductor, the average power delivered to a capacitor over a full cycle works out to exactly zero: energy is stored in the electric field on the charging quarter-cycle and returned on the discharging quarter-cycle. A simple memory aid ties the two together: in an inductor, voltage leads current ("ELI" — emf, then $L$ , then $I$ ); in a capacitor, current leads voltage ("ICE" — $I$ , then $C$ , then emf).

The series LCR circuit: impedance and phase angle

Wire a resistor, inductor and capacitor in series across the same AC source. Because they share one loop, the same current $i = i_m\sin(\omega t + \varphi)$ flows through all three at every instant — only the phase $\varphi$ relative to the source voltage is still unknown, and only the voltage across each element differs, since each element has its own opposition to that shared current.

Figure

Phasor diagram of a series LCR circuit showing the current phasor as reference, the resistor voltage in phase with it, the inductor and capacitor voltages at right angles in opposite directions, their combined phasor, and the source voltage as the hypotenuse making phase angle phi with the current

Since $\mathbf{V_R}$ stays parallel to $\mathbf{I}$ , $\mathbf{V_L}$ sits $\pi/2$ ahead of $\mathbf{I}$ , and $\mathbf{V_C}$ sits $\pi/2$ behind it, $\mathbf{V_L}$ and $\mathbf{V_C}$ point along the same vertical line in opposite directions and can be combined into one net phasor before the final addition. The full construction (carried out step by step in the derivation below) yields:

i_m = \frac{v_m}{Z}, \qquad Z = \sqrt{R^2 + (X_C - X_L)^2}, \qquad \tan\varphi = \frac{X_C - X_L}{R}

where: $Z$ is the impedance of the circuit (ohms), the AC analogue of resistance for a mixed circuit, and $\varphi$ is the phase by which the current leads ( $\varphi>0$ , when $X_C$ is greater than $X_L$ , a capacitance-dominated circuit) or lags ( $\varphi<0$ , when $X_L$ is greater than $X_C$ , an inductance-dominated circuit) the source voltage. Only $R$ ever appears alone in $Z$ without a frequency-dependent partner — a signal that only the resistor is responsible for whatever energy is actually dissipated.

Resonance and the quality factor

Hold $R$ , $L$ and $C$ fixed but let the driving frequency $\omega$ vary, and something striking happens: because $X_L = \omega L$ grows with frequency while $X_C = 1/\omega C$ shrinks, there is exactly one frequency at which they become equal and cancel inside $Z$ .

X_C = X_L \;\Longrightarrow\; \frac{1}{\omega_0 C} = \omega_0 L \;\Longrightarrow\; \omega_0 = \frac{1}{\sqrt{LC}}

where: $\omega_0$ is the resonant angular frequency, fixed purely by $L$ and $C$ (not by $R$ ). At $\omega = \omega_0$ the impedance collapses to its smallest possible value, $Z = R$ , so the current amplitude is largest: $i_{m,\text{max}} = v_m/R$ .

Figure

Graph of current amplitude versus angular frequency for a series LCR circuit for two resistance values showing a sharp tall resonance peak for the smaller resistance and a shorter broader peak for the larger resistance both centred on the same resonant frequency

The peak's sharpness is captured by the quality factor:

Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}

where: a larger $Q$ (smaller $R$ , relative to $L$ and $C$ ) gives a taller, narrower resonance peak — the circuit responds strongly to a narrow band of frequencies around $\omega_0$ and weakly to everything else. This selectivity is what lets a radio tuner pick one station out of the many signals its antenna picks up: varying the capacitance shifts $\omega_0$ to match the desired station, and a high $Q$ rejects the neighbours. A metal detector uses the same principle — nearby metal changes a tuned coil's effective inductance, shifting it off resonance and producing a detectable current change. Resonance needs both $L$ and $C$ present together: an $R$ – $L$ or $R$ – $C$ circuit alone has no frequency at which the reactive voltages could cancel, so $Z$ never dips to a minimum of $R$ alone.

Power in an AC circuit: the power factor

The instantaneous power delivered by the source is $p = vi = (v_m\sin\omega t)[i_m\sin(\omega t+\varphi)]$ , which expands (using a product-to-sum identity) into a constant term plus a term oscillating at $2\omega$ that averages to zero over a full cycle. What survives the averaging is:

P = \frac{v_m i_m}{2}\cos\varphi = VI\cos\varphi = I^2 Z\cos\varphi

where: $\cos\varphi$ is the power factor, a pure number between $0$ and $1$ that measures how much of the apparent product $VI$ is actually delivered as usable power. Four cases are worth holding in mind together. In a purely resistive circuit, $\varphi = 0$ , so $\cos\varphi = 1$ and power delivery is maximum for the given $V$ and $I$ . In a purely inductive or capacitive circuit, $\varphi = \pm\pi/2$ , so $\cos\varphi = 0$ and no power is delivered even though current visibly flows — this current is called a wattless current. In a general series LCR circuit, $\varphi = \tan^{-1}\!\big(\frac{X_C-X_L}{R}\big)$ can take any value, and even then only the resistor actually dissipates energy — the power formula above is exactly $I^2R$ once $Z\cos\varphi$ is simplified, since $\cos\varphi = R/Z$ . At resonance, $X_C = X_L$ gives $\varphi = 0$ and $\cos\varphi = 1$ , so the resonant condition is also the condition of maximum power transfer. A low power factor matters commercially: to deliver a fixed real power $P$ at fixed voltage $V$ , a smaller $\cos\varphi$ demands a larger current $I = P/(V\cos\varphi)$ , and a larger current means larger $I^2R$ losses in the wires for no extra useful power. Industrial loads such as motors are usually inductive; power factor correction adds a parallel capacitor whose leading wattless current cancels the motor's lagging wattless current, pulling $\cos\varphi$ back toward $1$ .

Transformers

A transformer changes an AC voltage from one value to another using nothing but two coils and mutual induction — no moving parts, and (in the ideal case) no loss of energy. A primary coil of $N_p$ turns and a secondary coil of $N_s$ turns are wound on a common soft-iron core, so essentially all of the flux produced by the primary's current also threads the secondary.

\frac{V_s}{V_p} = \frac{N_s}{N_p}, \qquad \frac{I_s}{I_p} = \frac{N_p}{N_s}

where: $V_p, I_p$ are the (rms) primary voltage and current, $V_s, I_s$ the secondary values, and $N_p, N_s$ the number of turns in each coil — the full derivation (from Faraday's law plus energy conservation) is worked below. If $N_s$ is greater than $N_p$ , the device is a step-up transformer: it raises voltage and correspondingly lowers current. If $N_s$ is less than $N_p$ , it is a step-down transformer, lowering voltage and raising current. Because power is (ideally) conserved, $I_pV_p = I_sV_s$ at every instant — a transformer trades voltage for current, never creates energy. Real transformers fall short of $100\%$ efficiency (though well-designed ones exceed $95\%$ ) for four reasons: flux leakage (reduced by winding the coils tightly over one another), resistive heating in the copper windings ( $I^2R$ losses, reduced with thick wire in high-current windings), eddy currents in the iron core (reduced with a laminated core), and hysteresis losses from repeatedly reversing the core's magnetisation (reduced with a low-hysteresis material). This is precisely why power grids do not send electricity at the voltage a home appliance uses. The generator's output is stepped up to a very high voltage for transmission, so the current in the long wires stays small and $I^2R$ line loss stays small, and it is only stepped back down in stages, at area and local sub-stations, to the safe voltage that reaches a home socket.

Derivation AThe rms value of an alternating current

The plain time-average of a sinusoidal current is zero, because positive and negative half-cycles cancel exactly — yet a resistor carrying that current visibly heats up, so it is clearly absorbing energy on average. The goal here is to define a single number, the root-mean-square (rms) current, that captures this "effective" heating ability of an alternating current in a form that lets the ordinary DC power formula $P = I^2R$ keep working unchanged, and then to do the identical calculation for voltage.

Step 1. Write the instantaneous power dissipated in a resistor carrying $i = i_m\sin\omega t$ : $p = i^2R = i_m^2R\sin^2\omega t$ . This is never negative, since it depends on $i^2$ , so unlike the current itself, its average over a cycle need not be zero.

Step 2. Average $\sin^2\omega t$ over one complete cycle using the identity $\sin^2\omega t = \tfrac12(1-\cos2\omega t)$ . Averaging the constant $\tfrac12$ gives $\tfrac12$ ; averaging $\cos2\omega t$ over an integer number of its own cycles gives exactly zero, since its positive lobes and negative lobes are equal and opposite. So $\langle\sin^2\omega t\rangle = \tfrac12$ .

Step 3. Substitute back to get the average power: $\overline{p} = i_m^2R\langle\sin^2\omega t\rangle = \tfrac12 i_m^2R$ . This is a fixed, non-zero number — the actual, physically-measured heating rate averaged over any whole number of cycles.

Step 4. Define the rms current $I$ as the steady (DC-equivalent) current that would give the same average power in the same resistor: demand $I^2R = \overline{p} = \tfrac12 i_m^2R$ , cancel $R$ , and take the square root: $I = i_m/\sqrt2 \approx 0.707\,i_m$ . Exactly the same reasoning applied to the source voltage, using $p = v_m i_m\sin^2\omega t$ and its own average, gives $V = v_m/\sqrt2$ .

Step 5. Check consistency: since $v_m = i_mR$ for a resistor, dividing both sides by $\sqrt2$ gives $V = IR$ — the AC equation reproduces Ohm's law in exactly the DC form, and $P = I^2R = IV$ follows immediately. This is why "220 V, 50 Hz mains" quotes an rms value: it is defined precisely so that a $220\ \text{V}$ AC supply delivers the same power to a resistor as a genuine $220\ \text{V}$ steady battery would, even though the instantaneous voltage swings all the way up to $v_m = \sqrt2(220) \approx 311\ \text{V}$ and down to $-311\ \text{V}$ every cycle.

Go deeper · Derivation B · 2 minImpedance and phase angle of a series LCR circuit

Derivation BImpedance and phase angle of a series LCR circuit

The goal is to find the amplitude and phase of the single current that flows through a resistor, inductor and capacitor wired in series, using only the phasor amplitudes already established for each element separately — $v_{Rm}=i_mR$ , $v_{Lm}=i_mX_L$ , $v_{Cm}=i_mX_C$ — combined geometrically rather than by solving the circuit's differential equation directly. The payoff is the impedance formula $Z=\sqrt{R^2+(X_C-X_L)^2}$ and the resonance condition that follows immediately from it.

Step 1. Since the three elements share one loop, the same current $i = i_m\sin(\omega t+\varphi)$ flows through all of them, with the same amplitude and phase everywhere. Represent it by a reference phasor $\mathbf I$ . From the single-element results already derived, the resistor's voltage phasor $\mathbf{V_R}$ is parallel to $\mathbf I$ (amplitude $i_mR$ ), the inductor's voltage phasor $\mathbf{V_L}$ is rotated $\pi/2$ ahead of $\mathbf I$ (amplitude $i_mX_L$ ), and the capacitor's voltage phasor $\mathbf{V_C}$ is rotated $\pi/2$ behind $\mathbf I$ (amplitude $i_mX_C$ ).

Step 2. Because $\mathbf{V_L}$ and $\mathbf{V_C}$ point along the same line but in opposite directions (one $+90°$ from $\mathbf I$ , the other $-90°$ ), they combine directly, by ordinary subtraction, into a single net phasor of magnitude $i_m|X_C-X_L|$ lying along that same perpendicular line. This is the one simplification that makes the LCR problem tractable by geometry alone: two of the three voltage phasors collapse into one before any triangle needs to be drawn.

Step 3. Kirchhoff's loop rule requires $v_L+v_R+v_C=v$ at every instant, which as a phasor statement is $\mathbf{V_L}+\mathbf{V_R}+\mathbf{V_C}=\mathbf V$ . Geometrically this is a right triangle: one leg is $\mathbf{V_R}$ (along $\mathbf I$ ), the other leg is the net $(\mathbf{V_C}-\mathbf{V_L})$ (perpendicular to $\mathbf I$ ), and the hypotenuse is $\mathbf V$ , the source voltage phasor. The Pythagorean theorem gives $v_m^2 = v_{Rm}^2+(v_{Cm}-v_{Lm})^2 = i_m^2R^2 + i_m^2(X_C-X_L)^2$ .

Step 4. Factor out $i_m^2$ and take the square root to isolate the current amplitude: $i_m = \dfrac{v_m}{\sqrt{R^2+(X_C-X_L)^2}} = \dfrac{v_m}{Z}$ , which defines the impedance $Z=\sqrt{R^2+(X_C-X_L)^2}$ — a single ohm-valued quantity that plays exactly the role $R$ played for a bare resistor, but built from all three elements together.

Step 5. The phase angle $\varphi$ between $\mathbf V$ and $\mathbf I$ is read off the same right triangle: $\tan\varphi = \dfrac{v_{Cm}-v_{Lm}}{v_{Rm}} = \dfrac{X_C-X_L}{R}$ . Finally, set $X_C=X_L$ (the impedance triangle's second leg vanishes, so $Z$ shrinks to its minimum possible value $R$ , and $\varphi\to0$ ): this is resonance, $\dfrac{1}{\omega_0C}=\omega_0L$ , giving $\omega_0=\dfrac{1}{\sqrt{LC}}$ and a maximum current amplitude $i_m=v_m/R$ .

Go deeper · Derivation C · 2 minVoltage and current transformation in an ideal transformer

Derivation CVoltage and current transformation in an ideal transformer

The goal is to derive the two defining transformer equations, $V_s/V_p = N_s/N_p$ and $I_s/I_p = N_p/N_s$ , starting only from Faraday's law of induction applied to the flux shared by the primary and secondary windings, plus conservation of energy for an ideal, loss-free device. This shows that a transformer's turns ratio alone fixes how it trades voltage for current, with power (not voltage, and not current alone) the conserved quantity.

Step 1. Let $\phi(t)$ be the magnetic flux through one turn of the shared iron core at time $t$ , assumed (for an ideal transformer) identical for every turn of both windings, since essentially no flux leaks out of a well-designed core. By Faraday's law, the emf induced in the secondary, with $N_s$ turns, is $\varepsilon_s = -N_s\,\dfrac{d\phi}{dt}$ , and with negligible secondary current drawn this equals the secondary terminal voltage in magnitude, $v_s = N_s\,\dfrac{d\phi}{dt}$ .

Step 2. The same changing flux also induces a back-emf in the primary itself, $\varepsilon_p = -N_p\,\dfrac{d\phi}{dt}$ . Because the primary is assumed to have negligible resistance, Kirchhoff's loop rule forces the applied primary voltage to balance this back-emf exactly (otherwise an unopposed voltage would drive an unphysical infinite current through zero resistance): $v_p = N_p\,\dfrac{d\phi}{dt}$ .

Step 3. Divide the Step 1 result by the Step 2 result. The common factor $d\phi/dt$ cancels completely, leaving a relation that depends only on the turns counts: $\dfrac{v_s}{v_p} = \dfrac{N_s}{N_p}$ . Since primary and secondary oscillate at the same frequency with the same waveform, this ratio holds equally for peak values and for rms values, so $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ .

Step 4. Now invoke energy conservation for an ideal (loss-free) transformer: input power equals output power at every instant, $i_pv_p = i_sv_s$ . Rearranging gives $\dfrac{i_p}{i_s} = \dfrac{v_s}{v_p} = \dfrac{N_s}{N_p}$ , and inverting, $\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}$ — the current ratio is the exact reciprocal of the voltage ratio, so their product, the power, stays fixed.

Step 5. Check with the textbook's own numbers: a primary of $100$ turns and secondary of $200$ turns has $N_s/N_p = 2$ , so a $220\ \text{V}$ , $10\ \text{A}$ primary supply is transformed to $V_s = 2(220) = 440\ \text{V}$ at $I_s = 10/2 = 5.0\ \text{A}$ . Power in is $(220)(10) = 2200\ \text{W}$ ; power out is $(440)(5.0) = 2200\ \text{W}$ — identical, exactly as energy conservation demands, even though voltage doubled and current halved.

Worked examples

Board level - Resistance, peak voltage and rms current of a light bulb

A light bulb is rated $100\ \text{W}$ for a $220\ \text{V}$ AC supply. Find its resistance, the peak voltage of the source, and the rms current through it.

Step 1. The rated power and rms voltage give the resistance directly from $P = V^2/R$ : $R = \dfrac{V^2}{P} = \dfrac{(220\ \text{V})^2}{100\ \text{W}} = \dfrac{48400}{100}\ \Omega = 484\ \Omega$ .

Step 2. The peak (amplitude) of the supply voltage follows from $v_m = \sqrt2\,V$ : $v_m = \sqrt2\,(220\ \text{V}) = 1.414\times220\ \text{V} \approx 311\ \text{V}$ .

Step 3. Using $P = IV$ (valid for a resistive load, where voltage and current are in phase), the rms current is $I = \dfrac{P}{V} = \dfrac{100\ \text{W}}{220\ \text{V}} \approx 0.454\ \text{A}$ . The bulb's resistance is 484 Ω, the peak supply voltage is 311 V, and the rms current through it is 0.454 A.

Board level - A resistor and capacitor in series, and the voltage-addition paradox

A resistor of $200\ \Omega$ and a capacitor of $15.0\ \mu\text{F}$ are connected in series to a $220\ \text{V}$ , $50\ \text{Hz}$ AC source. Find the current in the circuit and the rms voltages across each element, and explain why their arithmetic sum exceeds the source voltage.

Step 1. Find the capacitive reactance: $X_C = \dfrac{1}{2\pi\nu C} = \dfrac{1}{2\times3.14\times50\times15.0\times10^{-6}}\ \Omega \approx 212\ \Omega$ .

Step 2. Find the impedance of this series RC circuit (no inductor, so it is the $R$ – $X_C$ case of the general formula with $X_L=0$ ): $Z = \sqrt{R^2+X_C^2} = \sqrt{(200)^2+(212)^2}\ \Omega \approx 291.5\ \Omega$ .

Step 3. The rms current, common to both elements, is $I = \dfrac{V}{Z} = \dfrac{220\ \text{V}}{291.5\ \Omega} \approx 0.755\ \text{A}$ .

Step 4. The rms voltage across each element is $V_R = IR = (0.755)(200) \approx 151\ \text{V}$ and $V_C = IX_C = (0.755)(212) \approx 160\ \text{V}$ . Their plain arithmetic sum is $151+160 = 311\ \text{V}$ , which is larger than the $220\ \text{V}$ source — an apparent paradox, resolved by recognising that $V_R$ and $V_C$ are $90°$ out of phase and must be combined as phasors, not ordinary numbers: $V_{R+C} = \sqrt{V_R^2+V_C^2} = \sqrt{151^2+160^2}\ \text{V} \approx 220\ \text{V}$ , which correctly reproduces the source voltage. The current is 0.755 A; the resistor and capacitor carry 151 V and 160 V respectively, and their correctly-phased (Pythagorean) sum equals the source voltage of 220 V, not their simple arithmetic total.

Competitive level (JEE / NEET) - Impedance, phase, power, and resonance in a series LCR circuit

A sinusoidal voltage of peak value $283\ \text{V}$ and frequency $50\ \text{Hz}$ is applied to a series LCR circuit with $R = 3\ \Omega$ , $L = 25.48\ \text{mH}$ , $C = 796\ \mu\text{F}$ . Find the impedance, the phase difference between voltage and current, the power dissipated, and the power factor. Then find the frequency at which the same circuit would resonate, and the power dissipated there.

Step 1. Compute both reactances at $50\ \text{Hz}$ : $X_L = 2\pi\nu L = 2\times3.14\times50\times25.48\times10^{-3}\ \Omega \approx 8\ \Omega$ , and $X_C = \dfrac{1}{2\pi\nu C} = \dfrac{1}{2\times3.14\times50\times796\times10^{-6}}\ \Omega \approx 4\ \Omega$ .

Step 2. Since $X_L$ is greater than $X_C$ here, recognise the impedance triangle has legs $3$ and $4$ — a $3$ – $4$ – $5$ right triangle, a pattern worth spotting instantly rather than grinding through the square root: $Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{3^2+4^2}\ \Omega = 5\ \Omega$ .

Step 3. The phase angle is $\varphi = \tan^{-1}\!\left(\dfrac{X_C-X_L}{R}\right) = \tan^{-1}\!\left(\dfrac{4-8}{3}\right) = \tan^{-1}(-1.33) \approx -53.1°$ . The negative sign, with $X_L$ greater than $X_C$ , means the circuit is net inductive and the current lags the source voltage by about $53.1°$ .

Step 4. The peak current is $i_m = v_m/Z = 283/5 \approx 56.6\ \text{A}$ , so the rms current is $I = i_m/\sqrt2 \approx 40\ \text{A}$ . The power dissipated is $P = I^2R = (40)^2(3) = 4800\ \text{W}$ , and the power factor is $\cos\varphi = R/Z = 3/5 = 0.6$ .

Step 5. For resonance, adjust the source frequency until $X_L=X_C$ : $\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(25.48\times10^{-3})(796\times10^{-6})}} \approx 222\ \text{rad/s}$ , giving $\nu_0 = \omega_0/2\pi \approx 35.4\ \text{Hz}$ . At resonance $Z=R=3\ \Omega$ , so $I = V/R = (283/\sqrt2)/3 \approx 66.7\ \text{A}$ , and $P = I^2R = (66.7)^2(3) \approx 13{,}350\ \text{W} \approx 13.35\ \text{kW}$ — nearly triple the off-resonance power for the same peak voltage, because at resonance the full source voltage appears across $R$ alone. The impedance at $50\ \text{Hz}$ is 5 Ω, the current lags the voltage by 53.1 degrees, the power dissipated is 4800 W with power factor 0.6; the circuit resonates near 35.4 Hz, where the current rises to 66.7 A and the power dissipated rises to about 13.35 kW. Timing note: spot the $3$ – $4$ – $5$ (or $5$ – $12$ – $13$ ) impedance triangle before reaching for a calculator — competitive papers choose $R$ , $X_L$ , $X_C$ deliberately to give clean Pythagorean triples, and recognising one turns a slow square root into instant mental arithmetic. Budget under two minutes for the off-resonance part and under a minute for the resonance follow-up, since $\omega_0=1/\sqrt{LC}$ is a one-line plug-in once $L$ and $C$ are already in SI units.

Common traps

Mixing up peak and rms values. Always check whether a given number ( $220\ \text{V}$ , an ammeter reading) is rms or peak before substituting; remember $v_m=\sqrt2\,V$ is a factor of about $1.41$ , not $2$ .
Adding $V_R$ , $V_L$ , $V_C$ arithmetically. They are out of phase, so they must be combined as phasors (Pythagorean sum), never added like plain numbers — Worked Example 2 above is built entirely around this trap.
Forgetting which element leads and which lags. Use ELI (inductor: voltage leads, current lags) and ICE (capacitor: current leads voltage) to fix the sign instantly under exam pressure.
Believing a pure inductor or capacitor dissipates power. Average power in either, over a full cycle, is exactly zero; only resistance dissipates energy. A current can be large and still deliver zero real power (wattless) if $\varphi = \pi/2$ .
Assuming resonance can occur without both $L$ and $C$ . An $R$ – $L$ or $R$ – $C$ circuit alone has no frequency at which the reactive voltages cancel, so it cannot resonate.
Getting the sign of $\varphi$ backwards. $\tan\varphi = (X_C-X_L)/R$ : if $X_C$ is greater than $X_L$ , current leads; if $X_L$ is greater than $X_C$ , current lags. Swapping the subtraction order flips the conclusion.
Confusing power factor with efficiency. Power factor $\cos\varphi$ is about the phase relationship between voltage and current; it says nothing by itself about wasted heat versus useful work.
Expecting a transformer to work on DC. A transformer needs a changing flux to induce an emf; a steady DC current gives a steady flux and induces nothing, so transformers are useless on pure DC.

Common trapReactive voltages and reactive elements are not "extra" energy sources or sinks — they never add like DC quantities, and they never dissipate power

The single biggest source of lost marks in this chapter is treating an inductor's or capacitor's voltage like a resistor's: adding $V_R+V_L+V_C$ arithmetically to check against the source voltage, or assuming a large voltage across an inductor or capacitor implies power is being consumed there. Neither is true. Reactive voltages are $90°$ out of phase with the resistor's voltage and must be combined by the Pythagorean rule (or general phasor addition) to recover the actual source voltage; and however large the current or voltage across a pure $L$ or $C$ becomes, the average power delivered to it over a full cycle is exactly zero, because the energy it absorbs while charging or building a magnetic field is returned in full while discharging or collapsing that field. Only a resistor actually converts electrical energy to heat — a fact that underlies the entire power-factor discussion and the very possibility of a large "wattless" current that lights no lamp and turns no motor.

You should be able to…

Write down $v=v_m\sin\omega t$ and the corresponding current for a resistor, inductor, and capacitor separately, including the correct phase relationship in each case.
Convert confidently between peak and rms values for both voltage and current, and explain physically why rms values are the ones quoted in practice.
Define inductive reactance $X_L=\omega L$ and capacitive reactance $X_C=1/\omega C$ , and explain why each behaves oppositely with rising frequency.
Construct a phasor diagram for a series LCR circuit and read off the impedance $Z$ and phase angle $\varphi$ from it.
State the resonance condition $\omega_0=1/\sqrt{LC}$ , explain why both $L$ and $C$ are required, and connect resonance to real applications like radio tuning and metal detectors.
Compute average power in an AC circuit using $P=VI\cos\varphi$ , and explain what a wattless current is and why purely reactive elements dissipate no power.
Explain, in your own words, why a low power factor increases transmission losses and how a parallel capacitor can correct it.
Derive and apply the transformer relations $V_s/V_p=N_s/N_p$ and $I_s/I_p=N_p/N_s$ , and list the main real-world sources of transformer energy loss.
Justify, using $I^2R$ loss and the transformer equations together, why electrical power is transmitted at high voltage and low current over long distances.

Ask the tutor: if you would like a walkthrough of how the LCR phasor triangle changes shape as frequency sweeps through resonance, or a hand-worked power-factor-correction problem, just ask.

Why doesn't a capacitor completely block AC current the way it blocks DC current?

With DC, the capacitor charges up until its voltage matches the source, and then current stops entirely — a steady voltage can only push a fixed amount of charge onto the plates once. With AC, the source keeps reversing before the capacitor can finish charging in one direction, so charge keeps flowing back and forth every half-cycle. The capacitor still opposes the current through its reactance $X_C = 1/\omega C$ , limiting the current's amplitude, but never reducing it all the way to zero as a constant DC voltage does.

If a pure inductor or capacitor absorbs zero average power, why do power companies still care about the reactive current they carry?

Because the wires and generators still have to carry the full current amplitude even though the average power delivered by that current is zero. A larger current always means larger $I^2R$ heating losses in the lines and bigger capacity requirements for generators and cables, regardless of whether that current does useful work or is "wattless." This is exactly why a low power factor is penalised commercially: it forces a bigger current for the same useful power, and utilities often charge industrial customers extra when their power factor drops too low.

What is the actual difference between rms current and average current, and why is average current always zero for a sine wave?

The plain time-average of $i = i_m\sin\omega t$ over a full cycle is genuinely zero, because the positive half-cycle and the negative half-cycle are mirror images and cancel exactly when summed. The rms current is a completely different kind of average: it first squares the current (making every value positive, since $i^2\geq0$ always), averages that, and then takes a square root, giving $I=i_m/\sqrt2$ . It is not "the average current" in the everyday sense — it is defined specifically so that $I^2R$ reproduces the true average heating power, which is the physically meaningful quantity for anything that dissipates energy.

Why can't resonance happen in a circuit that has only a resistor and inductor, or only a resistor and capacitor?

Resonance is fundamentally about two opposing reactive voltages, $V_L$ and $V_C$ , cancelling each other so that the full source voltage appears across the resistor alone, driving the maximum possible current. With only $L$ present, or only $C$ present, there is nothing for that single reactance to cancel against — increasing or decreasing the frequency only ever makes $X_L$ (or $X_C$ ) monotonically bigger or smaller, so the impedance $\sqrt{R^2+X^2}$ never dips to a minimum equal to $R$ at some special frequency; it simply increases or decreases steadily. Both an inductor and a capacitor, together, are structurally required for the cancellation that defines resonance.

Why does a transformer stop working, or even get damaged, if connected to a DC supply?

A transformer works entirely through Faraday's law: a changing flux in the shared iron core is what induces an emf in the secondary. With AC, the flux is constantly changing, so an emf is continuously induced. With DC, once the current settles to a steady value the flux also becomes steady, and a constant flux induces zero emf, so the secondary gives no output. Worse, the primary coil has very low resistance (it relies on AC back-emf, not resistance, to limit current), so a steady DC voltage applied directly drives a dangerously large current through that low resistance, overheating and potentially burning out the winding.