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Chapter 4 — Electromagnetic Induction and Alternating Currents

Class 12 · Physics

Overview

This unit studies how changing magnetic fields produce electric effects and how alternating voltages and currents behave. It covers the principles of electromagnetic induction, including Faraday's law and Lenz's law, motional emf, induced current, self and mutual induction, and energy considerations. The second major part treats alternating current (AC): sinusoidal sources, phasor representation, reactance and impedance of resistors, inductors and capacitors, resonance in RLC circuits, power factor and average power, and practical devices such as transformers and AC generators. Understanding this unit explains how electric generators, transformers, induction motors, and many sensing devices work. It links electricity and magnetism dynamically and introduces time-varying fields that form the basis of modern power systems and electronics. For ISC students, rigorous use of calculus and circuit analysis is included for derivations, while numerical problems develop skill in using RMS values, phasors, and resonance conditions. Mastery of this unit is essential for engineering, electronics and physics careers and for understanding technologies that power homes and industries.

Learning Objectives

  • Explain Faraday's law and Lenz's law and use them to predict the direction and magnitude of induced emf.
  • Calculate motional emf and induced current in moving conductors within magnetic fields using appropriate formulas.
  • Derive expressions for self-inductance and mutual inductance, and compute energy stored in magnetic fields.
  • Analyze transient behaviour in LR circuits and solve first-order differential equations for current and emf.
  • Describe sinusoidal alternating current, represent AC quantities using phasors, and compute RMS values.
  • Compute reactance and impedance for R, L and C elements and combine them to analyse series and parallel AC circuits.
  • Determine resonance conditions and bandwidth in a series RLC circuit and calculate the quality factor.
  • Calculate real, reactive and apparent power in AC circuits and explain power factor and its correction.
  • Explain working principles and applications of transformers and AC generators including turns ratio and efficiency.

Topics in this chapter

18 topics · tap a topic title to jump straight to it.

🧲1

Magnetic flux

Magnetic flux is a central concept connecting magnetic fields and circuits. It quantifies how many magnetic field lines penetrate a chosen surface. For a uniform magnetic field B passing through a flat surface of area A at angle θ between the field and the area normal vector, the flux is Φ = B A cos θ. The cosine factor captures how the effective area 'seen' by the field changes with orientation: when the field is parallel to the normal (θ = 0°) flux is maximum, and when perpendicular to the normal (θ = 90°) flux is zero.

In practical situations the magnetic field is often non-uniform or the surface is curved. Then magnetic flux is defined by the surface integral Φ = ∫ B · dA, where dA is an infinitesimal vector area element. This integral sums contributions of B over the entire surface, accounting for variations in magnitude and direction. Mathematics of surface integrals is not required in every problem, but the idea allows treatment of coils, solenoids and complex geometries.

For a coil of N turns, what matters is flux linkage NΦ, the total flux linked with all turns. Flux linkage appears in Faraday's law and determines induced emf when flux changes. If each turn has the same flux Φ, total linkage is NΦ; if flux per turn differs, sum the contributions from each turn. The SI unit of magnetic flux is the weber (Wb), where 1 Wb = 1 T·m2. A more practical subunit is the milliweber (mWb).

Visualising flux helps: draw field lines and the surface bounded by a loop. In experiments, moving the loop or changing field strength changes flux. A loop rotated in a uniform field has time-varying flux Φ(t) = B A cos(ωt) if the angle changes as ωt; this results in alternating induced emf. Magnetic materials (iron cores) concentrate flux because they increase magnetic permeability μ and thereby increase Φ for given magnetising current. Thus core material choice directly affects inductance and flux in transformers and inductors.

When solving problems, check units, signs and orientation. Choose a direction for the area normal consistently (right-hand rule for loop current). For composite loops or solenoids break surfaces into parts where B is approximately uniform, compute flux contributions and add. Understanding flux and flux linkage prepares you to apply Faraday's law, analyse induced emf, and design magnetic circuits such as transformers, inductors and generators.

📌 Examples
  • A square loop of side 0.1 m in a uniform field 0.5 T with plane perpendicular to field: Φ = BA = 0.5×(0.1)2 = 0.005 Wb.
  • A circular loop of radius 0.05 m at angle 60° to B = 0.2 T: Φ = BA cosθ = 0.2×π×(0.05)2×cos60° ≈ 3.93×10−4 Wb.
  • A coil of N = 200 turns with single-turn flux Φ = 10−5 Wb has flux linkage NΦ = 2×10−3 Wb.
🧮 Formulas
  1. Φ = B A cosθ
  2. Φ = ∫ B · dA
  3. Flux linkage = N Φ
📊 Visual ideas
Diagram of a rectangular loop in a uniform magnetic field showing field lines, area vector normal and angle θ between B and normal.
Circle representing a coil with radius labelled and angle between plane and field indicated.
🧲2

Faraday's law of electromagnetic induction

Faraday's law provides the quantitative link between changing magnetic flux and the induced electromotive force (emf). It states that the emf induced in a circuit equals the negative rate of change of magnetic flux linkage through the circuit. For a coil of N turns this is written ε = − d(NΦ)/dt. The negative sign embodies Lenz's law and ensures energy conservation: the induced emf opposes the change that produced it.

There are several physical ways the flux through a circuit can change: the magnetic field B may vary with time, the area A of the circuit may change, or the orientation angle θ between the field and the area normal may change — for example when a loop rotates. Each causes Φ to be time-dependent. In the simple rotating-coil generator, Φ(t) = B A cos(ωt) and so ε(t) = −N dΦ/dt = N B A ω sin(ωt). The amplitude grows with rotation rate ω, number of turns N, field strength B and coil area A.

Faraday's law applies both to closed circuits and to moving conductors as part of a circuit: if a conductor moves through a magnetic field, charges inside experience magnetic force q(v × B) and separate, producing a motional emf consistent with the flux-change viewpoint. For continuous fields, Faraday's law in integral form relates the line integral of electric field around a closed loop to the negative rate of change of magnetic flux through the loop; differential form is one of Maxwell's equations: ∇ × E = − ∂B/∂t. This differential form shows that a time-varying magnetic field produces a non-conservative electric field in space, unlike static electrostatic fields.

When using the law in problems, proceed in steps: (1) compute flux Φ(t) through the circuit (consider geometry and variation), (2) differentiate to get dΦ/dt, (3) multiply by −N to obtain induced emf, and (4) apply circuit laws (Ohm's law, Kirchhoff's laws) if current determination is required. Keep careful track of sign and orientation; define the area normal and current direction using right-hand rules. Faraday's law also handles mutual induction where changing current in one coil changes flux in another: ε2 = −M dI1/dt with mutual inductance M determined by geometry.

Understanding Faraday's law is essential to explain generators, transformers, induced currents producing heating or braking, and the back emf in inductors. For ISC-level rigor, show derivations using calculus for rotating coils, integrate to find motional emf in non-uniform motion, and relate the integral and differential forms to complete the physical picture of time-varying electromagnetic phenomena.

📌 Examples
  • A coil with N = 100 turns has flux changing from 5×10−4 Wb to 1×10−4 Wb in 0.2 s. ε = −N ΔΦ/Δt = −100×(−4×10−4)/0.2 = +0.2 V.
  • A loop rotating uniformly in B: Φ = BA cos(ωt) so ε = −dΦ/dt = B A ω sin(ωt) for single turn.
  • A solenoid’s current decreases linearly causing B to drop; the change in flux through a nearby loop induces emf calculable via dΦ/dt.
🧮 Formulas
  1. ε = − d(NΦ)/dt
  2. ε = −N dΦ/dt
  3. ∇ × E = −∂B/∂t
📊 Visual ideas
A coil of N turns with magnetic field lines through it and a graph showing Φ(t) decreasing and induced ε(t) plotted as its time derivative.
A loop rotating with angle θ = ωt; show B, area vector and sinusoidal ε vs time.
🔌3

Lenz's law and direction of induced current

Lenz's law

To apply Lenz's law in practice follow these steps: first identify the change in magnetic flux through the loop (is flux increasing or decreasing and in which direction?). Second determine the direction of the induced magnetic field needed to oppose that change. Third use the right-hand rule to find the induced current direction that would produce that magnetic field. Consistent choice of the area normal vector and orientation is helpful: define the positive direction and stick to it while evaluating dΦ/dt and the resulting sign of ε.

Consider a bar magnet approaching a conducting loop with its north pole facing the loop. The magnetic flux into the loop (choose normal pointing towards you) increases. To oppose the increase, the induced current must produce a magnetic field pointing outwards towards the approaching pole; this corresponds to creating a north pole on the near face of the loop, which repels the approaching north pole of the magnet. The induced current direction found by the right-hand rule will be anticlockwise when viewed from the magnet side. If the magnet is pulled away, the induced current reverses so that the induced field attracts the magnet, opposing the decrease of flux.

Lenz's law also explains back emf in inductors: when circuit current tries to rise, the inductor generates an emf opposing the increase; when current falls the inductor produces an emf that attempts to maintain current. This underlies smoothing of current changes, transient behaviour and energy storage in magnetic fields. In complex circuits with multiple coils or changing geometries, use dot convention to track relative signs of induced voltages in coupled inductors; the dot marking indicates whether induced voltages add or subtract depending on current direction.

Remember that Lenz's law is a consequence of energy conservation: if the induced emf aided the change instead of opposing it, the system would produce energy without input. Always perform a sign check and visualize field directions to avoid mistakes when determining current direction in induced-emf problems.

📌 Examples
  • A magnet’s north pole moves towards a stationary loop: induced current is anticlockwise when viewed from the approaching pole so that the loop produces a north pole facing the magnet.
  • When a bar magnet is pulled away, the induced current reverses to create a south pole that attracts the receding magnet.
  • In an RL circuit when current is switched off, the inductor’s induced emf tries to keep current flowing in the same direction.
🧮 Formulas
  1. Direction of ε given by ε = − dΦ/dt (negative sign represented by Lenz's law)
📊 Visual ideas
Diagram showing a magnet approaching a loop labelled 'increase in flux' and arrows indicating induced current direction using right-hand rule.
Schematic of inductor with current decreasing and arrow showing induced emf polarity opposing decrease.
🏃4

Motional emf and moving conductors

Motional emf

This effect is the basic mechanism of many generators and devices like railguns. When the moving conductor forms part of a closed circuit, the induced emf drives a current determined by the circuit resistance. For example, a rod sliding on parallel conducting rails in a uniform magnetic field forms a loop; as the rod moves, the area of the loop changes and flux through the loop changes, producing an emf consistent with Faraday's law and equal to Bℓv for perpendicular motion.

More generally, motional emf along a path C moving with velocity field v in a magnetic field B is given by ε = ∮C (v × B) · dl where dl is an element along the conductor. This form is useful for complex shapes and rotating conductors. For a rotating conductor like a disc (Faraday disc), linear speed varies with radius r′ as v = ω r′, and the emf between centre and rim is found by integrating dε = B v dr′ giving ε = (1/2) B ω r2 for a disc of radius r.

Motional emf connects neatly with Faraday's law: a moving conductor frequently changes the area of a circuit and hence the magnetic flux, so both viewpoints agree. In many real situations magnetic flux change and v × B viewpoint are dual descriptions; choose the one that simplifies calculation. Motional emf also leads to forces on current-carrying conductors in magnetic fields (F = Iℓ × B). When current flows due to induced emf, magnetic forces may oppose motion (magnetic braking) and require external mechanical work to sustain motion, illustrating conversion between mechanical and electrical energy.

Applications include DC generators where conductors move in a magnetic field and supply current to external circuit via commutators; rail systems and electromagnetic launchers where large motional emf and forces appear; and velocity sensors that measure speed by induced voltage. In solving problems, draw the geometry, identify relative directions of v and B, compute v × B, and if needed use circuit resistance to find current and forces from Iℓ×B.

📌 Examples
  • A rod ℓ = 0.5 m moves at v = 2 m/s perpendicular to B = 0.3 T. ε = Bℓv = 0.3×0.5×2 = 0.3 V.
  • Rod sliding on rails with R = 5 Ω from above example: I = 0.3/5 = 0.06 A; magnetic force F = IℓB = 0.06×0.5×0.3 ≈ 0.009 N opposing motion.
  • A disc of radius 0.1 m rotating at ω = 100 rad/s in B = 0.02 T: ε between centre and rim = 1/2 B ω r2 = 0.01 V.
🧮 Formulas
  1. ε = B ℓ v (for rod moving perpendicular to B)
  2. ε = ∮ (v × B) · dl
  3. For rotating rod: ε = (1/2) B ω r2 between centre and rim
📊 Visual ideas
Rod of length ℓ sliding on parallel conducting rails in a magnetic field showing B into page, velocity right, induced emf across rod ends and current path through resistor.
Rotating disc with radial line showing small element and v = ωr', integrate to get emf between centre and rim.
5

Induced current, heating and energy considerations

When an emf is induced in a closed conducting path, an induced current flows. The magnitude of this current is given by Ohm's law I = ε / R where R is the total resistance of the circuit. Induced currents convert energy: part is dissipated as heat (Joule heating) in resistive elements and part may be stored temporarily in magnetic fields (in inductors). Energy considerations reveal how mechanical work or changing fields feed electrical energy into circuits or vice versa.

Consider a conducting loop moved out of a region of magnetic field while connected to a resistor. As the loop exits the field, magnetic flux through it decreases and an emf is induced that tries to maintain the original flux. An induced current flows whose magnetic effect opposes the change (Lenz's law). The induced current heats the resistor; the energy for this heating comes from the mechanical work done by the external agent pulling the loop. The external agent feels a magnetic drag force opposing its motion; performing work against this force supplies the electrical energy dissipated as heat, preserving energy conservation.

In inductors, when current increases energy is stored in the magnetic field. The energy stored in an inductor of inductance L carrying current I is U = 1/2 L I2. When the current decreases, this stored energy is released back into the circuit and may be dissipated or used elsewhere. For coupled coils, changes in one coil can transfer energy to another via mutual inductance. In transformers idealised as lossless, energy is transferred from primary to secondary via changing flux; in real transformers losses (copper resistance, core hysteresis, eddy currents) reduce efficiency.

Eddy currents deserve special mention: they are induced currents circulating inside bulk conductors exposed to changing fields. They produce heating and magnetic damping. In many devices eddy currents are undesirable and are minimised by laminating cores or using high-resistivity materials; in some applications (induction heating, eddy-current braking) they are exploited for practical effect.

When computing power quantities in circuits with induced emf, instantaneous power delivered by the induced emf to the circuit is p(t) = ε(t) i(t). For steady sinusoidal AC, average power over a cycle is P = Vrms Irms cosφ where φ is phase difference. For transient processes calculate energy exchange by integrating power over time and include energy stored in magnetic fields. Clear energy accounting—identifying sources, sinks, and stored energy—prevents conceptual mistakes and helps interpret phenomena like electromagnetic damping, generator work output and losses in practical machines.

📌 Examples
  • A loop has induced emf ε = 0.2 V and R = 2 Ω, so I = 0.1 A and power dissipated P = I2R = 0.01×2 = 0.02 W.
  • An inductor L = 0.5 H carrying I = 2 A stores U = 1/2 L I2 = 1 J.
  • A metal plate moving through non-uniform B has eddy currents; kinetic energy converts to heat — used in eddy current brakes.
🧮 Formulas
  1. Power dissipated P = I2R
  2. Energy stored in inductor U = 1/2 L I2
  3. Instantaneous power P = ε I
📊 Visual ideas
Loop exiting magnetic field region with arrows indicating induced current and external force doing work against magnetic braking.
Circuit symbol for inductor with stored energy U = 1/2 L I2 and arrow showing energy flow during increase of current.
🔬6

Self induction and inductance

Self-induction

Inductance depends on geometry and magnetic properties: number of turns N, cross-sectional area A, length ℓ of the coil and the permeability μ of the core. For a long solenoid with N turns, length ℓ and area A in air, approximate L ≈ μ0 N2 A / ℓ. If a magnetic core of relative permeability μr is inserted, L increases roughly by factor μr (until core saturation). For complex coil shapes or non-uniform fields, calculate L by finding flux Φ per unit current and using L = N Φ / I or use tabulated formulae for standard geometries.

Physically, inductance causes circuits to oppose changes in current. When a switch closes connecting a DC source to an LR circuit, the current cannot jump instantly to its final value because the induced back emf −L dI/dt limits the rate. The transient response for a step input is I(t) = (E/R) (1 − e−t/τ) with τ = L/R. The inductor thus stores energy in its magnetic field; the energy at current I is U = 1/2 L I2. During current decrease this energy is returned to the circuit and may be dissipated in resistances or transferred to other elements.

In AC circuits an inductor presents inductive reactance XL = ωL which grows with frequency, causing the current to lag the voltage by 90° in an ideal inductor. Real inductors have winding resistance and parasitic capacitance; at high frequencies these parasitics determine behaviour, sometimes forming resonant circuits. Practical design involves trade-offs: increasing L usually increases size and series resistance, so choices depend on application (filters, chokes, energy storage).

Measurement techniques include determining L from impedance at known frequency (L = XL/ω) or from transient response measuring τ = L/R. Inductors are rated for current and core saturation: above a certain current the core’s permeability drops and L reduces. Understanding self-inductance helps analyse back emf in motors, design smoothing chokes in power supplies, and manage transient voltages with protective circuits like snubbers or flyback diodes.

📌 Examples
  • A solenoid with N = 500 turns, A = 1×10−4 m2, ℓ = 0.2 m: L ≈ μ0 N2 A / ℓ ≈ 0.157 H (approx).
  • If L = 0.2 H and dI/dt = 10 A/s, induced emf ε = −L dI/dt = −2 V.
  • Energy stored in inductor with L = 0.5 H and I = 3 A: U = 1/2 L I2 = 2.25 J.
🧮 Formulas
  1. ε = − L dI/dt
  2. L = N Φ / I
  3. U = 1/2 L I2
  4. Approx. solenoid: L ≈ μ0 N2 A / ℓ
📊 Visual ideas
A solenoid drawn with N turns, labelled length ℓ and area A, showing magnetic field inside and flux linkage NΦ.
Plot of current vs time showing smooth change due to inductance and induced back emf opposing rapid change.
🔬7

Mutual induction and coefficient of mutual inductance

Mutual induction

The mutual inductance M depends on geometry, separation, orientation, numbers of turns and the magnetic medium between coils. When coils are tightly coupled on the same core such that most flux produced by one links the other, coupling coefficient k approaches 1 and M ≈ √(L1 L2). For loosely coupled coils k is smaller and M = k √(L1 L2). The dot convention on winding diagrams marks relative polarities: voltages at dotted ends have the same sign when currents enter dotted ends simultaneously.

In circuit analysis, mutual inductance introduces terms in loop equations: v1 = L1 dI1/dt + M dI2/dt and v2 = L2 dI2/dt + M dI1/dt with sign depending on winding senses. Correct sign handling is essential when solving coupled circuits such as transformers, coupled resonators or inductive sensors. Mutual coupling can cause transfer of energy between circuits, resonant interactions, and sometimes unwanted crosstalk in crowded wiring environments.

Computing M from first principles requires calculating flux from coil 1 that links coil 2. For simple coaxial circular coils analytic expressions exist; for complex geometries numerical integration or finite-element methods are used. Experimentally M can be measured by driving a known dI/dt in coil 1 and measuring induced emf in coil 2, giving M = −ε2/(dI1/dt). Applications include transformers (high M), wireless chargers (moderate M), and inductive sensors (designed M). Managing coupling—either maximising it for energy transfer or minimising it to avoid interference—is a practical engineering task.

📌 Examples
  • Two coils with L1 = 0.1 H, L2 = 0.4 H and k = 0.5 have M = k √(L1 L2) = 0.5×√(0.04) = 0.1 H.
  • If M = 0.02 H and dI1/dt = 100 A/s then induced emf in coil 2 is ε2 = −M dI1/dt = −2 V.
  • Using flux method: coil 1 produces flux 5×10−5 Wb per amp that links coil 2 of N2 = 200 turns, M = N2×(5×10−5) = 0.01 H.
🧮 Formulas
  1. ε2 = − M dI1/dt
  2. M = k √(L1 L2)
  3. M = N2 Φ21 / I1
📊 Visual ideas
Two coaxial coils with shared magnetic field lines and labelled N1, N2, showing flux linkage and dot convention for polarities.
Circuit diagram of two coupled inductors with mutual inductance M and arrows indicating sign convention.
🔌8

LR circuits and transients

LR circuits are simple first-order circuits combining an inductor L and resistor R. They exhibit transient behaviour when switching events change applied voltages or connections. The key differential equation for a series LR circuit with applied voltage E(t) is L (dI/dt) + R I = E(t). This linear first-order ODE has solutions that combine a particular (steady-state) part and a homogeneous (transient) part. For a constant DC step E0 applied at t = 0 the solution is I(t) = (E0/R) (1 − e−t/τ) with time constant τ = L/R.

The time constant τ measures how fast the current reaches steady state: after time τ the current attains about 63% of its final value E0/R. For t ≫ 5τ the transient term e−t/τ becomes negligible and current is essentially steady. When the supply is removed at t = 0 and the initial current is I0, the current decays as I(t) = I0 e−t/τ and energy stored in the inductor 1/2 L I02 is dissipated in the resistor over time.

For time-dependent inputs other than a step, superposition or Laplace transforms become useful. The Laplace method transforms differential equations into algebraic ones in the s-domain, making it straightforward to handle arbitrary driving functions and initial conditions. The presence of L makes current continuous at switching instants; ideal inductors do not allow instantaneous changes in current because that would require infinite voltage (L dI/dt would be infinite).

Transient analysis is important practically: when switching inductive loads large voltage spikes may appear as inductors try to oppose sudden current changes. Protection devices like snubber circuits or flyback diodes are used to limit voltage and protect switching transistors. Understanding LR transients also helps in designing chokes for smoothing ripples in power supplies and in timing circuits where τ sets characteristic delay or filtering behaviour.

In AC steady-state the same LR elements contribute reactance XL = ωL and form impedance Z = R + jXL; transient behaviour then reduces to sinusoidal steady-state with phase shifts. When solving exam problems, identify initial conditions, compute τ, form general solution, and interpret physical meaning: energy storage, dissipation, and the role of back emf in resisting sudden changes.

📌 Examples
  • Series LR with L = 0.2 H, R = 10 Ω, E0 = 20 V: τ = L/R = 0.02 s, final current E0/R = 2 A, I(t) = 2(1 − e−50t) A.
  • If supply is cut when current is 2 A, current decays I(t) = 2 e−50t A and energy dissipated = 1/2 L I02 = 0.4 J.
  • Time to reach 99% of steady current ≈ 5τ = 0.1 s for given values.
🧮 Formulas
  1. L dI/dt + R I = E(t)
  2. Time constant τ = L / R
  3. I(t) = (E0/R) (1 − e−t/τ) for step input
  4. Decay: I(t) = I0 e−t/τ
📊 Visual ideas
Plot of I(t) vs t showing exponential rise from 0 to steady value with time constant τ labelled.
Series circuit diagram with L and R and switch, showing current path and initial conditions.
🔌9

Alternating current: sinusoidal source and RMS values

Alternating current (AC) commonly uses sinusoidal voltages and currents because linear circuits respond in a simple way to sinusoids: the steady-state response is another sinusoid at the same frequency but possibly shifted in phase and changed in amplitude. A sinusoidal voltage can be written v(t) = Vmax sin(ωt + φ) where ω = 2πf is the angular frequency and φ the phase. For power systems f is typically 50 Hz or 60 Hz, giving ω = 100π or 120π rad/s respectively.

RMS (root-mean-square) values are central in AC power calculations: Vrms = Vmax/√2 and Irms = Imax/√2 for pure sinewaves. RMS values give the equivalent DC values that produce the same heating in a resistor: power dissipated in a resistor R with sinusoidal current is Pavg = Irms2 R. For general periodic waves RMS is defined by Vrms = √(1/T ∫0T v2(t) dt). This definition ensures correct calculation of average power even for non-sinusoidal waveforms commonly seen in electronics.

Phasors convert time-domain sinusoids to complex numbers representing amplitude and phase, making circuit algebra simpler. Replace v(t) = Vmax sin(ωt + φ) by phasor V = Vmax ∠ φ (or Vrms ∠ φ if working in rms). Circuit elements become impedances allowing algebraic addition instead of solving differential equations. For example, Ohm's law in phasor form is V = I Z where Z is complex impedance. Phasors also help visualise phase relationships using vector diagrams.

Instantaneous power p(t) = v(t) i(t) oscillates at twice the frequency. Average power over a cycle is P = Vrms Irms cosφ, where φ is the phase difference between voltage and current. This leads to the separation of power into real and reactive parts and motivates the power factor concept. For practical work, converting between peak and rms, tracking phase, and using phasors avoids tedious trig manipulations and clarifies how reactive elements store and return energy without consuming it on average.

When solving numerical problems, always note whether values given are peak or RMS; convert appropriately. Use phasors for AC circuit analysis and express final time-domain answers as sinusoidal functions with amplitude and phase. Understanding sinusoidal steady-state, RMS and phasors is a foundation for analysing filters, resonance, power systems and AC machines.

📌 Examples
  • If Vmax = 311 V for mains 50 Hz then Vrms = 311/√2 ≈ 220 V.
  • For sinusoidal current Imax = 10 A, Irms = 10/√2 ≈ 7.07 A. Power in R = 10 Ω is P = I rms2 R ≈ 500 W.
  • A voltage v(t) = 100 sin(100π t) V has frequency f = 50 Hz and Vrms = 100/√2 ≈ 70.71 V.
🧮 Formulas
  1. v(t) = Vmax sin(ωt + φ)
  2. Vrms = Vmax / √2, Irms = Imax / √2
  3. Average power P = Vrms Irms cosφ
📊 Visual ideas
Sine wave labelled Vmax and Vrms with one full cycle showing period T and angular frequency ω.
Phasor diagram with voltage phasor leading or lagging current phasor by phase angle φ.
🔬10

Reactance and impedance of L and C

Reactance

To combine resistance and reactance we use complex impedance Z. Impedances allow using Ohm's law in phasor form: V = IZ where Z = R + jX for series combinations. For a resistor ZR = R (real), for an inductor ZL = jωL (purely imaginary positive), and for a capacitor ZC = − j/(ωC) (purely imaginary negative). The imaginary parts represent stored energy oscillating between field and circuit; the sign indicates whether that energy flow leads or lags the applied voltage.

In series circuits impedances add, while in parallel circuits admittances Y = 1/Z add. Use complex algebra to compute total impedance, then find current phasors and voltages across components. The magnitude of impedance is |Z| = √(R2 + X2) and the phase angle of current relative to voltage is φ = arctan(X/R) for series circuits with net reactance X = XL − XC. Keep sign conventions consistent: if X positive circuit is inductive (current lags), if X negative circuit is capacitive (current leads).

Reactance dependence on frequency makes circuits frequency selective. At low frequency capacitors block (XC large) and inductors pass (XL small); at high frequency the reverse happens. This is the principle behind filters and coupling networks. Calculations often require converting between rectangular and polar forms of complex numbers; practice with complex multiplication and division is essential. When solving problems, compute ω from given frequency, evaluate XL and XC, form impedance using complex notation, and then compute magnitudes and phases for currents and voltages.

Practical inductors and capacitors have losses (resistance, dielectric loss) that introduce small real parts to their impedance; for many idealised problems these are ignored. Understanding reactance and impedance prepares students to analyse resonance, phasor diagrams, power factor and the behaviour of circuits across frequencies. Many examination questions require stepwise computation of XL, XC then Z and finally current or voltage using phasor arithmetic, so practise complex arithmetic carefully.

📌 Examples
  • L = 0.05 H at f = 50 Hz: ω ≈ 314.16 rad/s, XL = ωL ≈ 15.7 Ω.
  • C = 10 μF at f = 1 kHz: ω ≈ 6283 rad/s, XC ≈ 15.9 Ω.
  • Series circuit R = 10 Ω, L = 0.02 H at 50 Hz: XL ≈ 6.283 Ω, |Z| ≈ 11.8 Ω, phase ≈ 32°.
🧮 Formulas
  1. XL = ω L
  2. XC = 1 / (ω C)
  3. ZL = j ω L, ZC = − j / (ω C), ZR = R
  4. |Z| = √(R2 + (XL − XC)2) for series RLC
📊 Visual ideas
Phasor diagram showing voltage across R in phase with current, across L leading by 90° and across C lagging by 90° relative to current.
Plot of XL and XC versus frequency showing XL increasing linearly with ω and XC decreasing as 1/ω; intersection marks resonance in series RLC when XL = XC.
🔌11

Series RLC circuit and resonance

A series RLC circuit contains a resistor R, inductor L and capacitor C connected in series and driven by an AC source. The total impedance is Z = R + j(ωL − 1/(ωC)). The imaginary part (reactive part) depends on frequency; at a special frequency called resonance ω0 the reactances cancel: ω0 L = 1/(ω0 C), giving ω0 = 1/√(LC). At resonance the impedance is purely real and equals R, which means the current amplitude is maximum for a given supply voltage because reactive opposition vanishes.

The resonant frequency in hertz is f0 = 1/(2π√(LC)). Quality factor Q measures sharpness of the resonance peak: for a series circuit Q = ω0 L / R = 1/R √(L/C). A large Q indicates a narrow bandwidth and a high peak current at resonance. Bandwidth Δω is related to Q by Δω = ω0 / Q. In frequency-selective applications like radio tuning, a high Q helps select a narrow band of frequencies while rejecting others.

At frequencies below resonance the circuit behaves capacitively (XC > XL) and current leads voltage; above resonance it behaves inductively (XL > XC) and current lags. Near resonance, voltages across L and C individually can be much larger than the source voltage because they are equal and opposite and cancel in the sum, while the resistor sees the net small voltage. This phenomenon is called voltage magnification and explains why components must be rated appropriately in high-Q circuits.

Mathematically, current amplitude I = Vmax / |Z| where |Z| = √(R2 + (ωL − 1/(ωC))2). Use phasor diagrams to visualise that at resonance VL and VC are equal in magnitude and 180° out of phase. For practical designs consider non-idealities: series resistance in inductors and ESR in capacitors reduce Q and widen bandwidth. When solving exam problems derive ω0, compute Q, find current at resonance and off-resonance, and interpret energy exchange between L and C: they continuously trade energy while R dissipates some each cycle.

📌 Examples
  • L = 10 mH, C = 100 nF ⇒ ω0 ≈ 1×104 rad/s and f0 ≈ 1591.5 Hz.
  • For above with R = 10 Ω, Q ≈ 10 and bandwidth Δf ≈ 159.15 Hz.
  • At resonance with V = 10 V and R = 10 Ω, I = 1 A and voltages across L and C may be ≈ Q×V = 100 V each (opposite in phase).
🧮 Formulas
  1. Resonance: ω0 = 1 / √(L C), f0 = 1 / (2π √(L C))
  2. Z = R + j(ωL − 1/(ωC))
  3. Quality factor Q = ω0 L / R = 1/R √(L/C)
  4. Bandwidth: Δω = ω0 / Q
📊 Visual ideas
Bode-type plot of current amplitude vs frequency showing a peak at f0; bandwidth Δf at half-power points marked.
Phasor diagram at resonance with V L and V C equal and opposite, and V R in phase with current.
🔌12

Parallel RLC circuits and resonance

Parallel RLC circuits have resistor R, inductor L and capacitor C connected in parallel across an AC source. Instead of adding impedances directly, analysis uses admittances Y = 1/Z. The total admittance is Y = 1/R + 1/(jωL) + jωC. The imaginary part of Y represents net susceptance; at resonance this imaginary part vanishes: ωC − 1/(ωL) = 0 giving ω0 = 1/√(LC) the same resonant frequency as a series circuit formed from the same L and C.

The behaviour at resonance is complementary to series resonance: in a parallel circuit at resonance the total current drawn from the source is minimum and the circuit presents a high impedance approximately equal to R (assuming ideal cancellation of reactive currents). However, the currents in the L and C branches individually can be large and opposite in phase, cancelling each other in the supply. This leads to current magnification within branches while source current is small, making parallel resonance useful in tank circuits and oscillators.

Quality factor Q for a parallel circuit (in the high-Q approximation) is given by Q ≈ R / (ω0 L). A high Q produces a sharp peak in impedance vs frequency. Bandwidth Δω is again ω0 / Q. In tuning and filtering use series resonance to pass a frequency (low impedance at resonance) and parallel resonance to block or trap a frequency (high impedance at resonance).

Analysis of parallel circuits often requires converting branch impedances to admittances, adding complex numbers, and then inverting to find net impedance. For practical circuits consider losses: finite R in coils, dielectric losses in capacitors, and coupling to other circuits all reduce Q. Use phasor diagrams of branch currents to visualise cancellation at resonance. Typical applications include radio frequency tuning circuits, impedance matching networks and RF filters where selective high impedance at particular frequencies is desired.

📌 Examples
  • L = 10 mH and C = 100 nF gives f0 ≈ 1591.5 Hz as in the series example; at resonance supply current is minimal.
  • If R = 1 kΩ in parallel with L and C then Q ≈ R/(ω0 L) ≈ 10 (approx) indicating sharp tuning.
  • At resonance branch currents IL and IC are equal in magnitude and opposite in phase causing cancellation and low supply current.
🧮 Formulas
  1. Admittance Y = 1/R + 1/(jωL) + jωC
  2. Resonance condition ω0 = 1/√(LC)
  3. Parallel Q ≈ R / (ω0 L) (for high Q)
📊 Visual ideas
Plot of |Z| vs frequency for parallel RLC showing peak at f0 where impedance is maximum.
Phasor diagram of branch currents IL and IC equal and opposite at resonance, small supply current through R.
🔋13

Power in AC circuits: real, reactive and apparent power

In AC circuits power divides naturally into three related quantities: real power P (measured in watts), reactive power Q (measured in vars) and apparent power S (measured in volt-amperes, VA). For sinusoidal steady-state with Vrms and Irms and phase difference φ between voltage and current, apparent power is S = Vrms Irms, real power is P = Vrms Irms cosφ and reactive power is Q = Vrms Irms sinφ. These relationships form the power triangle with S as hypotenuse, P adjacent and Q opposite the angle φ.

Real power is the average power actually delivered to resistive elements and converted to work or heat. Reactive power represents energy that alternately flows into and out of reactive elements (inductors and capacitors) each cycle; it does not perform net work over a cycle but affects current magnitude and hence losses. Apparent power simply multiplies voltage and current magnitudes without considering phase and is the quantity that determines conductor and transformer sizing.

Power factor pf = cosφ is the ratio of real power to apparent power and indicates how effectively current is being converted into useful work. A low power factor (lagging for inductive loads) increases current for a given real power, raising I2R losses and requiring larger conductors and equipment ratings. Industries improve power factor using capacitor banks or synchronous condensers to offset inductive reactive power, reducing electricity costs and line losses.

Complex power S in phasor form is S = P + jQ = Vrms I*r where I* is the complex conjugate of current phasor. This compact expression allows calculation of P and Q from phasor voltages and currents in circuits with arbitrary impedances. Instantaneous power p(t) = v(t) i(t) oscillates around the mean P and includes terms at twice the supply frequency. Average power is obtained by integrating instantaneous power over a cycle.

When solving numerical problems, convert peak values to RMS if needed, determine phase difference using impedances (φ = arg(Z)), compute Vrms Irms and then P, Q and S. For power factor correction determine required reactive compensation Qc to shift pf to a desired value: Qc = P (tanφ1 − tanφ2) where φ1 and φ2 are initial and final load angles. Understanding these concepts is essential for power system design, energy billing, and efficient electrical equipment operation.

📌 Examples
  • Vrms = 230 V, Irms = 10 A, φ = 30° ⇒ P ≈ 1993 W, S = 2300 VA, Q = 1150 var, pf = 0.866.
  • A purely inductive load: φ = +90°, P = 0, Q positive indicating inductive reactive power drawn from source.
  • To correct pf of 0.8 (lagging) to unity for load drawing P = 8 kW, calculate required capacitor reactive power using Qc = P(tanφ1 − tanφ2).
🧮 Formulas
  1. P = Vrms Irms cosφ
  2. Q = Vrms Irms sinφ
  3. S = Vrms Irms
  4. Complex power S = P + jQ = Vrms I*
📊 Visual ideas
Power triangle showing S as hypotenuse, P adjacent and Q opposite with pf = cosφ.
Phasor diagram with voltage and current separated by phase angle φ and labeled Vrms, Irms.
🧬14

AC generator and emf produced by rotating coil

AC generators

Real alternators are more sophisticated: they use multiple coils, pole-pairs and shaped cores to produce desired waveform and voltage levels at practical rotation speeds. The rotor may carry the field winding (excitation) and the stator carries armature windings delivering power to external circuits. Slip rings provide continuous electrical connection for AC; in DC machines commutators convert AC induced in rotating windings to DC at the terminals.

Voltage amplitude depends on B, A, N and ω. Changing field excitation (varying field current) changes B and therefore the generated emf; prime mover speed control adjusts ω and thus frequency. Synchronising multiple alternators on the grid requires matching voltage amplitude, frequency and phase before paralleling. When load is applied the electromagnetic torque increases opposing rotation; the prime mover must supply extra mechanical power to maintain speed and frequency.

From a theoretical viewpoint, derive the time dependence ε(t), compute RMS voltage Vrms = εmax/√2, and use power relations to find load currents and power delivered. Consider armature reaction and internal impedance: winding resistance and leakage reactance lower terminal voltage under load. Large generators include cooling systems and are designed to limit core and copper losses. Classroom problems often idealise the machine but exploring non-idealities deepens understanding of how generators behave in real power systems.

📌 Examples
  • Single coil with N = 100 turns, A = 0.01 m2, B = 0.5 T, rotating at 3000 rpm (50 rps ⇒ ω ≈ 314.16 rad/s) gives εmax ≈ 157.08 V and Vrms ≈ 111.1 V.
  • If rotation speed doubles, both frequency and εmax double, so both amplitude and frequency scale with ω.
  • For a coil with 200 turns at 60 Hz compute induced emf amplitude using εmax = N B A ω.
🧮 Formulas
  1. Φ = B A cos(ωt)
  2. ε = − N dΦ/dt = N B A ω sin(ωt)
  3. εmax = N B A ω, Vrms = εmax / √2
📊 Visual ideas
A coil rotating in a uniform magnetic field with angle θ = ωt and arrows showing instantaneous flux and induced emf wave plotted as sine over time.
Sketch of generator with slip rings connecting rotating coil to external circuit and labelled parameters N, A, B.
🔬15

Transformers and their principles

Transformers

Real transformers are subject to non-ideal effects. Winding resistance causes copper losses (I2R) and heating. The core experiences hysteresis and eddy current losses which depend on core material, frequency and flux density; lamination and choice of silicon-steel or ferrite reduce these losses. Leakage flux is flux that does not link both windings and results in leakage reactance; it limits voltage transfer under load and affects regulation. Equivalent circuit models include magnetising inductance, core-loss resistance, series winding resistances and leakage reactances.

Transformers are central in power systems: step-up transformers at generation increase voltage for efficient long-distance transmission (reducing current and I2R losses), and step-down transformers near consumption reduce voltage to safe usable levels. Transformer rating in VA indicates maximum apparent power it can supply. Efficiency η = output power/input power is high for large units but depends on load because copper losses scale with current while core losses depend mainly on voltage and frequency.

Practical use requires correct connections (e.g., star/delta in three-phase systems), respecting polarity (dot convention), ensuring cooling and insulation, and protecting against overloads and short-circuits. Tests such as open-circuit (to measure core parameters) and short-circuit (to measure series impedance) characterise transformer performance. In classroom problems ideal transformer relations Vp/Vs = Np/Ns and Ip/Is = Ns/Np are often used for simplicity, but awareness of losses and limitations helps apply theory to real machines.

📌 Examples
  • Primary Np = 1000, secondary Ns = 100 ⇒ turns ratio 10:1. With Vp = 230 V, Vs = 23 V (step-down). If Is = 2 A, Ip = Is×(Ns/Np) = 0.2 A (ideal).
  • Step-up transformer example: Np = 200, Ns = 2000 and Vp = 110 V ⇒ Vs = 1100 V (ideal relation).
  • Open-circuit test measures magnetising current and core loss while short-circuit test measures equivalent series impedance.
🧮 Formulas
  1. Vp/Vs = Np/Ns
  2. Ip/Is = Ns/Np (ideal transformer)
  3. ε = −N dΦ/dt applies to each winding
📊 Visual ideas
Schematic of ideal transformer with primary and secondary coils on common core, arrows indicating flux Φ and turns N1, N2.
Phasor representation showing voltages and currents with turns ratio relation and inverse current scaling.
🔌16

Eddy currents and electromagnetic damping

Eddy currents

Eddy currents are an important loss mechanism in devices with alternating magnetic fields, such as transformers, motors and generators. In a solid metal core, eddy currents form large loops and produce significant heating, reducing efficiency and potentially causing overheating. To reduce these losses cores are made from thin laminations insulated from one another; laminations interrupt large loop paths and force eddy currents into many small loops with much smaller magnitude, drastically reducing heating. For high-frequency applications ferromagnetic materials with high resistivity or ferrites are used to keep eddy losses low.

While often undesirable, eddy currents are exploited in useful applications. Induction heating systems deliberately create strong eddy currents in workpieces using high-frequency magnetic fields to heat metal quickly and uniformly for forging, melting or cooking. Eddy current brakes generate smooth, contactless braking by inducing currents in moving metallic parts that produce magnetic forces opposing the motion—these are used in trains, amusement parks and some laboratory apparatus. Eddy current sensors detect the presence, position or flaws in conductive materials by measuring changes in induced currents and are widely used in non-destructive testing and proximity sensing.

The magnitude of eddy current losses depends on the square of the magnetic flux density amplitude, frequency, electrical conductivity, and characteristic thickness of the conductor. Detailed quantitative analysis requires solving Maxwell's equations with boundary conditions; however, practical rules guide engineers: reduce the thickness available for circulating currents (laminations), increase material resistivity or reduce the rate of change of flux. Electromagnetic damping due to eddy currents produces forces often proportional to velocity at moderate speeds, making it useful for smooth energy dissipation without wear.

Demonstrations include dropping a strong magnet through a conducting (non-magnetic) tube: the magnet falls much more slowly than in free fall due to eddy-current-induced opposing forces. Understanding eddy currents helps design efficient magnetic devices and also enables technologies that harness these currents for heating, braking and sensing.

📌 Examples
  • A copper plate moving into a magnetic field experiences eddy currents producing a retarding force proportional to velocity; used in eddy current brakes.
  • Transformer cores are laminated to reduce eddy current losses; a solid core would heat excessively under AC excitation.
  • Induction cooktops induce eddy currents in ferromagnetic cookware, heating it directly rather than the hob surface.
🧮 Formulas
  1. Eddy current magnitude scales with rate of change of flux and conductivity; detailed formulas depend on geometry and require solving Maxwell's equations.
📊 Visual ideas
Plate moving through magnetic field with circular eddy currents drawn inside plate and resulting opposing force shown.
Cross-section of laminated core showing thin insulated layers and reduced eddy current paths.
🔌17

AC circuits with combined elements and phasor analysis

Phasor analysis

For series combinations impedances add: Ztotal = ΣZk. The current phasor I = V/Ztotal is common to series elements; voltages across each element are Vk = I Zk and may have different phases. For parallel branches compute branch admittances Yk = 1/Zk and sum them to find Ytotal = ΣYk; the supply current phasor is I = V Ytotal and branch currents Ik = V Yk. Be careful with algebraic signs for imaginary parts when adding inductive and capacitive contributions.

Phasor diagrams help visualise relative phases and magnitudes. For example, in a series RLC circuit voltage across R is in phase with current, across L leads by 90°, and across C lags by 90°. The vector sum of voltages across components equals supply voltage. Use rectangular (a + jb) or polar (magnitude ∠ angle) forms of complex numbers as convenient; many calculators handle complex operations directly. When converting back to time domain, use i(t) = √2 |I| sin(ωt + arg(I)) for rms phasors converted appropriately.

For power calculations use complex power S = V I* where I* is the complex conjugate of current phasor; then P = Re(S) and Q = Im(S). This ensures correct signs for reactive power and consistent handling of phase. Phasor methods are not applicable during transients or for non-sinusoidal waveforms unless Fourier analysis is used; then decompose waveforms into sinusoids and apply superposition.

Practice problems combine R, L and C in various configurations: series, parallel and mixed networks. Work through stepwise: compute ω, form impedances, find Ztotal or Ytotal, compute I or V phasors, obtain magnitudes and phases, and convert back to time domain if required. Phasor fluency simplifies analysis of filters, resonance, power factor correction and many AC circuits encountered in exams and practical applications.

📌 Examples
  • Series R = 20 Ω, L = 50 mH, C = 10 μF at f = 50 Hz: compute ZL, ZC then Z total, current phasor for Vrms = 230 V and phase angle.
  • Parallel R and C: compute admittance Y = 1/R + jωC, then magnitude of total impedance |Z| = 1/|Y| and phase angle.
  • Use phasor V = 120∠0° V and Z = 10 + j5 Ω to compute I = V/Z = 120∠0 / (11.18∠26.6°) ≈ 10.74∠−26.6° A; convert to time domain.
🧮 Formulas
  1. Replace sinusoid by phasor: v(t) ↔ V∠φ, i(t) ↔ I∠ψ
  2. ZL = j ω L, ZC = − j / (ω C), ZR = R
  3. S = P + jQ = V I*
📊 Visual ideas
Phasor diagram with V, I and individual voltage drops across R, L, C shown as phasors and their vector sum equal to source phasor.
Circuit diagram of series RLC annotated with complex impedances and current phasor direction.
🧬18

Applications: generators, transformers, induction heating and motors

This topic gathers practical applications that use electromagnetic induction and AC theory and ties the material to real engineering. Generators

Transformers

Induction heating

Induction motors

Other applications include wireless power transfer via mutual inductance (chargers), magnetic levitation, metal detectors using eddy current response, and filters/tuned circuits in radios using resonance. In all applications practical constraints—losses (core, copper, eddy), heating, material limits, insulation, and safety—shape design choices. Laboratory demonstrations, such as dropping a magnet through a conducting tube, measuring transformer voltage ratios, or observing induction cooking, link theory to observable effects and reinforce understanding of how inductive phenomena power modern technology.

📌 Examples
  • Step-up transformers in power grid raise voltage from generator terminals to hundreds of kV for long-distance transmission to reduce I2R losses.
  • Induction cooktop uses a high-frequency oscillator and an induction coil under the hob to heat cookware via eddy currents.
  • A single-phase induction motor produces starting torque using a rotating magnetic field produced by auxiliary windings and phase shift elements.
🧮 Formulas
  1. Generator emf: εmax = N B A ω
  2. Transformer ideal: Vp/Vs = Np/Ns, Ip/Is = Ns/Np
📊 Visual ideas
Schematic of power distribution: generator → step-up transformer → transmission lines → step-down transformer → consumer.
Diagram of induction heating setup showing coil under workpiece and induced eddy current loops in the metal.

Key Concepts

Magnetic flux
Scalar measure of magnetic field passing through a surface, Φ = ∫ B · dA, unit weber (Wb).
Flux linkage
Total flux linked with a coil: N times the magnetic flux through one turn, NΦ.
Faraday's law
Induced emf in a coil equals negative rate of change of its flux linkage: ε = − d(NΦ)/dt.
Lenz's law
The induced emf acts to oppose the change in magnetic flux that produced it, giving the negative sign in Faraday's law.
Motional emf
Emf generated when a conductor moves in a magnetic field due to q(v × B), ε = B ℓ v for perpendicular motion.
Self-inductance (L)
Property of a coil where changing current induces emf in itself; ε = −L dI/dt; unit henry (H).
Mutual inductance (M)
Measure of flux in one coil due to current in another, ε2 = −M dI1/dt and M = k √(L1 L2).
Reactance
Frequency-dependent opposition by L or C: XL = ωL for inductors, XC = 1/(ωC) for capacitors.
Impedance
Complex generalisation of resistance in AC circuits: Z = R + jX, magnitude |Z| and phase φ = arg(Z).
RMS value
Effective value of AC equal to DC value delivering same heating effect: Vrms = Vmax/√2 for sine wave.
Resonance
Condition in RLC circuit where inductive and capacitive reactances are equal and cancel, ω0 = 1/√(LC).
Quality factor (Q)
Measure of sharpness of resonance: Q = ω0 L / R (series) or Q ≈ R / (ω0 L) (parallel high-Q).
Apparent, real and reactive power
S = Vrms Irms (apparent), P = Vrms Irms cosφ (real), Q = Vrms Irms sinφ (reactive).
Transformer turns ratio
Ideal transformer relation Vp/Vs = Np/Ns and Ip/Is = Ns/Np linking voltages and currents to turns.
Eddy currents
Circulating currents induced in conductors by changing magnetic fields that cause heating and damping.
Phasor
Complex representation of a sinusoidal quantity encoding amplitude and phase for AC analysis.

Practice Questions

  1. A coil of 200 turns has a magnetic flux through each turn changing from 4.0×10−4 Wb to 1.0×10−4 Wb in 0.1 s. Calculate the magnitude of the average induced emf. / एक 200 वाइंड के कॉइल में हर घूम पर चुंबकीय फ्लक्स 4.0×10−4 Wb से 1.0×10−4 Wb तक 0.1 s में बदलता है। औसत प्रेरित इएमएफ का परिमाण ज्ञात कीजिए।
    Show answer

    Change in flux ΔΦ = (1.0−4.0)×10−4 = −3.0×10−4 Wb. Magnitude of average emf ε = N |ΔΦ/Δt| = 200×(3.0×10−4 / 0.1) = 200×3.0×10−3 = 0.6 V. / फ्लक्स परिवर्तन ΔΦ = −3.0×10−4 Wb. औसत इएमएफ का परिमाण ε = N |ΔΦ/Δt| = 200×(3.0×10−4 / 0.1) = 0.6 V.

  2. A rod of length 0.4 m moves at 5 m/s perpendicular to a magnetic field of 0.2 T. If the rod is part of a closed circuit of resistance 4 Ω, find the induced emf and current. / एक 0.4 m लंबा रॉड 5 m/s की गति से चुंबकीय क्षेत्र 0.2 T में लम्बवत चल रहा है। यदि रॉड एक 4 Ω प्रतिरोध पर बने बंद परिपथ का हिस्सा है, प्रेरित इएमएफ और धारा ज्ञात कीजिए।
    Show answer

    Motional emf ε = B ℓ v = 0.2×0.4×5 = 0.4 V. Current I = ε / R = 0.4 / 4 = 0.1 A. / मोशनल इएमएफ ε = Bℓv = 0.4 V. धारा I = ε/R = 0.1 A.

  3. A coil with inductance 0.1 H carries current that increases at 20 A/s. Calculate the self-induced emf and state its polarity relative to the change. / 0.1 H की स्व-इन्डक्टेंस वाली कॉइल में धारा 20 A/s की दर से बढ़ रही है। स्व-प्रेरित इएमएफ ज्ञात करें और वृद्धि के सापेक्ष उसकी ध्रुवता बताइए।
    Show answer

    ε = − L dI/dt = −0.1×20 = −2.0 V. The induced emf opposes the increase, so its polarity is such as to oppose the rise in current (it will be negative with respect to the assumed positive direction of current). / ε = −2.0 V. इएमएफ धारा में वृद्धि का विरोध करती है; इससे वह उस दिशा में धैर्य रखेगी जो वृद्धि को रोकने वाली ध्रुवता दर्शाती है।

  4. Find the resonant frequency f0 of a series RLC circuit with L = 50 mH and C = 2 μF. / L = 50 mH और C = 2 μF के सीरीज़ RLC परिपथ का अनुनादी आवृत्ति f0 ज्ञात कीजिए।
    Show answer

    ω0 = 1/√(LC) = 1/√(50×10−3 × 2×10−6) = 1/√(1.0×10−7) = 1×104 rad/s. Therefore f0 = ω0/(2π) ≈ 10,000/(2π) ≈ 1591.5 Hz. / ω0 = 1/√(LC) ≈ 1×104 rad/s, अतः f0 ≈ 1591.5 Hz.

  5. An AC source v = 311 sin(100π t) V supplies a series circuit R = 10 Ω, L = 50 mH. Calculate the current amplitude and phase angle. / v = 311 sin(100π t) V वाला AC स्रोत R = 10 Ω, L = 50 mH वाले सीरीज़ परिपथ को सप्लाई करता है। धारा की परिमाण और चरण कोण ज्ञात कीजिए।
    Show answer

    ω = 100π rad/s ≈ 314.16 rad/s. XL = ωL ≈ 314.16×0.05 = 15.708 Ω. Impedance magnitude |Z| = √(R2 + XL2) = √(100 + 246.74) ≈ √346.74 ≈ 18.62 Ω. Current amplitude Imax = Vmax / |Z| = 311 / 18.62 ≈ 16.7 A. Phase angle φ = − arctan(XL / R) (current lags voltage) = − arctan(15.708/10) ≈ −57.5°. In time domain i(t) = 16.7 sin(100π t − 57.5°) A. / ω ≈ 314.16 rad/s, XL ≈ 15.708 Ω, |Z| ≈ 18.62 Ω. Imax ≈ 16.7 A, φ ≈ −57.5°, अतः i(t) ≈ 16.7 sin(100π t − 57.5°) A.

  6. A transformer has 500 turns on the primary and 50 turns on the secondary. If the primary is connected to 240 V AC, find the secondary voltage and the secondary current when the load draws 2 A. / एक ट्रांसफ़ॉर्मर के प्राथमिक में 500 और द्वितीयक में 50 वाइंड हैं। प्राथमिक पर 240 V AC दिया जाता है; द्वितीयक वोल्टेज क्या होगा और यदि लोड 2 A खींचता है तो प्राथमिक धारा क्या होगी? (दी गई वाइंड अनुपात के साथ)।
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    Turns ratio Np/Ns = 500/50 = 10. Secondary voltage Vs = Vp × (Ns/Np) = 240 × (50/500) = 24 V. For ideal transformer power conservation: Vp Ip = Vs Is ⇒ Ip = (Vs Is) / Vp = (24×2)/240 = 0.2 A. Alternatively using current ratio Ip/Is = Ns/Np = 50/500 = 0.1 ⇒ Ip = 0.1×2 = 0.2 A. / Ns/Np = 0.1, V s = 24 V, प्राथमिक धारा Ip = 0.2 A (आदर्श ट्रांसफॉर्मर के लिए)।

  7. Define power factor and explain why power factor correction is used in industries. / पावर फैक्टर को परिभाषित कीजिए और बताइए कि औद्योगिक क्षेत्रों में पावर फैक्टर सुधार क्यों किया जाता है।
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    Power factor is the cosine of phase difference between voltage and current: pf = cosφ, equal to real power divided by apparent power. Low pf (lagging for inductive loads) means higher current for given real power, increasing I2R losses and requiring larger conductors and transformers. Power factor correction (usually by adding capacitors) reduces phase difference, lowers current draw for same real power, reduces losses and electricity charges, and improves voltage regulation. / पावर फैक्टर वोल्टेज और धारा के चरण अंतर का कोसाइन है: pf = cosφ. यह वास्तविक शक्ति और एपेरेंट शक्ति के अनुपात के रूप में भी व्यक्त होता है। निचला pf अतिरिक्त करंट और हानि बढ़ाता है; पावर फैक्टर सुधार कैपेसिटर डालकर चरण अंतर घटाकर करंट और हानियों को कम करता है तथा उपकरणों की दक्षता और आर्थिकता बढ़ाता है।

  8. Explain why laminating transformer cores reduces heating due to eddy currents. / बताइए कि ट्रांसफ़ॉर्मर कोर को लेमिनेट करने से एड्डी करंट्स के कारण होने वाली गर्मी कैसे कम होती है।
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    Laminations divide the core into thin insulated sheets, interrupting large closed paths for eddy currents so that each sheet supports much smaller circulating currents. Because eddy current heating scales with the square of loop area and conductivity, reducing loop area and increasing path resistance drastically reduces eddy losses. Thus laminations lower induced eddy currents and heating while maintaining magnetic properties. / लेमिनेशन कोर को पतली अलग-थलग शीट में विभाजित करता है जिससे एड्डी करंट्स के बड़े वृत्त टूट जाते हैं और केवल छोटे करंट्स बनते हैं; इससे करंट की मात्रा और इसलिए ऊष्मा हानि कम हो जाती है।

  9. A series circuit has R = 20 Ω, L = 0.1 H and is connected to DC source 100 V at t = 0. Find current after 0.05 s. / एक सीरीज़ परिपथ R = 20 Ω, L = 0.1 H पर DC स्रोत 100 V से t = 0 पर जोड़ा जाता है। 0.05 s के बाद धारा ज्ञात कीजिए।
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    Time constant τ = L/R = 0.1 / 20 = 0.005 s. Final steady current I∞ = E/R = 100 / 20 = 5 A. Transient: I(t) = I∞ (1 − e−t/τ). For t = 0.05 s, t/τ = 10 so e−10 ≈ 4.54×10−5. Thus I(0.05) ≈ 5 (1 − 4.54×10−5) ≈ 4.99977 A ≈ 4.9998 A (≈ 5.0 A to three significant figures). / τ = 0.005 s, I∞ = 5 A. I(0.05) ≈ 5(1 − e−10) ≈ 4.9998 A ≈ 5.0 A.

  10. Calculate the reactance of a 20 μF capacitor at 50 Hz and state whether the circuit is capacitive or inductive if connected with an inductor of reactance 10 Ω in series. / 50 Hz पर 20 μF कैपेसिटर का रिएक्टेंस ज्ञात कीजिए और यदि यह 10 Ω के रिएक्टेंस वाले इंडक्टिव तत्व के साथ सीरीज़ में जोड़ा गया है तो क्रमिक परिपथ कैपेसिटिव होगा या इंडक्टिव?
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    ω = 2π×50 ≈ 314.16 rad/s. XC = 1/(ωC) = 1/(314.16×20×10−6) = 1/(0.0062832) ≈ 159.15 Ω. Series net reactance X = XL − XC = 10 − 159.15 = −149.15 Ω (negative) so circuit is capacitive (current leads voltage). / XC ≈ 159.15 Ω, कुल X = −149.15 Ω (नकारात्मक) अतः परिपथ कैपेसिटिव है।

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