Alternating Current | ISC Class 12 Physics Notes

This note covers sinusoidal alternating current and voltage, peak and mean values, root mean square values, phasors, resistive and reactive circuits, series resonance, bandwidth, power factor, wattless current, choke coils, LC oscillations and the working of an alternating-current generator.

What describes a sinusoidal alternating current?

Alternating current (AC) changes direction periodically. A sinusoidal current varies with time according to a sine function. Direct current (DC) does not change direction with time. A steady direct current also has constant magnitude; direction and constancy of magnitude are distinct ideas.

Voltage means potential difference between two points, or energy transferred per unit charge. Current is the rate of flow of charge. In the International System of Units (SI), current is measured in amperes, symbol A. The SI unit of voltage is the volt, symbol V.

Let v be instantaneous voltage, vₘ its peak value, t time and ω angular frequency. The subscript m denotes maximum magnitude. Taking the voltage to pass through zero towards positive values at time zero gives:

v = vₘ sin(ωt)

Here sin denotes the sine function. The angle ωt is the phase, specifying the position within a cycle. Angles in these equations are in radians; π radians correspond to 180°. A complete cycle corresponds to 2π radians.

How are period and frequency related?

The time period T is the time for one complete cycle. The frequency f is the number of cycles per second. The SI unit of frequency is the hertz, symbol Hz. Angular frequency measures the change of phase per second, in radians per second.

f = 1/T

ω = 2πf

For a current in phase with this voltage, write i = iₘ sin(ωt), where i is instantaneous current and iₘ is peak current. In phase means that corresponding zero crossings and positive and negative peaks occur together. Other circuit elements can shift current relative to voltage.

Note: Unless specified otherwise, a quoted AC voltage or current is ordinarily its root mean square value. A peak value must be identified explicitly. The sine-wave conversion factors below apply to sinusoidal quantities.

How do mean and RMS values differ from peak values?

The peak value is the largest magnitude reached during a cycle. The mean value is an average over a stated interval. A sinusoidal current has equal positive and negative contributions over one full cycle, so its full-cycle mean is zero.

Derivation: mean value over a positive half-cycle

Let Iₐᵥ denote the mean current over the positive half-cycle, and let θ = ωt denote phase. The integral sign ∫ represents accumulation over the stated interval; dθ is an infinitesimal change of phase.

1. The positive half-cycle extends from θ = 0 to θ = π, with current i = iₘ sin θ.
2. Divide the accumulated current over this interval by its phase width: Iₐᵥ = (iₘ/π) × the integral of sin θ dθ from 0 to π.
3. The integral of sin θ between zero and π is 2, giving the positive half-cycle mean below.

Iₐᵥ = 2iₘ/π

Likewise, the positive half-cycle mean voltage Vₐᵥ is 2vₘ/π. Each is approximately 0.637 times its peak value. The negative half-cycle mean has the opposite sign. A full-cycle mean must not be substituted for the half-cycle mean in these relations.

Derivation: root mean square current

The root mean square (RMS) current I is the square root of the mean of i² over a complete cycle. RMS voltage is denoted by V. Angle brackets ⟨ ⟩ mean a time average over one complete cycle.

1. Square i = iₘ sin(ωt) to obtain i² = iₘ² sin²(ωt).
2. Use sin²(ωt) = [1 − cos(2ωt)]/2, where cos denotes the cosine function. The cosine term averages to zero over a cycle.
3. Therefore ⟨i²⟩ = iₘ²/2. Taking the positive square root gives the RMS current.

I = iₘ/√2

V = vₘ/√2

RMS current is the equivalent DC current producing the same average heating in the same resistance. If R denotes constant electrical resistance and P denotes average power, or energy transferred per unit time, then P = I²R. The SI unit of resistance is the ohm, symbol Ω; the SI unit of power is the watt, symbol W.

A zero mean current therefore does not imply zero heating. Heating depends on the square of current, so reversing its direction does not reverse the heating effect. RMS values retain this energy-transfer information while the signed full-cycle mean does not.

How do a resistor and a phasor represent AC behaviour?

A pure resistor is treated as having resistance without inductive or capacitive effects. For a fixed resistance, Ohm's law gives v = iR at each instant. Applying a sinusoidal voltage therefore produces a sinusoidal current of the same frequency and phase.

iₘ = vₘ/R

V = IR

The peak relation connects peak quantities, while the RMS relation connects RMS quantities. Mixing a peak voltage with an RMS current introduces an incorrect factor of √2. Both expressions assume the same constant resistance.

What is a phasor?

A phasor is a rotating vector used to represent the amplitude and phase of a sinusoidal quantity. Its vertical projection gives the instantaneous value. Peak-value phasors have lengths proportional to vₘ and iₘ and rotate with angular speed ω.

For a resistor, the voltage and current phasors point in the same direction. Their phase difference is zero. Voltage and current themselves are scalar quantities, without a spatial direction; the rotating vectors are a mathematical representation that makes phase relationships and addition easier to follow.

The voltage and current curves are plotted against ωt. Both cross zero together and reach their positive and negative peaks together. Their drawn amplitudes differ, but their phase is the same.

See Fig. 7.2 in your NCERT textbook

Worked example 1. A resistive light bulb is rated at 100 W on a 220 V RMS supply. Find its operating resistance, the peak supply voltage and the RMS current.

Formula: R = V²/P; vₘ = √2V; I = P/V.

Substitute: R = 220²/100; vₘ = 1.414 × 220; I = 100/220.

Answer: R = 484 Ω, vₘ = 311 V and I = 0.454 A, using the quoted rounding.

The bulb's rating describes average power consumption. Its current changes sign during each cycle, yet energy is dissipated throughout the cycle except at instants when the current is zero. This follows from the instantaneous heating rate p = i²R, where p denotes instantaneous power.

Why does current lag voltage in a pure inductor?

An inductor is a circuit element whose changing current produces a self-induced electromotive force. Electromotive force (emf) is energy supplied per unit charge by a source, or the induced voltage associated with changing magnetic flux. Magnetic flux measures the magnetic field passing through an oriented surface. Self-induction opposes changes in the inductor's current.

The self-inductance L relates the magnitude of induced emf to the rate of change of current. The SI unit of inductance is the henry, symbol H. Usually, inductors have appreciable resistance in their windings; a pure inductor here has negligible resistance.

Derivation: current through a pure inductor

Kirchhoff's loop law states that the algebraic sum of potential changes around a circuit loop is zero. Here di/dt denotes the rate of change of current with time. We seek the sinusoidal response with no constant current component.

1. The applied voltage balances the inductive voltage: v = L(di/dt).
2. Substitute v = vₘ sin(ωt), then integrate with respect to time to obtain i = −[vₘ/(ωL)] cos(ωt).
3. Use −cos(ωt) = sin(ωt − π/2). The current amplitude is vₘ/(ωL), and its phase is behind the voltage by π/2.

i = iₘ sin(ωt − π/2)

The inductive reactance Xₗ is the opposition offered by an inductance to sinusoidal current. Its unit is the ohm. It limits current without the average energy dissipation associated with resistance in this ideal case.

Xₗ = ωL

Thus iₘ = vₘ/Xₗ and I = V/Xₗ. Current reaches its positive maximum one-quarter period after voltage. Increasing frequency at fixed inductance increases reactance, so current decreases if the applied RMS voltage remains the same.

The voltage phasor is a quarter-turn ahead of the current phasor. In the accompanying curves against ωt, the dashed current curve reaches its positive peak after the solid voltage curve.

See Fig. 7.6 in your NCERT textbook

Worked example 2. A pure 25.0 mH inductor is connected to a 220 V RMS, 50 Hz source. The prefix milli means 10⁻³. Find its reactance and RMS current, using π = 3.14.

Formula: Xₗ = 2πfL; I = V/Xₗ.

Substitute: Xₗ = 2 × 3.14 × 50 × 25.0 × 10⁻³; I = 220/7.85.

Answer: Xₗ = 7.85 Ω and I = 28 A.

Why does current lead voltage in a capacitor?

A capacitor stores charge and electric-field energy. Its capacitance C is the charge stored per unit potential difference. The SI unit of capacitance is the farad, symbol F. If q denotes the instantaneous charge on one plate, then q = Cv.

With a DC source, current flows while the capacitor charges; after it is fully charged, the current falls to zero. With an AC source, repeated charging and discharging allow alternating current in the connecting circuit. The capacitor limits the current without completely preventing this alternating flow.

Derivation: current through a pure capacitor

For an ideal capacitor of constant capacitance, current equals the rate of change of charge, written i = dq/dt.

1. Substitute the applied voltage into q = Cv to get q = Cvₘ sin(ωt).
2. Differentiate with respect to time: i = ωCvₘ cos(ωt).
3. Since cos(ωt) = sin(ωt + π/2), the current amplitude is ωCvₘ and the current leads voltage by π/2.

i = iₘ sin(ωt + π/2)

The capacitive reactance X꜀ is the opposition offered by capacitance to sinusoidal current. Like inductive reactance, it is measured in ohms. It decreases when either capacitance or frequency increases.

X꜀ = 1/(ωC)

Consequently, iₘ = vₘ/X꜀ and I = V/X꜀. Current reaches its maximum one-quarter period before voltage. This occurs because current depends on how fast charge changes, while voltage depends on the amount of charge already stored.

The current phasor is a quarter-turn ahead of the voltage phasor. The dashed current curve starts at its positive maximum when the solid voltage curve crosses zero towards positive values.

See Fig. 7.8 in your NCERT textbook

Worked example 3. A 15.0 µF capacitor is connected to a 220 V RMS, 50 Hz source. The prefix micro, written µ, means 10⁻⁶. Find its reactance, RMS current and peak current.

Formula: X꜀ = 1/(2πfC); I = V/X꜀; iₘ = √2I.

Substitute: X꜀ = 1/(2π × 50 × 15.0 × 10⁻⁶); I = 220/212; iₘ ≈ 1.41 × 1.04.

Answer: X꜀ = 212 Ω, I = 1.04 A and iₘ = 1.47 A. Doubling frequency at the same voltage halves reactance and doubles current.

How are impedance and phase found in a series LCR circuit?

A series LCR circuit contains an inductor, capacitor and resistor connected in one current path. Their instantaneous currents are identical. Their voltages differ in phase, so their voltage magnitudes cannot generally be added as ordinary positive numbers.

Let Vᵣ, Vₗ and V꜀ denote the RMS voltages across the resistor, inductor and capacitor respectively. Taking current as the phase reference, Vᵣ is in phase with current, Vₗ leads by π/2 and V꜀ lags by π/2.

Derivation: the series impedance

Impedance Z is the ratio of voltage amplitude to current amplitude, combining resistive and reactive opposition. Its unit is the ohm. RMS values give the same ratio.

1. Write the component voltage magnitudes as Vᵣ = IR, Vₗ = IXₗ and V꜀ = IX꜀.
2. The inductive and capacitive voltage phasors point in opposite directions. Their resultant has magnitude |Vₗ − V꜀|, where vertical bars mean absolute magnitude.
3. This resultant is perpendicular to Vᵣ. Therefore V² = Vᵣ² + (Vₗ − V꜀)².
4. Substitute the component relations and divide by I² to obtain the impedance expression.

Z = √[R² + (Xₗ − X꜀)²]

I = V/Z

Define φ as the phase angle by which voltage leads current. With this convention, i = iₘ sin(ωt − φ). A positive φ means current lags voltage; a negative φ means current leads voltage. The tangent function, written tan, gives:

tan φ = (Xₗ − X꜀)/R

If Xₗ exceeds X꜀, the circuit is predominantly inductive. If X꜀ exceeds Xₗ, it is predominantly capacitive. These are steady-state relations, describing behaviour after the transient, or temporary initial response, has died away in a circuit with resistance.

The resistor voltage phasor lies along the current phasor. The inductor and capacitor voltage phasors lie on a perpendicular line in opposite directions. Their resultant combines with the resistor voltage to give the source voltage.

See Fig. 7.11 in your NCERT textbook

Worked example 4. A series circuit has R = 3 Ω, L = 25.48 mH and C = 796 µF, driven by a sinusoidal source of peak voltage 283 V and frequency 50 Hz. Find the impedance, RMS current and whether current leads or lags.

Formula: Xₗ = 2πfL; X꜀ = 1/(2πfC); Z = √[R² + (Xₗ − X꜀)²].

Substitute: Xₗ ≈ 8 Ω; X꜀ ≈ 4 Ω; Z = √[3² + (8 − 4)²]; tan φ = 4/3.

Answer: Z = 5 Ω and I = 283/(√2 × 5) ≈ 40 A. Current lags voltage by 53.1°, so φ = +53.1°.

How do RL and RC circuits and their frequency graphs compare?

An RL circuit contains resistance and inductance in series. An RC circuit contains resistance and capacitance in series. Both follow from the series impedance relation by retaining the reactance of the component actually present.

For RL, Z = √(R² + Xₗ²) and tan φ = Xₗ/R. Current lags voltage. For RC, Z = √(R² + X꜀²) and tan φ = −X꜀/R. Current leads voltage. The phase angle identifies the reactive behaviour.

What changes when frequency changes?

For fixed L, inductive reactance Xₗ = 2πfL increases linearly with frequency. For fixed C, capacitive reactance X꜀ = 1/(2πfC) decreases inversely with frequency. Resistance is treated as constant in these ideal circuit comparisons.

Put frequency on the horizontal axis and reactance on the vertical axis. Draw the inductive-reactance graph as a straight line through the origin. Draw capacitive reactance as a decreasing curve for positive frequency, approaching the axes without meeting them.

In RC, the resistor and capacitor voltage phasors are perpendicular. Thus V² = Vᵣ² + V꜀², even though the same current flows through both. Their arithmetic sum can exceed source voltage without violating Kirchhoff's law, which applies to signed instantaneous voltages.

Series resonance requires both inductive and capacitive reactances to cancel. An RL circuit or an RC circuit alone cannot provide this cancellation. The presence of resistance does not supply the missing opposite reactance.

What happens at resonance, and what do bandwidth and Q measure?

Resonance in a driven series LCR circuit occurs when inductive and capacitive reactances are equal. Let ω₀ denote resonant angular frequency and f₀ resonant frequency in hertz. At this frequency, ω₀L = 1/(ω₀C).

ω₀ = 1/√(LC)

f₀ = 1/[2π√(LC)]

The reactive voltage phasors cancel. Impedance has its minimum value Z = R, the phase angle is zero, and current is in phase with source voltage. For a fixed source voltage and fixed resistance, current and average power are maximum at resonance.

The individual inductor and capacitor voltages are equal in magnitude and opposite in phase. Their resultant is zero; neither individual voltage must be zero. Both energy-storage elements remain active even though the source sees a purely resistive impedance.

Both plotted curves peak at the same resonant angular frequency. Curve (i), for 100 Ω, has a taller and narrower peak than curve (ii), for 200 Ω. The shared inductance is 1.00 mH.

See Fig. 7.14 in your NCERT textbook

For a series LCR circuit, draw impedance falling to its minimum R at f₀ and rising afterwards. On a separate graph at fixed RMS source voltage, draw current rising to its maximum V/R at f₀ and falling afterwards.

How are bandwidth and quality factor stated?

The half-power frequencies f₁ and f₂ are the lower and upper frequencies where average power is half its resonant maximum. Current there is 1/√2 times its resonant value. The bandwidth Δf is the frequency interval f₂ − f₁; Δ denotes a difference.

Δf = R/(2πL)

The dimensionless quality factor Q measures resonance sharpness. For the series circuit, Q = f₀/Δf = ω₀L/R = 1/(ω₀CR). These bandwidth and Q relations are stated without derivation.

At fixed L and C, a smaller resistance gives a higher Q and narrower bandwidth. This helps distinguish nearby signal frequencies. In radio tuning, capacitance is varied until the resonant frequency becomes nearly equal to the frequency of the desired radio signal.

How are average power and power factor calculated?

Instantaneous power p is the product of instantaneous voltage and current: p = vi. In a circuit with resistance and reactance, this product varies during the cycle. Its average depends on the phase difference as well as the voltage and current magnitudes.

Derivation: average power in a sinusoidal circuit

Use v = vₘ sin(ωt) and i = iₘ sin(ωt − φ), with φ defined as the voltage lead over current.

1. Multiply the two expressions: p = vₘiₘ sin(ωt) sin(ωt − φ).
2. Use the product identity to obtain p = (vₘiₘ/2)[cos φ − cos(2ωt − φ)].
3. The time-dependent cosine averages to zero over a complete cycle, leaving P = (vₘiₘ/2) cos φ.
4. Replace peak quantities using vₘ = √2V and iₘ = √2I to express the result in RMS quantities.

P = VI cos φ

The power factor is cos φ. For a series LCR circuit, cos φ = R/Z, so P = I²R. Energy is dissipated in the resistance; ideal inductance and capacitance store energy and return it to the circuit.

Current through a pure inductor or capacitor is sometimes referred to as wattless current. Current exists, but the average power over a full cycle is zero. This does not imply that instantaneous power is zero at every instant.

Worked example 5. A series LCR circuit has R = 3 Ω, L = 25.48 mH and C = 796 µF. A 50 Hz source supplies peak voltage 283 V. Its reactances are 8 Ω and 4 Ω, and its impedance is 5 Ω. Find RMS current, average power and power factor.

Formula: I = vₘ/(√2Z); P = I²R; cos φ = R/Z.

Substitute: I = 283/(√2 × 5) ≈ 40; P = 40² × 3; cos φ = 3/5.

Answer: I = 40 A, P = 4800 W and power factor = 0.6, using the stated rounded current.

How does a choke coil control current, and why does power factor matter?

A choke coil is an inductive element used to limit alternating current. Its action is described by Xₗ = ωL. In the ideal pure-inductance limit, its phase difference is π/2, its power factor is zero and its average power consumption is zero.

In contrast, a resistor limits current while dissipating energy as heat. A real coil has winding resistance, so its average heating is not exactly zero. The wattless conclusion belongs to the pure-inductor approximation, not to every physical coil.

How does inductance affect current control?

Increasing inductance at fixed frequency raises the coil's reactance. In a series bulb-and-coil circuit, inserting an iron rod increases inductance and reactance. A larger fraction of the applied voltage then appears across the inductor, leaving less across the bulb, so its glow decreases.

For a given average power delivered at a given RMS voltage, P = VI cos φ shows that a smaller power factor requires a larger current. If Rₗᵢₙₑ denotes the resistance of the transmission line, its heating loss is I²Rₗᵢₙₑ.

Power factor can often be improved by adding a capacitor of appropriate capacitance. For an inductive load, its leading current can compensate the lagging reactive part of the load current. The useful in-phase current then forms a larger fraction of the supply current.

This compensation concerns phase as well as magnitude. It does not mean that every capacitor value gives the desired result. The capacitance must be appropriate to the circuit, and the comparison of transmission loss assumes the same delivered power and supply voltage.

How does energy oscillate in an LC circuit?

An LC circuit consists of an inductor and capacitor. If an initially charged capacitor is connected to an inductor, energy can alternate between the capacitor's electric field and the inductor's magnetic field. These are free oscillations when no periodic external source drives them.

What happens during an oscillation?

1. Initially, the capacitor has maximum charge and the current is zero. Energy is stored in its electric field.
2. As it discharges through the inductor, current increases. Electric-field energy decreases while magnetic-field energy increases.
3. When capacitor charge becomes zero, current has its maximum magnitude. Energy is then stored in the inductor's magnetic field.
4. The current continues and charges the capacitor with opposite polarity. It falls to zero when this reversed charge reaches its maximum magnitude.
5. The sequence reverses, transferring energy back through the inductor until the original charge arrangement is restored.

If Uₑ denotes capacitor electric energy and Uᵦ denotes inductor magnetic energy, the storage expressions are Uₑ = q²/(2C) and Uᵦ = Li²/2. Energy is measured in joules, symbol J. The total Uₑ + Uᵦ remains constant in the ideal loss-free circuit.

The natural angular frequency is ω₀ = 1/√(LC), with natural frequency f₀ = 1/[2π√(LC)]. This natural frequency also sets series resonance.

Free oscillation and resonance differ. Free oscillation uses energy initially stored in the circuit. Driven resonance involves an external alternating source supplying energy. Resistance dissipates stored energy, so oscillations in a practical undriven circuit diminish instead of continuing indefinitely.

How does a simple AC generator produce an alternating emf?

An AC generator converts mechanical energy into electrical energy by electromagnetic induction, the production of emf when magnetic flux linked with a circuit changes.

The rotating coil is called the armature. Its axis is perpendicular to a uniform magnetic field. Slip rings connected to the coil ends rotate with it, while stationary brushes maintain electrical contact with the external circuit.

A coil on an axle lies between magnetic poles labelled N and S, meaning north and south. The diagram labels slip rings and carbon brushes. The brushes connect to the two output leads labelled alternating emf.

See Fig. 6.13 in your NCERT textbook

Derivation: the induced emf

Let N be the number of coil turns, A its area and B the uniform magnetic field magnitude. Here A denotes area, not the ampere unit. Let Φ be the magnetic flux through one turn and ε the instantaneous induced emf.

1. With the coil's normal initially parallel to the field, the angle between that normal and the field is θ = ωt. Thus Φ = BA cos(ωt).
2. Faraday's law gives ε = −N(dΦ/dt), where dΦ/dt is the rate of flux change. The minus sign expresses Lenz's law: induced effects oppose the flux change causing them.
3. For constant B, A and angular speed ω, differentiation gives ε = NBAω sin(ωt).
4. Define ε₀ as peak emf. Its value is NBAω, so the emf changes sign in successive half-cycles.

ε₀ = NBAω

At maximum flux, the flux is momentarily changing at zero rate and emf is zero. When flux passes through zero, the rate of change has maximum magnitude and emf has an extremum. Emf therefore depends on flux change, not simply flux magnitude.

Worked example 6. A 100-turn coil of area 0.10 m² rotates at 0.5 revolutions per second in a uniform field of 0.01 T perpendicular to its rotation axis. Here T as a unit means tesla. Find peak emf using π = 3.14.

Formula: ω = 2πf; ε₀ = NBAω.

Substitute: ε₀ = 100 × 0.01 × 0.10 × 2 × 3.14 × 0.5.

Answer: The peak generated emf is 0.314 V.

Generators provide electrical power from mechanical rotation. Water falling from a height can drive hydroelectric generators; high-pressure steam can drive thermal generators. In most generators, the coils are held stationary and the electromagnets rotate. The rotating-coil model explains the induction principle.

Glossary

• Alternating current — Electric current whose direction reverses periodically, with sinusoidal current following a sine function of time.
• Peak value — Greatest magnitude reached by an alternating current or voltage during a cycle.
• Time period — Time required for one complete repetition of the alternating waveform.
• RMS current — Equivalent direct current producing the same average heating in the same resistance as the alternating current.
• Phasor — Rotating vector representing the amplitude and phase of a sinusoidally varying scalar quantity.
• Inductive reactance — Opposition to sinusoidal current due to inductance, increasing with frequency for a fixed inductance.
• Capacitive reactance — Opposition to sinusoidal current due to capacitance, decreasing with frequency for a fixed capacitance.
• Impedance — Combined resistive and reactive opposition, equal to the ratio of voltage amplitude to current amplitude.
• Series resonance — Condition where inductive and capacitive reactances cancel, leaving the series circuit impedance equal to resistance.
• Bandwidth — Interval between the lower and upper half-power frequencies of a resonant circuit.
• Quality factor — Dimensionless measure of resonance sharpness, equal to resonant frequency divided by bandwidth for the series circuit.
• Power factor — Cosine of the phase difference between sinusoidal voltage and current, determining average power for given RMS values.
• Wattless current — Current with zero average power over a complete cycle in a pure inductive or capacitive circuit.
• Choke coil — Inductive circuit element that limits alternating current through its inductive reactance.

Common errors and misconceptions

• Misconception: Zero average sinusoidal current means zero heating. Correct: Heating depends on current squared. The RMS current describes the average heating effect even when the signed full-cycle mean is zero.
• Misconception: The mean of a sinusoidal current is 2iₘ/π over any interval. Correct: This is its positive half-cycle mean. The signed mean over one complete cycle is zero.
• Misconception: Current leads voltage in an inductor. Correct: It lags by π/2 in a pure inductor and leads by π/2 in a pure capacitor.
• Misconception: Component voltage magnitudes simply add in a series AC circuit. Correct: Add the voltage phasors with their phases, or add signed instantaneous voltages at the same instant.
• Misconception: Resonance makes each reactive voltage zero. Correct: Inductor and capacitor voltages have equal magnitudes and opposite phases, so their resultant is zero while each can remain non-zero.
• Misconception: Every coil consumes zero average power. Correct: That conclusion assumes a pure inductor. The resistance of a real winding dissipates energy as heat.
• Misconception: A generator's emf is greatest at maximum magnetic flux. Correct: Emf depends on the rate of flux change. Maximum flux and maximum emf occur at different coil orientations.

Exam-style questions with model answers

Q1. Define RMS current and give its relation to peak current for a sinusoidal alternating current. [2 marks]

1. RMS current is the equivalent direct current that produces the same average heating in the same resistance as the alternating current.
2. For a sinusoidal current, I = iₘ/√2, where I denotes RMS current and iₘ denotes peak current.

Q2. A pure 25.0 mH inductor is connected to a 220 V RMS, 50 Hz sinusoidal supply. Using π = 3.14, calculate its reactance and RMS current, and state the current-voltage phase relation. [3 marks]

1. The inductive reactance is Xₗ = 2πfL, where f is frequency and L is inductance. Substitution gives Xₗ = 2 × 3.14 × 50 × 25.0 × 10⁻³ = 7.85 Ω.
2. The RMS current is I = V/Xₗ = 220/7.85 = 28 A approximately, using the given RMS voltage V.
3. Because the inductor is pure, current lags the applied voltage by π/2 radians, or one-quarter of a cycle.

Q3. A pure 15.0 µF capacitor is connected to a 220 V RMS, 50 Hz sinusoidal supply. Calculate its capacitive reactance and RMS current. State what happens to both when frequency doubles at unchanged voltage. [3 marks]

1. The capacitive reactance is X꜀ = 1/(2πfC), where f is frequency and C is capacitance. With C = 15.0 × 10⁻⁶ F, this gives approximately 212 Ω.
2. The RMS current is I = V/X꜀ = 220/212 = 1.04 A approximately, where V is RMS supply voltage.
3. At unchanged capacitance, doubling frequency halves capacitive reactance. At unchanged RMS voltage, the RMS current therefore doubles because it is inversely proportional to reactance.

Q4. A series LCR circuit has resistance 3 Ω, inductive reactance 8 Ω and capacitive reactance 4 Ω at its operating frequency. Find its net reactance, impedance, phase relation and power factor. [4 marks]

1. The net reactance is Xₗ − X꜀ = 8 − 4 = 4 Ω. Its positive sign identifies a predominantly inductive circuit.
2. The impedance is Z = √[R² + (Xₗ − X꜀)²] = √(3² + 4²) = 5 Ω, where R is resistance.
3. With φ defined as the voltage lead over current, tan φ = 4/3. Thus φ = 53.1°, meaning current lags voltage.
4. The power factor is cos φ = R/Z = 3/5 = 0.6. This factor relates average power to the product of RMS voltage and current.

Q5. Derive the RMS current of a sinusoidal current i = iₘ sin(ωt), defining the symbols used. Explain why its zero full-cycle mean does not imply zero heating in a constant resistance R. [5 marks]

1. Here i is instantaneous current, iₘ is peak current, ω is angular frequency and t is time. Define RMS current I by I² = ⟨i²⟩, with brackets indicating a full-cycle time average.
2. Squaring the current gives i² = iₘ² sin²(ωt). The peak value is constant, so it can be taken outside the averaging operation.
3. Using sin²(ωt) = [1 − cos(2ωt)]/2, and the zero full-cycle mean of the cosine term, obtain ⟨i²⟩ = iₘ²/2.
4. Taking the positive square root gives I = iₘ/√2. This is the direct current producing the same average heating in resistance R.
5. The signed mean of i is zero, but the mean of i² is not. Average heating power is therefore P = I²R = iₘ²R/2.

Q6. For a driven series LCR circuit with fixed positive resistance R, inductance L, capacitance C and fixed RMS supply voltage V, obtain the resonant frequency and explain the impedance, current, phase and reactive voltages at resonance. [5 marks]

1. At resonance, inductive reactance equals capacitive reactance: ω₀L = 1/(ω₀C). Here ω₀ is resonant angular frequency, so ω₀ = 1/√(LC) and resonant frequency f₀ = 1/[2π√(LC)].
2. The difference of reactances is zero. Hence impedance Z = √[R² + (Xₗ − X꜀)²] reduces to its minimum value R.
3. The RMS current is I = V/Z = V/R. With source voltage and resistance fixed, this is the maximum current as frequency varies.
4. The phase difference is zero, so current and source voltage are in phase. The power factor, which is the cosine of their phase difference, is one.
5. The inductor and capacitor voltages have equal magnitudes but opposite phases. Their phasor sum is zero; the individual voltages need not be zero.

Q7. A sinusoidal source of peak voltage 283 V supplies a series LCR circuit with impedance 5 Ω and resistance 3 Ω. Calculate RMS current, power factor and average power, and identify where power is dissipated. [4 marks]

1. The RMS current is I = vₘ/(√2Z), where vₘ is peak voltage and Z is impedance. Thus I = 283/(√2 × 5) ≈ 40 A.
2. The power factor is cos φ = R/Z = 3/5 = 0.6, where R is resistance and φ is the phase difference.
3. Average power is P = I²R = 40² × 3 = 4800 W, using the rounded current.
4. Power is dissipated in the resistor. The ideal inductor and capacitor store and return energy, each having zero average power over a complete cycle.

Q8. A simple AC generator has an N-turn coil of area A rotating at constant angular speed ω in uniform magnetic field B, with its rotation axis perpendicular to the field. At time zero the coil's normal is parallel to the field. Derive its emf and explain the output connection. [5 marks]

1. The coil's normal turns through angle ωt after time t. Magnetic flux through each turn is Φ = BA cos(ωt), where Φ denotes flux.
2. Faraday's law gives the induced emf ε = −N(dΦ/dt). The negative sign expresses opposition to the change of flux, as required by Lenz's law.
3. Differentiating for constant B, A and ω gives ε = NBAω sin(ωt). The peak emf is therefore ε₀ = NBAω.
4. The sine function changes sign in successive half-cycles, so the output emf alternates. The generator converts the mechanical energy maintaining rotation into electrical energy.
5. The two coil ends connect to slip rings rotating with the coil. Stationary brushes touch these rings and connect the rotating coil to the external circuit.

Key takeaways

• A sinusoidal current has zero signed mean over a full cycle, but its positive half-cycle mean is 2iₘ/π.
• RMS current equals peak current divided by √2 and gives the equivalent direct-current heating effect in the same resistance.
• Current is in phase with voltage in a resistor, lags in a pure inductor and leads in a pure capacitor.
• Inductive reactance increases with frequency, while capacitive reactance decreases; both reactances are measured in ohms.
• Series circuit voltages combine through phasors, giving impedance Z = √[R² + (Xₗ − X꜀)²] and RMS current I = V/Z.
• At series resonance, reactances cancel, impedance is R, current is maximum at fixed voltage and power factor is one.
• Average sinusoidal power is VI cos φ; pure inductance and capacitance have zero average power despite carrying current.
• An LC circuit exchanges electric and magnetic energy, while an AC generator produces emf through changing magnetic flux.

Test yourself

Why must the averaging interval be stated when giving mean AC current?

The signed mean over a complete sinusoidal cycle is zero. Over its positive half-cycle, the mean is 2iₘ/π, so the interval changes the result.

Which quantity determines heating when current reverses direction?

Heating depends on current squared. Its full-cycle mean determines average power in a resistor, so reversing current does not cancel the heating.

What happens to current through a pure inductor when frequency rises at fixed RMS voltage?

Inductive reactance increases because Xₗ = 2πfL. The RMS current therefore decreases, provided inductance and applied RMS voltage remain unchanged.

Why is a capacitor voltage maximum not a current maximum?

Voltage depends on stored charge, while current is its rate of change. At maximum charge, that rate is momentarily zero.

Can an RC series circuit show the series resonance described here?

No. Both inductance and capacitance are needed for opposite reactive voltage phasors to cancel at a finite resonant frequency.

What does a narrower resonance peak indicate when inductance and capacitance are fixed?

It indicates smaller bandwidth and higher quality factor. In the series circuit, reducing resistance sharpens the peak and increases resonant current.

Does wattless current mean that no energy is exchanged?

No. Energy is stored and returned during different parts of the cycle. Zero average power means no net energy absorption over the complete cycle.

When does a rotating-coil generator have maximum emf magnitude?

Maximum emf magnitude occurs when magnetic flux passes through zero, because the magnitude of the rate of flux change is then greatest.